Class 12

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 13 Probability Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 13 Probability Miscellaneous Exercise

Question 1.
A and B are two events such that P(A) ≠ 0. Find P(B/A), if
(i) A is a subset of B
(ii) A ∩ B = Φ
Solution.
It is given that, P(A) ≠ 0
A is a subset of B.
⇒ A ∩ B = A
∴ P(A ∩ B) = P(B ∩ A) = P(A)
∴ P(B/A) = \(\frac{P(B \cap A)}{P(A)}=\frac{P(A)}{P(A)}\) = 1

(ii) A ∩ B = Φ
⇒ P(A ∩ B) = 0
⇒ P(B/A) = \(\frac{P(A \cap B)}{P(A)}\) = 0.

Question 2.
A couple has two children,
(i) Find the probability that both children are males, if it is known that at least one of the children is male.
(ii) Find the probability that both children are females, if it is known that the elder child is a female.
Solution.
If a couple has two children, then the sample space is
S = {(b, b), (b, g), (g, b), (g, g)}
(i) Let E and P respectively denote the events that both children are males and atleast one of the children is a male.
E ∩ F = {(b, b)}
⇒ P(E ∩ F) = \(\frac{1}{4}\)

P(E) = \(\frac{1}{4}\);

P(F) = \(\frac{3}{4}\)

P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}\)

(ii) Let A and B respectively denote the events that both children are females and the elder child is a female.
A = {(g, g)}
⇒ P(A) = \(\frac{1}{4}\)
B = {(g, b), (g, g)}
⇒ P(B) = \(\frac{2}{4}\)
A ∩ B = {(g, g)}
⇒ P(A ∩ B) = \(\frac{1}{4}\)
P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{4}}{\frac{2}{4}}=\frac{1}{2}\)

Question 3.
Suppose that 5% of men and 0.25% of women have grey hair. A gray haired person is selected at random. What is the probability of this person being male?
Assume that there are equal number of males and females.
Solution.
Probability of selecting a male = P(E₁)
= \(\frac{1}{2}\)

Probability of selecting a female = P(E₂)
= \(\frac{1}{2}\)

5% men are grey haired P(A/E₁) = 5% = 0.05
0.25% women have grey hair
i.e., P(A/E₂) = 0.25% = 0.0025

Now, P(E₁/A) = \(\frac{P\left(E_{1}\right) \cdot P\left(A / E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(A / E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A / E_{2}\right)}\)

= \(\frac{\frac{1}{2} \times 0.05}{\frac{1}{2} \times 0.05+\frac{1}{2} \times 0.0025}\)

= \(\frac{0.05}{0.0525}=\frac{500}{525}=\frac{20}{21}\)

Question 4.
Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?
Solution.
A person can be either right-handed or left-handed.
It is given that 90% of the people are right-handed.
∴ p = P (right-handed) = \(\frac{1}{10}\)
q = P (left-handed)
= 1 – \(\frac{1}{10}\)
= \(\frac{9}{10}\)

Using binomial distribution, the probability that more than 6 people are right-handed is given by,
\(\sum_{r=7}^{10}{ }^{10} C_{r} p^{r} q^{n-r}=\sum_{r=7}^{10}{ }^{10} C_{r}\left(\frac{9}{10}\right)^{r}\left(\frac{1}{10}\right)^{10-r}\)

Therefore, the probability that at most 6 people are right-handed
= 1 – P (more than 6 are right-handed)
= 1 – \(\sum_{r=7}^{10}{ }^{10} C_{r}(0.9)^{r}(0.1)^{10-r}\)

Question 5.
An urn contains 25 balls of which 10 balls bear a mark ‘X’ and the remaining 15 bear a mark T\ A ball is drawn at random from the urn, its mark is noted down and it is replaced.
If 6 balls are drawn in this way, find the probability that
(i) all will bear ‘X’ mark.
(ii) not more than 2 will bear ‘Y’ mark.
(iii) at least one ball will bear ‘Y’ mark.
(iv) the number of balls with ‘X’ mark and ‘Y’ mark will be equal.
Solution.
Total number of balls in the urn = 25
Balls bearing mark ‘X’ = 10
Balls bearing mark ‘Y’ = 15
p = P (ball bearing mark ‘X’)
= \(\frac{10}{25}=\frac{2}{5}\)

q = P (ball bearing mark ‘Y’)
= \(\frac{15}{25}=\frac{3}{5}\)

Six balls are drawn with replacement.
Therefore, the number of trials are Bernoulli trials.
Let Z be the random variable that represents the number of balls with ‘Y’ mark on them in the trials.
Clearly, Z has a binomial distribution with n = 6 and p = \(\frac{2}{5}\)
∴ p(Z = z) = \({ }^{n} C_{z} p^{n-z} q^{z}\)

(i) P(all will bear ‘X’ mark) = P(Z = 0)
= \({ }^{6} C_{0}\left(\frac{2}{5}\right)^{6}=\left(\frac{2}{5}\right)^{6}\)

(ii) P (not more than 2 ear ‘Y’ mark)
= P(Z ≤ 2) = P(Z = 0) + P(Z= 1) + P(Z = 2)

(iii) P (at least one ball bears ‘T mark) = P(Z ≥ 1)
= 1 – P(Z = 0) = 1 – \(\left(\frac{2}{5}\right)^{6}\)

(iv) P (equal number of balls with ‘X’ mark and ‘Y’ mark) = P(Z = 3)

= \({ }^{6} C_{3}\left(\frac{2}{54}\right)^{3}\left(\frac{3}{5}\right)^{3}\)

= \(\frac{20 \times 8 \times 27}{15625}=\frac{864}{3125}\)

Question 6.
In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is 5/6. What is the probability that he will knock down fewer than 2 hurdles?
Solution.
Let p and q respectively be the probabilities’that the player will clear and knock down the hurdle.
∴ p = \(\frac{5}{6}\)
⇒ q = 1 – p
= 1 – \(\frac{5}{6}\) = \(\frac{1}{6}\)
Let X be the random variable that represents the number of times the player will knock down the hurdle.
Therefore, by binomial distribution, we get
P(X = x) = \({ }^{n} C_{x} p^{n-x} q^{x}\)
P (player knocking down less than 2 hurdles) = P(X < 2) = P(X = 0) + P(X = 1)
= \({ }^{10} C_{0}(q)^{0}(p)^{10}+C_{1}(q)(p)^{9}\) = \(\left(\frac{5}{6}\right)^{10}+10 \cdot \frac{1}{6} \cdot\left(\frac{5}{6}\right)^{9}\)

= \(\left(\frac{5}{6}\right)^{9}\left[\frac{5}{6}+\frac{10}{6}\right]=\frac{5}{2}\left(\frac{5}{6}\right)^{9}\)

= \(\frac{(5)^{10}}{2 \times(6)^{9}}\)

Question 7.
A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.
Solution.
The probability of getting a six in a throw of die is \(\frac{1}{6}\) and not getting a six is \(\frac{5}{6}\).
Let p = \(\frac{1}{6}\) and
q = \(\frac{5}{6}\) The probability that the 2 sixes come in the first five throws of the die is \({ }^{5} C_{2}\left(\frac{1}{6}\right)^{2}\left(\frac{5}{6}\right)^{3}=\frac{10 \times(5)^{3}}{(6)^{5}}\)

Probability that third six comes m the sixth throw = \(\frac{10 \times(5)^{3}}{(6)^{5}} \times \frac{1}{6}\)
= \(\frac{10 \times 125}{(6)^{6}}=\frac{10 \times 125}{46656}=\frac{625}{23328}\)

Question 8.
If a leap year is selected at random, what is the chance that it will contain 53 Tuesdays?
Solution.
A leap year has 366 days which contain 52 full weeks and 2 extra days.
The extra days may occur as (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun), (Sun, Mon)
Now the favourable cases are (Mon, Tue), (Tue, Wed)
∴ Probability that a leap year will have 53 Tuesdays = \(\frac{2}{7}\).

Question 9.
An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be at least 4 successes.
Solution.
The probability of success is twice the probability of failure. Let the probability of failure be x. ∴ Probability of success = 2x x + 2x = 1
⇒ 3x = 1
⇒ x = \(\frac{1}{3}\)
∴ 2x = \(\frac{2}{3}\)
Let p = \(\frac{1}{3}\) and q = \(\frac{1}{3}\)
Let X be the random variable that represents the number of successes in six trials.
By Binomial distribution, we get
P(X = x) = \({ }^{n} C_{x} p^{n-x} q^{x}\) Probability of at least 4 successes
= P(X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6)

Question 10.
How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%? Solution.
Let the man toss the coin n times.
The n tosses are n Bernoulli trials.
Probability (p) of getting a head at the toss of a coin is \(\frac{1}{2}\).
It is given that, 90 P (getting at least one head) > \(\frac{90}{100}\)
P(x ≥ 1) > 0.9
∴ 1 – P(x = 0) > 0.9
1 – \({ }^{n} C_{0}\), \(\frac{1}{2^{n}}\) > 0.9
\({ }^{n} C_{0}\), \(\frac{1}{2^{n}}\) < 0.1
\(\frac{1}{2^{n}}\) < 0.1,
2ⁿ > \(\frac{1}{0.1}\),
2ⁿ > 10 …………….(i)
The minimum value of n that satisfies the given inequality is 4.
Thus, the man should toss the coin 4 or more than 4 times.

Question 11.
In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins/loses.
Solution.
In a throw of a die, the probability of getting a six is 1/6 and the probability of not getting a 6 is 5/6.
Three cases can occur.

(i) If he gets a six in the first throw, then the required probability is 1/6.
Amount he will receive = ₹ 1

(ii) If he does not get a six in the first two throws and gets a six in the second throw, then
Probability = \(\left(\frac{5}{6} \times \frac{1}{6}\right)=\frac{5}{36}\)
Amount he will receive = – ₹ 1 + ₹ 1 = 0

(iii) If he does not get a six in the first two throws and gets a six in the third throw, then
Probability = \(\left(\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\right)=\frac{25}{216}\)
Amount he will receive = – ₹ 1 – ₹ 1 + ₹ 1 = – ₹ 1
Expected value he can win = \(\frac{1}{6}(1)+\left(\frac{5}{6} \times \frac{1}{6}\right)(0)+\left[\left(\frac{5}{6}\right)^{2} \times \frac{1}{6}\right](-1)\)

= \(\frac{1}{6}=\frac{25}{216}=\frac{36-25}{216}=\frac{11}{216}\)

Question 12.
Suppose we have four boxes. A, B, C and D containing coloured marbles as given below.

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A? box B? box C?
Solution.
Let R be the event of drawing the red marble.
Let E_A, E_B and E_C respectively denote the events of selecting the box A, B and C.
Total number of marbles = 40
Number of red marbles = 15
∴ P(R) = \(\frac{15}{40}=\frac{3}{8}\)
Probability of drawing the red marble from box A is given by P(E_A/R).

∴ P(E_A/R) = \(\frac{P\left(E_{A} \cap R\right)}{P(R)}=\frac{\frac{1}{40}}{\frac{3}{8}}=\frac{1}{15}\)

Probability that the red marble is from box B is P{E_B/R)

⇒ P(E_B/R) = \(\frac{P\left(E_{B} \cap R\right)}{P(R)}=\frac{\frac{6}{40}}{\frac{3}{8}}=\frac{2}{5}\)

Probability that the red marble is from box C is P(E_C/R)

⇒ P(E_C/R) = \(\frac{P\left(E_{C} \cap R\right)}{P(R)}=\frac{\frac{8}{40}}{\frac{3}{8}}=\frac{8}{15}\).

Question 13.
Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities.

It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?
Solution.
Let A, E₁ and E₂ respectively denote the events that a person has a heart attack the selected person followed the course of yoga and meditation, and the person adopted the drug prescription.
∴ P(A) = 0.40, P(E₁) = P(E₂) = \(\frac{1}{2}\)
P(A|E₁) = 0.40 ×0.70 = 0.28,
P(A|E₂) = 0.40 × 0.75 = 0.30
Probability that the patient suffering a heart attack followed a course of meditation and yoga is given by P(E₁/A).

P(E₁/A) = \(\frac{P\left(E_{1}\right) \cdot P\left(A / E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(A / E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A / E_{2}\right)}\)

= \(\frac{\frac{1}{2} \times 0.28}{\frac{1}{2} \times 0.28+\frac{1}{2} \times 0.30}=\frac{14}{29}\).

Question 14.
If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability 1/2).
Solution.
The total number of determinants of second order with each element being 0 or 1 is (2)⁴ = 16
The value of determinant is positive in the following cases
\(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\left|\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right|\left|\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right|\)
∴ Required probability = \(\frac{3}{16}\)

Question 15.
An electronic assembly consists of two subsystems, say A and B. From previous testing procedures, the following probabilities are assumed to be known
P(A fails) = 0.2
P(B fails alone) = 0.15
P(A andB fail) = 0.15
Evaluate the following probabilities
(i) P(A fails / B has failed)
(ii) P(A fails alone)
Solution.
Let the event in which A fails and B fails be denoted by E_A and E_B.
P(E_A) = 0.2, P(E_A and E_B) = 0.15

P (B fails alone = P(E_B) – P(E_A and E_B)
0.15 = P(E_B) – 0.15
∴ P(E_B) = 0.3
(i) P(E_A/E_B) = \(\frac{P\left(E_{A} \cap E_{B}\right)}{P\left(E_{B}\right)}=\frac{0.15}{0.3}\)
= 0.5

(ii) P (A fails alone) = P (E_A) – P(E_A and E_B)
= 0.2 – 0.15 = 0.05.

Question 16.
Bag I contains 3 red and 4 black halls and Bag II contains 4 red and 5 black halls. One hall is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.
Solution.
Let E₁ and E₂ respectively denote the events that a red ball is transferred from bag I to II and a black ball is transferred from bag I to II.
P(E₁) = \(\frac{3}{7}\) and

P(E₂) = \(\frac{4}{7}\)

Let A be the event that the ball drawn is red.
When a red ball is transferred from bag I to II,

P(A/E₁) = \(\frac{5}{10}=\frac{1}{2}\)

When a black ball is transferred from bag I to II,
P(A/E₂) = \(\frac{4}{10}=\frac{2}{5}\)

Direction (17 – 19) : Choose the correct answer.
Question 17.
If A and B are two events such that P(A) ≠ 0 and P(B/A) = 1, then
(A) A ⊂ B
(B) B ⊂ A
(C) B = Φ
(D) A = Φ
Sol.
Given, P(A) ≠ 0 and P(B/A) = 1

⇒ P(B/A) = \(\frac{P(A \cap B)}{P(A)}\)

⇒ 1 = \(\frac{P(B \cap A)}{P(A)}\)

P(A) = P(A ∩ B) ⇒ A ⊂ B
Thus, the correct answer is (A).

Question 18.
If P(A/B) > P(A), then which of the following is correct
(A) P(B/A) < P(B)
(B) P(A ∩ B) < P(A) . P(B) (C) P(B/A) > P(B)
(D) P(B/A) = P(B)
Solution.
Given, P(A/B) > P(A)
⇒ \(\frac{P(A \cap B)}{P(B)}\) > p(A)
⇒ P(A ∩ B) > P(A) . P(B)

⇒ \(\frac{P(A \cap B)}{P(A)}\) p(B)
⇒ P(B/A) > P(B)
Thus, the correct answer is (C).

Question 19.
If A and B are any two events such that P(A) + P(B) – P (A and B) = P (A), then
(A) P(B/A) = 1
(B) P(A/B) = 1
(C) P(B/A) = 0
(D) P(A/B) = 0
Sol.
Given, P(A) + P(B) – P(A and B) = P(A)
⇒ P(A) + P(B) – P(A ∩ B) = P(A)
⇒ P(B) – P(A ∩ B) = 0
⇒ P(A ∩ B) = P(B)
∴ P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{P(B)}{P(B)}\) = 1
Thus, the correct answer is (B).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 13 Probability Ex 13.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 13 Probability Ex 13.5

Question 1.
A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of
(i) 5 successes?
(ii) at least 5 successes?
(iii) at most 5 successes?
Solution.
The repeated tosses of a die are Bernoulli trials.
Let X denote the number of successes of getting odd numbers in an experiment of 6 trials.
Probability of getting an odd number in a single throw of a die is,
P = \(\frac{3}{6}=\frac{1}{2}\)

q = 1 – p = \(\frac{1}{2}\)
X has a binomial distribution.

Question 2.
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two successes.
Solution.
The repeated tosses of a pair of dice are Bernoulli trails.
Let X denote the number of times of getting doublets in an experiment of throwing two dice simultaneously four times.
Probability of getting doublets in a single throw of the pair of dice is
p = \(\frac{6}{36}=\frac{1}{6}\)

q = 1 – \(\frac{1}{6}\)
= \(\frac{5}{6}\)
Clearly, X has the binomial distribution with, n = 4, p = \(\frac{1}{6}\) and q = \(\frac{5}{6}\)

Question 3.
There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item?
Solution.
Let X denote the number of defective items in a sample of 10 items drawn successively.
Since the drawing is done with replacement the trials are Bernoulli trials.

Question 4.
Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that
(i) all the five cards are spades?
(ii) only 3 cards are spades?
(iii) none is a spade?
Solution.
Let X represent the number of spade cards among the five cards drawn.
Since the drawing of card is with replacement the trials are Bernoulli trials.
In a well shuffled deck of 52 cards, there are 13 spade cards.

Question 5.
The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs
(i) none
(ii) not more than one
(iii) more than one
(iv) at least one will fuse after 150 days of use.
Solution.
Probability that a bulb gets fuse after 150 days of its use = 0.05
Probability that the bulb will not fuse after 150 days of its use
=1 – 0.05
= 0.95
(i) Probability that no bulb will fuse after 150 days of its use
= P(none) = (0.95)
(ii) P (not more than one)
= P(0) + P(1)
= (0.95)⁵ + \({ }^{5} C_{1}\) × (0.95)⁴ × (0.05)
= (0.95)⁴ [0.95 + 5 × 0.05]
= (0.95)⁴ (0.95 + 0.25)
=(0.95) × 1.2

(iii) P(more than one) = P(2) + P(3) + P(4) + P(5)
= [P(0) + P(1) + P(2) + P(3) + P(4) + P(5)] – [P(0) + P(1)]
= 1 – [P(0) + P(1)]
= 1 – (0.95) × 1..2 [see part (ii)]
(iv) A(at least one) = P(1) + P(2) + P(3) + P(4) + P(5)
= 1 – P(0)
= 1 – (0.95)⁵ [From part (i)].

Question 6.
A bag consists of 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0.
Solution.
Let X denote the number of balls marked with the digit 0 among the 4 balls drawn.
Since the balls are drawn with replacement, the trials are Bernoulli trials. X has a binomial distribution with n = 4 and p = \(\frac{1}{10}\)
∴ q = 1 – p
= 1 – \(\frac{1}{10}\) = \(\frac{9}{10}\)

Question 7.
In an examination, 20 questions of true-false type are asked Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads he answers, ‘ture’. If it falls tails he answers ‘false’. Find the probability that he answers at least 12 questions correctly.
Solution.
Let X represent the number of correctly answered questions out of 20 questions.
The repeated tosses of a coin are Bernoulli trails.
Since “head” on a coin represents the true answer and “tail” represents the false answer, the correctly answered questions are Bernoulli trials.

Question 8.
Suppose X has a binomial distribution B (6, \(\frac{1}{2}\)). Show that X = 3 is the most likely outcome.
[Hint : P(X = 3) is the P(X_i), X_i = 0, 1, 2, 3, 4, 5, 6]
Solution.
X is the random variable whose binomial distribution is B
Therefore, n = 6 and p = \(\frac{1}{2}\)
∴ q = 1 – p
= 1 – \(\frac{1}{2}\)
= \(\frac{1}{2}\)

The value of \({ }^{6} C_{3}\)is maximum.
Therefore, for x = 3, P(X = x) is maximum.
Thus, X = 3 is the most likely outcome.

Question 9.
On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?
Solution.
The repeated guessing of correct answer from multiple choice questions are Bernoulli trials.
Let X represent the number of correct answers by guessing the the set of 5 multiple choice questions.
Probability of getting a correct answer is, p = \(\frac{1}{3}\)
∴ q = 1 – p
= 1 – \(\frac{1}{3}\)
= \(\frac{2}{3}\)
Clearly, X has a binomial distribution with n = 5 and p = \(\frac{1}{3}\)

= \(5 \cdot \frac{2}{3} \cdot \frac{1}{81}+1 \cdot \frac{1}{243}\)

= \(\frac{10}{243}+\frac{1}{243}=\frac{11}{243}\)

Question 10.
A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is \(\frac{1}{100}\). What is the probability that he will in a prize
(a) at least once
(b) exactly once
(c) at least twice?
Solution.
Let X represent the number of wining prizes in 50 lotteries.
The trials are Bernoulli trials.
Clearly, X has a binomial distribution with n = 50 and p = \(\frac{1}{100}\).

Question 11.
Find the probability of getting 5 exactly twice in 7 throws of a die.
Solution.
The repeated tossing of a die are Bernoulli trials Let X, represent the number of times of getting 5 in 7 throws of the die.
Probability of getting 5 in a single throw of the die, p = \(\frac{1}{6}\)
q = 1 – p
= 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
Clearly, X has the probability distribution with n = 7 and p = \(\frac{1}{6}\)
∴ p(X = x) = \({ }^{n} C_{x} q^{n-x} p^{x}\)

= \({ }^{7} C_{x}\left(\frac{5}{6}\right)^{7-x} \cdot\left(\frac{1}{6}\right)^{x}\)
p(getting 5 exactly twice) = P(X = 2)
= \({ }^{7} C_{2}\left(\frac{5}{6}\right)^{5} \cdot\left(\frac{1}{6}\right)^{2}\)

= \(21 \cdot\left(\frac{5}{6}\right)^{5} \cdot \frac{1}{36}=\left(\frac{7}{12}\right)\left(\frac{5}{6}\right)^{5}\).

Question 12.
Find the probability of throwing at most 2 sixes in 6 throws of a single die.
Solution.
When a die is thrown, probability of getting a six = \(\frac{1}{6}\)
Probability of not getting a six = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
Probability of getting atmost 2 sixes in 6 throws of a single die.
= P(0) + P(1) + P(2)

Question 13.
It is known that 10% of a certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles, 9 are defective?
Solution.
The repeated selections of artides in a random sample space are Bernoulli trails.
Let X denote the number of times of selecting defective articles in a random sample space of 12 articles.
Clearly, X has a binomial distribution with n = 12 and p = 10%

Question 14.
In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is
(A) 10^-1

() \(\left(\frac{1}{2}\right)^{5}\)

(C) \(\left(\frac{9}{10}\right)^{5}\)

(D) \(\frac{9}{10}\)
Solution.
The repeated selections of defective bulbs from a box are Bernoulli trials. Let X denote the number of defective bulbs out of a sample of 5 bulbs.

Probability of getting a defective bulb, p = \(\frac{10}{100}=\frac{1}{10}\)
q = 1 – p
= 1 – \(\frac{1}{10}\)
= \(\frac{9}{10}\)
Clearly, X has a binomial distribution with n = 5 and p = \(\frac{1}{10}\)

Hence, the correct answer is (C).

Question 15.
The probability that a student is not a swimmer is \(\frac{1}{5}\). Then the probability that out of five students, four are swimmers is
(A) \({ }^{5} C_{4}\left(\frac{4}{5}\right)^{4} \frac{1}{5}\)

(B) \(\left(\frac{4}{5}\right)^{4} \frac{1}{5}\)

(C) \({ }^{5} C_{1} \frac{1}{5}\left(\frac{4}{5}\right)^{4}\)

(D) None of these
Solution.
The repeated selection of students who are swimmers are Bernoulli trials.
Let X denote the number of students, out of 5 students, who are swimmers.
Probability of students who are not swimmers, q = \(\frac{1}{5}\)
p = 1 – q
= 1 – \(\frac{1}{5}\)
= \(\frac{4}{5}\)
Clearly, X has a binomial distribution with n = 5 and p = \(\frac{4}{5}\)
P(X = x) = \({ }^{n} C_{x} q^{n-x} p^{x}\)

= \({ }^{5} C_{x} \cdot\left(\frac{1}{5}\right)^{5-x} \cdot\left(\frac{4}{5}\right)^{x}\)
P (four students are swimmers) = P(X = 4)
= \({ }^{5} C_{4}\left(\frac{1}{5}\right) \cdot\left(\frac{4}{5}\right)^{4}\)
Hence, the correct answer is (A).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 13 Probability Ex 13.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4

Question 1.
State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
Solution.
It is known that the sum of all the probabilities in a probability distribution is one.

(i)

Solution.
Sum of the probabilities = 0.4 + 0.4 + 0.2 = 1
Therefore, the given table is a probability distribution of random variables.

(ii)

Solution.
It can be seen that for X = 3, P(X) = – 0.1
It is known that probability of any observation is not negative.
Therefore, the given table is not a probability distribution of random variables.

(iii)

Solution.
Sum if the probabilities = 0.6 + 0.1 + 0.2 = 0.9 < 1. Therefore, the given table is not a probability distribution of random variables.

(iv)

Solution.
Sum of the probabilities = 0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1.05 > 1.
Therefore, the given table is not a probability distribution of random variables.

Question 2.
An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X? Is X a random variable?
Solution.
The two balls are selected can be represented as BB, BR, RB, RR, where B represents a black ball and R represents a red ball.
X represents the number of black balls
∴ X(BB) = 2; X(BR) = 1; X(RB) = 1 and X(RR) = 0
Therefore, the possible values of X are 0, 1, and 2.
Thus, X is a random variable.

Question 3.
Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X?
Solution.
A coin is tossed six times and X represents the difference between the number of heads and the number of tails.
∴ X (6H, 0T) = | 6 – 0 | = 6,
X (4H, 2T) = |4 – 2| = 2,
X (2H, 4T) = |2 – 4| = 2,
X (0H, 6T) = |0 – 6| = 6
X (5H, IT) = | 5 – 1 | = 4
X (0H, 3T) = |3 – 3 | = 0
X(1H, 5T) = |1 – 5| = 4
Thus, the possible values of X are 6, 4, 2 and 0.

Question 4.
Find the probability distribution of
(i) number of heads in two tosses of a coin.
(ii) number of tails in the simultaneous tosses of three coins.
(iii) number of heads in four tosses of a coin.
Solution.
(i) When one coin is tossed twice, the sample space is
S = {TT, HT, TH, HH}
Let X represent the number of heads
∴ X(HH) = 2, X(HT) = 1, X(TH) = 1 and X(7T) = 0
Therefore, X can take the value of 0, 1, or 2. It is known that,
P(HH) = P(HT) = P(TH) = P(TT) = \(\frac{1}{4}\)
∴ P(X = 0) – P(TT) = \(\frac{1}{4}\)
P(X = 1) = P(HT) + P(TH)
= \(\frac{1}{4}\) + \(\frac{1}{4}\) = \(\frac{1}{2}\)
and P(X = 2) = P(HH) = \(\frac{1}{4}\)
Thus, the required probability distribution is as follows

(ii) When three coins are tossed simultaneously, the sample space is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Let X represent the number of tails. It can be seen that X can take the value of 0, 1, 2, or 3. ]
P(X = 0) = P{HHH) = \(\frac{1}{8}\)

P(X = 1) = P(HHT) + P{HTH) + P(THH)
= \(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}\)

P(X = 2) = P(H7T) + P(THT) + P(TTH)
= \(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}\)

P(X = 3) = P(TTT) = \(\frac{1}{8}\)
Thus, the probability distribution is as follows.

(iii) When a coin is tossed four times, the sample space is

Let X be the random variable, which represents the number of heads.
It can be seen that X can take the value of 0, 1, 2, 3, or 4.
P(X = 0) = P(TTT) = \(\frac{1}{6}\)

P(X = 1) = P(TTTH) + P(TTHT) + P(THTT) + P(HTTT)
= \(\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{4}{16}=\frac{1}{4}\)

P(X = 2) = P(HHTT) + P(THHT) + P(TTHH) + P(HTTH) + P(HTHT)
= \(\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{6}{16}=\frac{3}{8}\)

P(X = 3) = P{HHHT) + P(HHTH) + P(HTHH) + P(THHH)
= \(\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{4}{16}=\frac{1}{4}\)

P(X = 4) = P(HHHH) = \(\frac{1}{16}\)
Thus, the probability distribution is as follows.

Question 5.
Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as
(i) number greater than 4
(ii) six appears on atleast one die
Solution.
When a die is tossed two times, we get (6 × 6) = 36 number of sample points.
(i) Let X be the random variable, which represents the number of successes. Here, success refers to the number greater than 4.
P(X = 0) = P (number less than or equal to 4 on both the tosses)
= \(\frac{4}{6} \times \frac{4}{6}=\frac{4}{9}\)

P(X = 1) = P (number less than or equal to 4 on first toss and greater than 4 on second toss) + P (number greater than 4 on first toss and less than or equal to 4 on second toss)
= \(\frac{4}{6} \times \frac{2}{6}+\frac{4}{6} \times \frac{2}{6}=\frac{4}{9}\)

P(X = 2) = P (number greater than 4 on both the tosses)
= \(\frac{2}{6} \times \frac{2}{6}=\frac{1}{9}\)

Thus, the probability distribution is as follows.

(ii) Here, success means six appears on at least one die.

P(X = 0) = P (six does not appear on any of the dice)
= \(\frac{5}{6} \times \frac{5}{6}=\frac{25}{36}\)

P(X = 1) = P (six appears on at least one of the die)
= \(\frac{11}{36}\)

Thus, the required probability distribution is as follows.

Question 6.
From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.
Solution.
It is given that out of 30 bulbs 6 are defective.
Number of non-defective bulbs = 30 – 6 = 24
4 bulbs are drawn from the lot with replacement
Let X be the random variable that denotes the number of defective bulbs in the selective bulbs.
∴ P(X = 0) = P (4 non-defective and 0 defective)
= \({ }^{4} C_{0}\)
= \(\frac{4}{5} \cdot \frac{4}{5} \cdot \frac{4}{5} \cdot \frac{4}{5}=\frac{256}{625}\)

P(X = 1) = P (3 non-defective and 1 defective)
= \({ }^{4} C_{1}=\left(\frac{1}{5}\right) \cdot\left(\frac{4}{5}\right)^{3}=\frac{256}{625}\)

P(X = 2) = P (2 non-defective and 2 defective)
= \({ }^{4} C_{2}=\left(\frac{1}{5}\right)^{2} \cdot\left(\frac{4}{5}\right)^{2}\)

= \(6 \times \frac{16}{25} \times \frac{1}{25}=\frac{96}{625}\)

P(X = 3) = P (1 non-defective and 3 defective)
= \({ }^{4} C_{3}=\left(\frac{1}{5}\right)^{3} \cdot\left(\frac{4}{5}\right)\)

= \(4 \times \frac{4}{5} \times \frac{1}{125}=\frac{16}{625}\)

P(X = 4) = P (0 non-defective and 4 defective)
= \({ }^{4} C_{4}=\left(\frac{1}{5}\right)^{4} \cdot\left(\frac{4}{5}\right)^{0}=\frac{1}{625}\)

Therefore, the required probability distribution is as follows.

Question 7.
A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.
Solution.
Let the probability of getting a tail in the biased coin be x.
∴ P(T) = x
⇒ P(H) = 3x
For a biased coin, P(T) + P(H) = 1
⇒ x + 3x = 1
⇒ 4x = 1
⇒ x = \(\frac{1}{4}\)
∴ P(T) = \(\frac{1}{4}\) and P(H) = \(\frac{3}{4}\)
When the coin is tossed twice the sample space is S = {HH,TT,HT,TH}
Let X be the random variable representing the number of tails
∴ P(X = 0) = P (no tail) = P(H) x P(H)
= \(\frac{3}{4} \times \frac{3}{4}=\frac{9}{16}\)

P(X = 1) = P (one tail) = P(HT) + P(TH)
= \(\frac{3}{4} \cdot \frac{1}{4}+\frac{1}{4} \cdot \frac{3}{4}=\frac{3}{16}+\frac{3}{16}=\frac{3}{8}\)

P(X = 2) = P (two tails) = P(TT)
= \(\frac{1}{4} \times \frac{1}{4}=\frac{1}{16}\)

Therefore, the required probability distribution is as follows.

Question 8.
A random variale X has the following proaility distribution.

Determine
(i) k
(ii) P(X < 3) (iii) P(X > 6)
(iv)P(0 < X < 3)
Solution.
(i) It is known that the sum of probabilities of a probability distribution of random variables is one
∴ 0 + k + 2k + 2k + 3k + k² + 2 k² + (7k² + k) = 1
⇒ 10k² + 9k – 1 = 0
(10k -1) (k + 1) = 0
k = – 1, \(\frac{1}{10}\)
k = – 1 is not possible as the probability of an event is never negative.
∴ k = \(\frac{1}{10}\)

(ii) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + k + 2k = 3k = 3 × \(\frac{1}{10}\) = \(\frac{3}{10}\) (iii) P(X > 6) = P(X = 7)
= 7k² + k
= 7 × (\(\frac{1}{10}\))² + \(\frac{1}{10}\)
= \(\frac{7}{100}+\frac{1}{10}=\frac{17}{100}\)

(iv) P(0 < X < 3) = P(X = 1)
= P(X = 1) + P(X = 2)
= k +2k = 3k
= 3 × \(\frac{1}{10}\) = \(\frac{3}{10}\)

Question 9.
The random variable X has a probability distribution P(X) of the following form, where k is some number.

(a) Determine the value of k.
() Find P(X < 2), P(X ≤ 2), P(X ≥ 2).
Solution.
(a) It is known that the sum of probabilities of a probability distribution of random variables is one.
∴ k + 2 k + 3k + 0 = 1
⇒ 6k = 1
⇒ k = \(\frac{1}{6}\)

(b) P(X ≤ 2) = P(X = 0) + P(X = 1)
= k + 2k = 3k
= 3 × \(\frac{1}{6}\)
= \(\frac{3}{6}=\frac{1}{2}/latex]

P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = k + 2k + 3k = 6k = [latex]\frac{6}{6}\) = 1 P(X ≥ 2) = P(X = 2) + P(X > 2)
= 3k + 0 = 3k
= \(\frac{3}{6}=\frac{1}{2}/latex]

Question 10.
Find the mean number of heads in three tosses of a fair coin.
Solution.
Let X denote the success of getting heads.
Therefore, the sample space is
S = {HHH, HHT, HTH, HIT, THH, THT, TTH, TTT}
It can be seen that X can take the value of 0, 1, 2 or 3.
∴ P(X = 0) = P(TTT)
= P(T) . P(T) . P(T)
= [latex]\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8}\)

∴ P(X = 1) = P(HHT) + P(HIH) + P(THH)
= \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{3}{8}\)

∴ P(X = 2) = P(HHT) + P(HTH) + P(THH)
= \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{3}{8}\)

P(X = 3) = P(HHH)
= \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8}\)

Therefore, the required probability distribution is as follows

Mean of Σ X P(X) = \(0 \times \frac{1}{8}+1 \times \frac{3}{8}+2 \times \frac{3}{8}+3 \times \frac{1}{8}\)

= \(\frac{3}{8}+\frac{3}{4}+\frac{3}{8}=\frac{3}{2}\) = 1.5

Question 11.
Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.
Solution.
Here, X represents the number of sixes obtained when two dice are thrown simultaneously.
Therefore, X can take the value of 0, 1 or 2.
∴ P(X = 0) = P (not getting six on any of the dice) = \(\frac{25}{36}\)

P(X = 1) = P (six on-first die and non six on second die) + P (non-six on the first die and six on the second die)
= \(2\left(\frac{1}{6} \times \frac{5}{6}\right)=\frac{10}{36}\)

P(X = 2) = P (six on both dice)
= \(\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}\)

Therefore, the required probability distribution is as follows.

Then, expectation of X = mean of the variable X = ΣX P(X)
= \(0 \times \frac{25}{36}+1 \times \frac{10}{36}+2 \times \frac{1}{36}\)

= \(\frac{10}{36}+\frac{2}{36}=\frac{1}{3}\)

Question 12.
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find E(X)
Solution.
There are six numbers 1, ,2, 3, 4, 5, 6.
One of them is selected in 6 ways.
When one of the number has been selected, 5 numbers are left.
One number out of 5 may be selected in 5 ways
∴ No. of ways of selecting two numbers without replacement out of 6 positive integers = 6 x 5 = 30

Question 13.
Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.
Solution.
When two fair dice are rolled, 6 × 6 = 36 observations are obtained.
P(X = 2) = P(1, 1) = \(\frac{1}{36}\)

P(X = 3) = P(1, 2) + P(2, 1)
= \(\frac{2}{36}=\frac{1}{18}\)

P(X = 4) = P(1, 3) + P(2, 2) + P(3, 1)
= \(\frac{3}{36}=\frac{1}{12}\)

P(X = 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(4, 1)
= \(\frac{4}{36}=\frac{1}{9}\)

P(X = 6) = P(1, 5) + P(2, 4) + P(3, 3) + P(4, 2) + P(5, 1)
= \(\frac{5}{36}\)

P(X = 7) = P(1, 6) + P(2, 5) + P(3, 4) + P(4, 3) + P(5, 2) + P(6, 1)
= \(\frac{6}{36}=\frac{1}{6}\)

P(X = 8) = P(2, 6) + P(3, 5) + P(4, 4) + P(5, 3) + P(6, 2)
= \(\frac{5}{36}\)

P(X = 9) = P(3, 6) + P(4, 5) +P (5, 4) + P(6, 3)
= \(\frac{4}{36}=\frac{1}{9}\)

P(X = 10) = P(4, 6) + P(5, 5) + P(6, 4)
= \(\frac{3}{36}=\frac{1}{12}\)

P(X = 11) = P(5, 6) + P(6, 5)
= \(\frac{2}{36}=\frac{1}{18}\)

P(X = 12) =P(6, 6)
= \(\frac{1}{36}\)
Therefore, the required probability distribution is as follows
Mean X = Σ X P(X)

Question 14.
A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean variance and standard deviation of X.
Solution.
There are 15 students in the class. Each students has the same chance to be chosen.
Therefore, the probability of each student to be selected is \(\frac{1}{15}\).
The given information can be compiled in the frequency table as follows.

Question 15.
In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed and X = 1, if he is in favour. Find E(X) and Var(X).
Solution.
It is given that P(X = 0) = 30%
= \(\frac{30}{100}\) = 0.3

P(X = 1) = 70%
= \(\frac{70}{100}\) = 0.7
Therefore, the probability distribution is as follows.

Mean of X = E(X) = ΣXP(X)
= 0 × 0.3 + 1 × 0.7 = 0.7

Variance of X = ΣX²P(X) – (Mean)²
= 0² × 0.3 + (1)² × 0.7 – (0.7)²
= 0.7 – 0.49 = 0.21

Direction (16 – 17): Choose the correct answer.

Question 16.
The mean of the number obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is
(A) 1
(B) 2
(C) 5
(D) \(\frac{8}{3}\)
Solution.
Let X be the random variable representing a number on the die.
The total number of observations is six.
P(X = 1) = \(\frac{3}{6}=\frac{1}{2}\)

P(X = 2) = \(\frac{2}{6}=\frac{1}{3}\) and

P(X = 5) = \(\frac{1}{6}\)

Therefore, the probability distribution is as follows.

Hence, the correct answer is (B).

Question 17.
Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E{X) is
(A) \(\frac{37}{221}\)

(B) \(\frac{5}{13}\)

(C) \(\frac{1}{13}\)

(D) \(\frac{2}{13}\)
Solution.
Let X denote the number of aces obtained.
Therefore, X can take any of the values of 0, 1 or 2.
In a deck of 52 cards, 4 cards are aces.
Therefore, there are 48 non-ace cards.
∴ P(X = 0) = P (0 ace and 2 non-ace cards)
= \(\frac{{ }^{4} C_{0} \times{ }^{48} C_{2}}{{ }^{52} C_{2}}\)

= \(\frac{1128}{1326}\)

P(X = 1) = P (1 ace and 1 non-ace cards)
= \(\frac{{ }^{4} C_{1} \times{ }^{48} C_{1}}{{ }^{52} C_{2}}\)

= \(\frac{192}{1326}\)

P(X – 2) = P (2 ace and 0 non-ace cards)
= \(\frac{{ }^{4} C_{2} \times{ }^{48} C_{0}}{{ }^{52} C_{2}}\)

= \(\frac{6}{1326}\)

Thus, the probability distribution is as follows.

img 22

Then, E(X) = ΣX P(X)
= \(0 \times \frac{1128}{1326}+1 \times \frac{192}{1326}+2 \times \frac{6}{1326}=\frac{204}{1326}\)

= \(\frac{2}{13}\)
Hence, the correct answer is (D).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 13 Probability Ex 13.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2

Question 1.
If P(A) = \(\frac{3}{5}\) and P(B) = \(\frac{1}{5}\), find P(A ∩ B) if A and B are independent events.
Solution.
It is given that P(A) = \(\frac{3}{5}\) and P(B) = \(\frac{1}{5}\)
As, A and B are independent events.
Therefore, P(A ∩ B) = P(A) . P(B)
= \(\frac{3}{5}\) × \(\frac{1}{5}\)
= \(\frac{3}{25}\).

Question 2.
Two cards are drawn at random and without replacement from a pack of 52 playIng cards. Find the probabifity that both cards are black.
Solution.
There are 26 black cards in a pack of 52 cards.
Let P(A) be the probability of getting a black card in the first draw
∴ P(A) = \(\frac{26}{52}=\frac{1}{2}\)
Let P(B) be the probability of getting a black card on the second draw Since, the card is not replaced
∴ P(B) = \(\frac{25}{51}\)
Thus, probability of getting both the cards black = \(\frac{1}{2}\) × \(\frac{25}{51}\)
= \(\frac{25}{102}\).

Question 3.
A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.
Solution.
Let A, B and C be the respective events that the first, second, and third drawn orange is good, P(A) = \(\frac{12}{15}\)

The oranges are not replaced

Therefore, probability of getting second orange good, P(B) = \(\frac{11}{14}\)
Similarly, probability of getting third orange good, P(C) = \(\frac{10}{13}\)
The box is approved for sale, if all the three oranges are good.
Thus, probability of getting all the oranges good = \(\frac{12}{15} \times \frac{11}{14} \times \frac{10}{13}=\frac{44}{91}\)
Therefore, the probability that the box is approved for sale is \(\frac{44}{91}\)

Question 4.
A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.
Solution.
If a fair coin and an unbiased die are tossed, then the sample space S is given by

Let A: Head appears on the coin
A = {(H, 1)(H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}
⇒ P(A) = \(\frac{6}{12}=\frac{1}{2}\)
B : 3 on die = {(H, 3),(T,3)}
P(B) = \(\frac{2}{12}=\frac{1}{6}\)
∴ A ∩ B = {(H, 3)}
⇒ P(A ∩ B) = \(\frac{1}{12}\)
P(A) . P(B) = \(\frac{1}{2} \times \frac{1}{6}=\frac{1}{12}\)
= P(A ∩ B)
Therefore, A and B are independent events.

Question 5.
A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even’, and B be the event, ‘the number is red’. Are A and B independent?
Solution.
When a die is thrown, the sample space
S = {1, 2, 3, 4, 5, 6}
Let A: number is even = {2, 4, 6}
⇒ P(A) = \(\frac{3}{6}=\frac{1}{2\)
B : number is red = {1, 2, 3}
P(B) = \(\frac{3}{6}=\frac{1}{2\) and A ∩ B = {2}
P(AB) = P(A ∩ B) = \(\frac{1}{6}\)
P(A) . P(B) = \(\frac{1}{2} \times \frac{1}{2}=\frac{1}{4} \neq \frac{1}{6}\).
⇒ P(A)P(B) ≠ P(A ∩ B)
Therefore, A and B are not independent.

Question 6.
Let E and F be the events with P(E) = \(\frac{3}{5}\), P(F) = \(\frac{3}{10}\) and P(E ∩ F) = \(\frac{1}{5}\). Are E and F independent?
Solution.
It is given that, P(E) = \(\frac{3}{5}\), P(F) = \(\frac{3}{10}\) and P(E ∩ F) = \(\frac{1}{5}\)
∴ P(E) . P(F) = \(\frac{3}{5} \times \frac{3}{10}=\frac{9}{50} \neq \frac{1}{5}\) = P(E ∩ F)
Therefore, A and B are nor independent.

Question 7.
Given that the events A and B are such that P(A) = \(\frac{1}{2}\), P(A ∪ B) = \(\frac{3}{5}\) and P(B) = p. Find p if they are
(i) mutually exclusive
(ii) independent
Solution.
It is given that P(A) = \(\frac{1}{2}\), P(A ∩ B) = \(\frac{3}{5}\) and P(B) = p
(i) When A and B are mutually exclusive, A ∩ B = Φ
∴ P(A ∩ B) = 0
It is known that, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
\(\frac{3}{5}=\frac{1}{2}+p-0\)

p = \(\frac{3}{5}-\frac{1}{2}=\frac{1}{10}\)

(ii) When A and B are independent, P(A ∩ B) = P(A) . P(B) = \(\frac{1}{2}\) p
It is known that, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ \(\frac{3}{5}=\frac{1}{2}+p-\frac{1}{2} p\)

⇒ \(\frac{3}{5}=\frac{1}{2}+\frac{p}{2}\)

⇒ \(\frac{p}{2}=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}\)

⇒ p = \(\frac{2}{10}=\frac{1}{5}\).

Question 8.
Let A and B be independent events with P(A) = 0.3 and P(B)= 0.4.Find
(I) P(A ∩ B)
(ii) P(A ∪ B)
(iii) P(A/B)
(iv) P(B/A)
Solution.
k is given that P(A) = 0.3 and P(B) 0.4
(i) If A and B are independent events, then
P(A ∩ B) = P(A) × P(B) = 0.3 × 0.4 = 0.12

(ii) We know that, P(A ∪ B) = P(A)+ P(B) – P(A ∩ B)
⇒ P(A ∪ B) = 0.3 + 0.4 – 0.12 = 0.58

(iii) We know that, P(A/B) = \(\frac{P(A \cap B)}{P(B)}\)
⇒ P(A/B) = \(\frac{0.12}{0.4}\) = 0.3

(iv) We know that, P(B/A) = \(\frac{P(B \cap A)}{P(A)}\)
⇒ P(B/A) = \(\frac{0.12}{0.3}\) = 0.4

Question 9.
If A and B are two events such that P(A) = \(\frac{1}{4}\), P(B) = \(\frac{1}{2}\) and P(A ∩ B) = \(\frac{1}{8}\) find P (not A and not B).
Solution.
We have, P(A) = \(\frac{1}{4}\), P(B) = \(\frac{1}{2}\), P(A ∩ B) = \(\frac{1}{8}\)
⇒ P(A’) = 1 – P(A)
= 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\) and
P(B’) = 1 – P(B)
= 1 – \(\frac{1}{2}\)
= \(\frac{1}{2}\)
∴ As P(A ∩ B) = \(\frac{1}{8}\)
= \(\frac{1}{4}\) × \(\frac{1}{2}\)
= P(A) × P(B)
Therefore, A and B are independent events.
⇒ A’ and B’ are also independent events
⇒ P(A’ ∩ B’) = P(A’) P(B’)
⇒ P(not A and not B) = P(A’ ∩ B’)
= P(A’ )P(B’)
= \(\frac{3}{4}\) × \(\frac{1}{2}\)
= \(\frac{3}{8}\).

Question 10.
Events A and B are such that P(A) = \(\frac{1}{2}\), P(B) = \(\frac{7}{12}\) and P (not A or
not B) = \(\frac{1}{4}\). State whether A and B are independent?
Solution.

Question 11.
Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6. Find
(i) P (A and B)
(ii) P(A and not B)
(iii) P(A or B)
(iv) P(neither A nor B)
Solution.
It is given that P(A) = 0.3 and P(B) = 0.6
Also, A and B are independent events
(i) P(A and B) = P(A) P(B)
⇒ P(A ∩ B) = 0.3 × 0.6 = 0.18

(ii) P(A and not B~) = P(A ∩ B’)
= P(A) – P(A ∩ B)
= 0.3 – 0.18 = 0.12

(iii) P(A or B) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= 0.3 + 0.6 – 0.18 = 0.72.

(iv) P(neither A nor B) = P(A’ ∩ B’)
= P(A ∪ B)
= 1 – P(A ∪ B)
= 1 – 0.72 = 0.28.

Question 12.
A die is tossed thrice. Find the probability of getting an odd number at least once.
Solution.
Probability of getting an odd number in a single throw of a die = \(\frac{3}{6}=\frac{1}{2}\)
Similarly, probability of getting an even number = \(\frac{3}{6}=\frac{1}{2}\)
Probability of getting an even number three times = \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8}\)
Therefore, probability of getting an odd number at least once.
= 1 – Probability of getting an odd number in none of the throws
= 1 – Probability of getting an even number thrice
= 1 – \(\frac{1}{8}\)
= \(\frac{7}{8}\)

Question 13.
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
(i) both balls are red.
(ii) first ball is black and second is red.
(iii) one of them is black and other is red.
Solution.
Total number of balls = 18
Number of red balls = 8
Number of black balls = 10

(i) Probability of getting a red ball in the first draw = \(\frac{8}{18}=\frac{4}{9}\)
The balls is replaced after the first draw
∴ Probability of getting a red ball in the second draw = \(\frac{8}{18}=\frac{4}{9}\)
Therefore, probability of getting both the balls red = \(\frac{4}{9} \times \frac{4}{9}=\frac{16}{81}\).

(ii) Probability of getting first ball black = \(\frac{10}{18}=\frac{5}{9}\)
The ball is replaced after the first draw.
8 4
Probability of getting second balls as red = \(\frac{8}{18}=\frac{4}{9}\)
Therefore, probability of getting first ball as black and second ball as red = \(\frac{5}{9} \times \frac{4}{9}=\frac{20}{81}\).

(iii) Probability of getting first ball as red = \(\frac{8}{18}=\frac{4}{9}\)
The ball is replaced after the first draw
Probability of getting second ball as black = \(\frac{10}{18}=\frac{5}{9}\)

Therefore, probability of getting first ball as black and second ball as red = \(\frac{4}{9} \times \frac{5}{9}=\frac{20}{81}\).

Therefore, probability that one of them is black and other is red = Probability of getting first ball black and second as red + Probability of getting first ball red and second ball black
= \(\frac{20}{81}+\frac{20}{81}=\frac{40}{81}\).

Question 14.
Probability of solving specific problem independently by A and B are \(\frac{1}{2}\) and \(\frac{1}{3}\) respectively. If both try to solve the problem independently, find the probability that
(i) the problem is solved
(ii) exactly one of them solves the problem.
Solution.
Probability of solving the problem by A, P(A) = \(\frac{1}{2}\)
Probability of solving the problem by B, P(B) = \(\frac{1}{3}\)
Since the problem is solved independently by A and B,
∴ P(AB) = P(A) . P(B)
= \(\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}\)

P(A’) = 1 – P(A)
= 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\);

P(B’) = 1 – P(B)
= 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)

(i) Probability that the problem is solved = P(A ∪ B)
= P(A) + P(B) – P(AB)
= \(\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{4}{6}=\frac{2}{3}\)

(ii) Probability that exactly one of them solves the problems is given by,
P(A)P(B’) + P(B)P(A’)
= \(\frac{1}{2} \times \frac{2}{3}+\frac{1}{2} \times \frac{1}{3}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}\)

Question 15.
One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?
(i) E: ‘the card drawn is a spade’.
F: ‘the card drawn is an ace’

(ii) E: ‘the card drawn is black.
F: ‘the card drawn is a king’.

(iii) E: ‘the card drawn is a king or queen’
F: ‘the card drawn is a queen or jack’
Solution.
(i) In a deck of 52 cards, 13 cards are spades and 4 cards are aces
∴ P(E) = P (the card drawn is a spade) = \(\frac{13}{52}=\frac{1}{4}\)

∴ P(F) = P (the card drawn is an ace) = \(\frac{4}{52}=\frac{1}{13}\)
In the deck of cards, only 1 card is an ace of spades
P(EF) = P (the card drawn is spade and an ace) = \(\frac{1}{52}\)

P(E) × P(F) = \(\frac{1}{4} \cdot \frac{1}{13}=\frac{1}{52}\) = P(EF)
⇒ P(E) × P(F) = P(EF)
Therefore, the events E, and F are independent.

(ii) In a deck of 52 cards, 26 cards are black and 4 cards are kings.
∴ P(E) = P(the card drawn is black) = \(\frac{26}{52}=\frac{1}{2}\)

∴ P(F) = P(the card drawn is a king) = \(\frac{4}{52}=\frac{1}{13}\)

In the pack of 52 cards, 2 cards are black as well as kings.
∴ P(EF) = P(the card drawn is a black king)
= \(\frac{2}{52}=\frac{1}{26}\)

⇒ P(E) × P(F) = \(\frac{1}{2} \cdot \frac{1}{13}=\frac{1}{26}\) = P(EF)
Therefore, the given events E and F are independent.

(iii) In a deck of 52 cards, 4 cards are kings, 4 cards are queens and 4 cards are jacks.
∴ P(E) = P(the card drawn is a king or a queen) = \(\frac{8}{52}=\frac{2}{13}\)

P(F) = P(the card drawn is a queen or a jack) = \(\frac{8}{52}=\frac{2}{13}\)

There are 4 cards which are king or queen and queen or jack.
∴ P(EF) = P(the card drawn is a king or a queen, or queen or a jack)
= \(\frac{4}{52}=\frac{1}{13}\)

P(E) × P(F) = \(\frac{2}{13} \cdot \frac{2}{13}=\frac{4}{169} \neq \frac{1}{13}\)
⇒ P(E) . P(F) ≠ P(EF)
Therefore, the given events E and F are not independent.

Question 16.
In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is a selected at random.
(a) Find the probability that she reads neither Hindi nor English newspapers.
(b) If she reads Hindi news paper, find the probability that she reads English newspaper.
(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.
Solution.
Let H denote the students who read Hindi newspaper and E denote the students who read English newspaper.
It is given that, P(H) = 60%
= \(\frac{6}{10}=\frac{3}{5}\)

P(E) = 40%
= \(\frac{40}{100}=\frac{2}{5}\) and

P(H ∩ E) = 20%
= \(\frac{20}{100}=\frac{1}{5}\)

(a) Probability that a student reads Hindi or English newspaper is (H ∪ E’) = 1 -P(H ∪ E)
= 1 – {P(H) + P(E) – P(H ∩ E)}
= 1 – \(\left(\frac{3}{5}+\frac{2}{5}-\frac{1}{5}\right)\)
= 1 – \(\frac{4}{5}\)
= \(\frac{1}{5}\)

(b) Probability that a randomly chosen student reads English newspaper, if she reads Hindi newspaper,is given by P(E|H)

P(E/H) = \(\frac{P(E \cap H)}{P(H)}=\frac{\frac{1}{5}}{\frac{3}{5}}=\frac{1}{3}\)

(c) Probability that a randomly chosen student reads Hindi newspaper, if she reads English newspapers, is given by P(H|E).

P(H/E) = \(\frac{P(H \cap E)}{P(E)}=\frac{\frac{1}{5}}{\frac{1}{2}}=\frac{1}{2}\).

Direction (17 – 18): Choose the correct answer.

Question 17.
The probability of obtaining an even prime number on each die, when a pair of dice is rolled is
(A) 0
(B) \(\frac{1}{3}\)

(C) \(\frac{1}{12}\)

(D) \(\frac{1}{36}\)
Solution.
When two dice are rolled, the number of out comes is 6 × 6 = 36.
The only even prime number is 2.
Let E be the event of getting an even prime number on each die.
∴ E = {(2, 2)}
⇒ P(E) = \(\frac{1}{36}\)
Therefore, the correct answer is (D).

Question 18.
Two events A and B will be independent, if
(A) A and B are mutually exclusive
(B) P(A’B’) = [1 – P(A)] [1 – P(B)]
(C) P(A) = P(B)
(D) P(A) + P(B) = 1
Solution.
Two events A and B are independent if P(A ∩ B) = P(A) P(B)
∴ P(A’ ∩ B’) = P(A’) P(B’)
= P(A ∪ B’) = 1 – P(A ∪ B)
= 1 – [P(A) + P(B) – P(A ∩ B)]
= 1 – P(A) – P(B) + P(A)P(B) [∵ P(A ∩ B) = P(A)P(B)]
= [1 – P(A)] [1 – P(B)]
Therefore, the correct answer is (B).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 13 Probability Ex 13.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 13 Probability Ex 13.3

Question 1.
An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
Solution.
The urn contains 5 red and 5 black balls.
Let a red ball be drawn in the first attempt
∴ P (drawing a red ball) = \(\frac{5}{10}=\frac{1}{2}\)
If two red balls are added to the urn, then the urn contains 7 red and 5 black balls
P (drawing a red ball) = \(\frac{7}{12}\)
Let a black ball be drawn in the first attempt
P (drawing a black ball in the first attempt) = \(\frac{5}{10}=\frac{1}{2}\)
If two black balls are added to the urn, then the urn contains 5 red and 7 black balls.
P (drawing a red ball) = \(\frac{5}{12}\)
Therefore, probability of drawing second ball as red is \(\frac{1}{2} \times \frac{7}{12}+\frac{1}{2} \times \frac{5}{12}\)
= \(\frac{1}{2}\left(\frac{7}{12}+\frac{5}{12}\right)\)

= \(\frac{1}{2}\) × 1
= \(\frac{1}{2}\).

Question 2.
A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that ball is drawn from the first bag.
Solution.
Let E₁ and E<sub2 be the events of selecting first bag and second bag respectively.
P(E₁) = P(E₂) = \(\frac{1}{2}\)
Let A be the event of getting a red ball
⇒ P(A/E₁) = P (drawing a red ball from first bag)
= \(\frac{4}{8}=\frac{1}{2}\)

⇒ P(A/E₂) = P (drawing a red ball from second bag)
= \(\frac{2}{8}=\frac{1}{4}\)

The probability of drawing a ball from the first bag, given that it is red, is given by P(E₂/A).
By using Bayes’ theorem, we get

P(E₁/A) = \([\frac{P\left(E_{1}\right) \cdot P\left(A / E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(A / E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A / E_{2}\right)}]\)

= \(\frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{2} \cdot \frac{1}{2}+\frac{1}{2} \cdot \frac{1}{4}}=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{8}}=\frac{\frac{1}{4}}{\frac{8}{3}}=\frac{2}{3}\).

Question 3.
Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous years results report that 30% of all students who reside in hostel attain. A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year one student is chosen at random from the college and he has an A garde, what is the probability that the student is a hostlier?
Solution.
Let E₁ and E₂ be the events that the student is a hostler and a day scholar respectively and A be the event that the chosen student gets grade A.
∴ P(E₁) = 60%
= \(\frac{60}{100}\) = 0.6;

P(E₂) = 40%
= \(\frac{40}{100}\) = 0.4

P(A/E₁) = P(student getting an A grade is a hostler)
= 30% = 0.3
P(A/E₁) = P(student getting an A grade is a day scholar)
= 20% = 0.2
The probability that a randomly chosen student is a hostler, given that he has an A grade, is given by P(E₁/A).

By using Bayes’ theorem, we get
P(E₁/A) = \(\frac{P\left(E_{1}\right) \cdot P\left(A / E_{1}\right) \quad \text { PSEBS }}{P\left(E_{1}\right) \cdot P\left(A / E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A / E_{2}\right)}\)

= \(\frac{0.6 \times 0.3}{0.6 \times 0.3+0.4 \times 0.2}\)

= \(\frac{0.18}{0.26}=\frac{18}{26}=\frac{9}{13}\).

Question 4.
In answering a question on a multiple choice test, a student either kpows the answer or guesses. Let \(\frac{3}{4}\) be the probability that he knows the answer and \(\frac{1}{4}\) be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability \(\frac{1}{4}\). What is the probability that the student knows the answer given that he answered it correctly?
Solution.
Let E₁ and E₂ be the respective events that the student knows the answer and he guesses the answer.
Let A be the event that the answer is correct
∴ P(E₁) = \(\frac{3}{4}\);

P(E₂) = \(\frac{1}{4}\)

The probability that the student answered correctly, given that he knows the answer, is 1.
∴ P(A/E₁)) = 1

Probability that the student answered correctly, given that he guessed, is \(\frac{1}{4}\).
P(A/E₂) = \(\frac{1}{4}\)

The probability that the student knows the answer given that he answered it correctly, is given by P(E₁/A)
By using Bayes’ theorem, we get
P(E₁/A) = \(\frac{P\left(E_{1}\right) \cdot P\left(A / E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(A / E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A / E_{2}\right)}\)

= \(\frac{\frac{3}{4} \cdot 1}{\frac{3}{4} \cdot 1+\frac{1}{4} \cdot \frac{1}{4}}=\frac{\frac{3}{4}}{\frac{3}{4}+\frac{1}{16}}\)

= \(\frac{\frac{3}{4}}{\frac{13}{16}}=\frac{12}{13}\).

Question 5.
A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e., if a healthy person is tested, then with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?
Solution.
Let E₁ and E₂ be the respective events that a person has a disease and a person has no disease.
Since E₁ and E₂ are events complimentary to each other,
∴ P(E₁) + P(E₂) = 1
=> P(E₂) = 1 – P(E₁)
= 1 – 0.001 = 0.999
Let A be the event that the blood test result it positive.
P(E₁) = 0.1%
= \(\frac{0.1}{100}\) = 0.001

P(A/E₂) = P (result is positive given the person has disease)
= 99% = 0.99

P(A/E₂) = P (result is positive given the person has no disease)
= 0.5% = 0.005.

Probability that a person has a disease, given that his test result is positive is given by P(E₁/A)
By using Bayes’ theorem, we get
P(E₁/A) = \(\frac{P\left(E_{1}\right) \cdot P\left(A / E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(A / E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A / E_{2}\right)}\)

= \(\frac{0.001 \times 0.99}{0.001 \times 0.99+0.999 \times 0.005}\)

= \(\frac{0.00099}{0.00099+0.004995}=\frac{0.00099}{0.005985}\)

= \(\frac{990}{5985}=\frac{110}{665}=\frac{22}{133}\).

Question 6.
There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?
Solution.
There are three coins. Probability that one of them is selected = \(\frac{1}{3}\).
If E₁, E₂, E₃ is the event of selecting a coin and A is the event of getting a head.
∴ P(E₁) = p(E₂) = P(E₃) = \(\frac{1}{3}\)

First coin is two headed ⇒ It will always show head
∴ i.e., P(A/E₁) = 1

The second coin is biased and head come up in 75% cases.
⇒ P(A/E₂) = 0.75 = \(\frac{3}{4}\)

Third coin is unbiased
P(A/E₂) = \(\frac{1}{2}\)
By using Bayes’ theorem, we get

Question 7.
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probabifity of accidents are 0.01, 0.03 and 0.15 respectively. One of the Insured persons meets with an accident. What is the probability that he is a scooter driver?
Solution.
Let E₁, E₂ and E₃ be the respective events that the driver is a scooter driver, a car driver, and a truck driver.
Let A be the event that the person meets with an accident.
There are 2000 scooter drivers, 4000 car drivers, and 6000 truck drivers.
Total number of drivers = 2000 + 4000 + 6000 = 12000

P(E₁) = P (driver is a scooter driver)
= \(\frac{2000}{12000}=\frac{1}{6}\)

P(E₂) = P (driver is a car driver)
= \(\frac{4000}{12000}=\frac{1}{3}\)

P(E₃) = P (driver is a truck driver)
= \(\frac{6000}{12000}=\frac{1}{2}\)

P(A/E₁) = P (scooter driver met with an accident)
= 0.01 = \(\frac{1}{100}\)

P(A/E₂) = P (car driver met with an accident)
= 0.03 = \(\frac{3}{100}\)

P(A/E₃) = P (truck driver met with an accident)
= 0.15 = \(\frac{15}{100}\)

The probability that the driver is a scooter driver, given that, he met with an accident, is given by P(E₁/A).
By using Bayes’ theorem, we get

Question 8.
A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that was produced by machine B?
Solution.
Let E₁ and E₂ be the respective events items produced by machines A and B.
Let X be the event that the produced item was found to be defective.
∴ Probability of items produced by machine A, P(E₁)
= 60% = \(\frac{3}{5}\)

Probability of items produced by machine B,P(E₂)
= 40% = \(\frac{2}{5}\)
Probability that machine A produced defective items,
P(X/E₁) = 2%
= \(\frac{2}{100}\)
Probability that machine B produced defective items,
P(X/E₂) = 1%
= \(\frac{1}{100}\)
The probability that the randomly selected item was from machine B, given that iris defective, is given by P(E₂/X)
By using Bayes’ theorem, we get

Question 9.
Two groups are competing for the position on the board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.
Solution.
Let E₁ and E₂ be the respective events that the first group and the second group win the competition. Let A be the event of introducing a new product.
P(E₁) = Probability that the first group wins the competition = 0.6
P(E₂) = Probability that the second group wins the competition = 0.4
P(A/E₁) = Probability of introducing a new product if the first group wins = 0.7
P(A/E₂) = Probability of introducing a new product if the second group wins = 0.3
The probability that the new product is introduced by the second group is given by P(E₂/A).
By using Bayes’ theorem, we get

Question 10.
Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?
Solution.
Let E₁ be the event that the outcome on the die is 5 or 6 and E₂ be the event that the outcome on the die is 1, 2, 3 or 4.
Let A be the event of getting exactly one head.
P(A/E₁)= Probability of getting exactly one head by tossing the coin three times if she get 5 or 6 = \(\frac{3}{8}\)

P(A/E₂) = Probability of getting exactly one head in a single throw of coin if she gets 1, 2, 3 or 4 = \(\frac{1}{2}\)
The probability that the girl threw 1, 2, 3 or 4 with the die if she obtained exactly one head, is given by P(E₂/A).
By using Bayes’ theorem, we get
P(E₂/A) = \(\frac{P\left(E_{2}\right) \cdot P\left(A / E_{2}\right)}{P\left(E_{1}\right) \cdot P\left(A / E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A / E_{2}\right)}\)

= \(=\frac{\frac{2}{3} \cdot \frac{1}{2}}{\frac{1}{3} \cdot \frac{3}{8}+\frac{2}{3} \cdot \frac{1}{2}}\)

= \(\frac{\frac{1}{3}}{\frac{1}{3}\left(\frac{3}{8}+1\right)}\)

= \(\frac{1}{\frac{11}{8}}=\frac{8}{11}\)

Question 11.
A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5t and 7% defective items respectively. A is on the job for 50% of the time. B is on the job for 30 of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that was produced by A?
Solution.
Let E₁, E₂ and E₃ be the respective events of the time consumed by machines A, B and C for the job.
P(E₁) = 50%
= \(\frac{50}{100}=\frac{1}{2}\)

P(E₂) = 30%
= \(\frac{30}{100}=\frac{3}{10}\)

P(E₃) = 20%
= \(\frac{20}{100}=\frac{1}{5}\)

Let X be the event of producing defective items.
P(X/E₁) = 1%
= \(\frac{1}{100}\)

P(X/E₂) = 5%
= \(\frac{5}{100}\)

P(X/E₃) = 7%
= \(\frac{7}{100}\)

The probability that the defective item was produced by A is given by P(E₁/A).
By using Bayes’ theorem, we get

Question 12.
A card from x a pack of 52 cardsislost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probabifity of the lost card being a diamond.
Solution.
Let E₁ and E₂ be the respective events of choosing a diamond card and a card which is not diamond.
Let A denote the lost card
Out of 52 cards, 13 cards are diamond and 39 cards are not diamond
∴ P(E₁) = \(\frac{13}{52}=\frac{1}{4}\);

P(E₂) = \(\frac{39}{52}=\frac{3}{4}\)

When one diamond card is lost, there are 12 diamond cards out of 51 cards.
Two cards can be drawn out of 12 diamond cards in \({ }^{12} C_{2}\) ways.
Similarly, 2 diamond cards can be drawn out of 51 cards in \({ }^{51} C_{2}\) ways.
The probability of getting two cards, when one diamond card is lost, is given by P(A/E₁)

P(A/E₁) = \(\frac{{ }^{12} C_{2}}{{ }^{51} C_{2}}=\frac{12 !}{2 ! \times 10 !} \times \frac{2 ! \times 49 !}{51 !}\)

= \(\frac{11 \times 12}{50 \times 51}=\frac{22}{425}\)

When the lost card is not a diamond, there are 13 diamond cards out of 51 cards.
Two cards can be drawn out of 13 diamond cards in \({ }^{13} C_{2}\) ways whereas 2 cards can be drawn out of 51 cards in \({ }^{51} C_{2}\) ways.
The probability of getting two cards, when one card is lost which is not diamond, is given by P(A/E₂).

P{A/E₂) = \(\frac{{ }^{13} C_{2}}{{ }^{51} C_{2}}=\frac{13 !}{2 ! \times 11 !} \times \frac{2 ! \times 49 !}{51 !}\)

= \(\frac{12 \times 13}{50 \times 51}=\frac{26}{425}\)

The probability that the lost card is diamond is given by P(E1/A). By using Bayes’ theorem, we get

Question 13.
Probability that A speaks truth is \(\frac{4}{5}\). A coin is tossed. A reports that a head appears. The probability that actually there was head is
(A) \(\frac{4}{5}\)

(B) \(\frac{1}{2}\)

(C) \(\frac{1}{5}\)

(D) \(\frac{2}{5}\)
Solution.
Let E₁ and E₂ be the events such that
E₁ : A speaks truth
E₂ : A does not speak truth
Let X be the event that a head appears.
P(E₁) = \(\frac{4}{5}\)
∴ P(E₂) = 1 – P(E₁)
= 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
If a coin is tossed, then it may result in either head (H) or tail (T).
The probability of getting a head is \(\frac{1}{2}\) whether A speaks truth or not
P(X/E₁) = P(X/E₂) = \(\frac{1}{2}\)
The probability that there is actually a head is given by P(E₁/X)

Hence, the correct answer is (A).

Question 14.
If A and B are two events such that A ⊂ Band P(B) ≠ 0 then which of the following is correct?
(A P(A/B) = \(\frac{P(B)}{P(A)}\)
(B)P(A/B) < P(A)
(C) P(A/B) ≥ P(A)
(D) None of these
Solution.
If A ⊂ B then A ∩ B = A
⇒ P(A ∩ B) = P(A)
P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{P(A)}{P(B)}\)
But P(B) ≤ 1
⇒ P(A/B) ≥ P(A)
Hence, the correct answer is (C).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 13 Probability Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1

Question 1.
Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P(E/F) and P(F/E).
Solution.
It is given that, P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2
⇒ P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{0.2}{0.3}=\frac{2}{3}\)

and P(F/E) = \(\frac{P(E \cap F)}{P(E)}=\frac{0.2}{0.6}=\frac{1}{3}\)

Question 2.
Compute P(A/B), if P(B) 0.5 and P(A ∩ B) = 0.32.
Solution.
It is given that, P(B) = 0.5 and P(A ∩ B)= 0.32
⇒ P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{0.32}{0.5}=\frac{16}{25}\).

Question 3.
If P(A) = 0.8, P(B)= 0.5 and P(B/A)= 0.4, find
(i) P(AB)
(ii) P(A/B)
(iii) P(A ∪ B)
Solution.
k is given that, P(A) = 0.8, P(B) 0.5 and P(B/A) = 0.4
(i) P(B|A) = 0.4
∴ \(\frac{P(A \cap B)}{P(A)}\) = 0.4

⇒ \(\frac{P(A \cap B)}{0.8}\) = 0.4

⇒ P(A ∩ B) = 0.32

(ii) P (A/B) = \(\frac{P(A \cap B)}{P(B)}\)
⇒ P(A/B) = \(\frac{0.32}{0.5}\) = 0.64

(iii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ P(A ∪ B) = 0.8 + 0.5 – 0.32 = 0.98

Question 4.
Evaluate P(A ∪ B), if 2P(A) = P(B) = \(\frac{5}{13}\) and P(A/B) = \(\frac{2}{5}\).
Solution.
It is given that, 2P(A) = P(B) = \(\frac{5}{13}\)

Question 5.
If P(A) = \(\frac{6}{11}\), P(B) = \(\frac{5}{11}\) and P(A uB) = \(\frac{7}{11}\), find
(i) P(A ∩ B)
(ii) P(A/B)
(iii) P(B/A)
Solution.

Direction (6 – 9): Determine P(E/F):

Question 6.
A coin is tossed three times, where
(i) E : head on third toss, F : heads on first two tosses.
(ii) E : at least two heads, F : at most two heads
(iii) E : at most two tails, F : at least on tail
Solution.
When a coin is tossed three times, then the sample space S is
S = {HHH, HHT, HTH, HIT, THH, THT, TTH, TTT}
It, can be seen that the sample space has 8 elements.

(i) E = {HHH, HTH, THH, TTH}
F = {HHH, HHT}
∴ E ∩ F = {HHH}
P(F) = \(\frac{2}{8}=\frac{1}{4}\) and P(E ∩ F) = \(\frac{1}{8}\)
P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{8}}{\frac{1}{4}}=\frac{4}{8}=\frac{1}{2}\)

(ii) E = {HHH, HHT, HTH, THH}
F = {HHT, HTH, HTT, THH, THT, TTH, TTT}
∴ E ∩ F – {HHT, HTH, THH}
Clearly, P(E ∩ F) = \(\frac{3}{8}\) and P(F) = \(\frac{7}{8}\)
P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{3}{8}}{\frac{7}{8}}=\frac{3}{7}\)

(iii) E = {HHH, HHT, HTT, HTH, THH, THT, TTH}
F = {HHT, HTT, HTH, THH, THT TTH, TTT}
∴ E ∩ F = {HHT, HTT, HTH, THH, THT, TTH} v
P(F) = \(\frac{7}{8}\) and P(E ∩ F) = \(\frac{6}{8}\)

Therefore, P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{6}{8}}{\frac{7}{8}}=\frac{6}{7}\)

Question 7.
Two coins are tossed once, where
(i) E : tail appears on one coin, F : one coin shows head
(ii) E : no tail appears, F : no head appears
Solution.
If two coins are tossed once, then the sample space S is,
S = {HH, HT, TH, TT}
E : tail appears on one coin – {TH, HT}
F : one coin shows head = {HT, TH}
E ∩ F : {TH, HT}
∴ P(E ∩ F) = \(\frac{2}{4}=\frac{1}{2}\),

P(F) = \(\frac{2}{4}=\frac{1}{2}\)

P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{1 / 2}{1 / 2}\) = 1

(ii) E = set of events having no tail = {HH}
F = set of events having no head = {TT} and E ∩ F = Φ
P(F) = 1 and P(E ∩ F) = 0
∴ P(E / F) = \(\frac{P(E \cap F)}{P(F)}=\frac{0}{1}\) = 0

Question 8.
A die is thrown three times,
E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses.
Solution.
If a die is thrown three times, then the number of elements in the sample space will be 6 × 6 × 6 = 216

and F = {(6, 5, 1) (6, 5, 2), (6, 5, 3) (6, 5, 4) (6, 5, 5), (6, 5, 6)}

∴ E ∩ F = {(6, 5, 4)}

P(F) = \(\frac{6}{216}\) and P(E ∩ F) = \(\frac{1}{216}\)
[∵ From three dice, number of exhaustive cases = 6 × 6 × 6 = 216]
∴ P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{216}}{\frac{6}{216}}=\frac{1}{6}\)

Question 9.
Mother, father and son line up at random for a family picture
E : son on one end, F : father in middle.
Solution.
If mother (M), father (P), and son (S) line up for family picture, then the sample space will be
S = {MFS, MSF, FMS, FSM, SMF, SFM}
⇒ E = {MFS, FMS, SMF, SFM}
F = {MFS, SFM}
E ∩ F = {MFS, SFM}
P(E ∩ F) = \(\frac{2}{6}=\frac{1}{3}\)
P(F) = \(\frac{2}{6}=\frac{1}{3}\)
P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{1 / 3}{1 / 3}\) = 1

Question 10.
A black and a red dice are rolled.
(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
Solution.
Let the first observation be from the black die and second from the red die.
When two dice (one black and another red) are rolled the sample space S has 6 × 6 = 36 number of elements
(a) Let
A : Obtaining a sum greater than 9 = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)
B : Black die results in a 5 = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
∴ A ∩ B = {(5, 5), (5, 6)}
The conditional probability of obtaining a sum is greater than 9, given that the black die resulted in a 5 is given by P(A/B).
∴ P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{\frac{2}{36}}{\frac{6}{36}}=\frac{2}{6}=\frac{1}{3}\)

(b) E : Sum of the observations is 8 = {2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
F : Red die resulted in a number less than 4
=
⇒ E ∩ F – {(5, 3), (6, 2)}
Similarly, P(F) = \(\frac{18}{36}\) and P(E ∩ F) = \(\frac{2}{36}=\frac{1}{18}\)
The conditional probability of obtaining the sum equal to 8 given that the red die resulted in a number less then 4 is given by P(E/F).
Therefore, P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{2}{36}}{\frac{18}{36}}=\frac{2}{18}=\frac{1}{9}\).

Question 11.
A fair die is rolled. Consider events E = {1, 3, 5}, F (2, 3) and G = {2, 3, 4, 5}. Find
(i) P(E/F) and P(F/E)
(ii) P(E/G) and P(G/E)
(iii) P((E ∪ F)/G) and P((E ∩ F)/G)
Solution.
When a fair die is rolled, the sample space S will be S = {1, 2, 3, 4, 5, 6}

(i)

(ii) E ∩ G = {3, 5}

(iii) E ∪ F = {1, 2, 3, 5}
(E ∪ F) ∩ G = {12, 3, 5} ∩ {2, 3, 4, 5} = {2, 3, 5} £ ∩ F = {3}
(E ∩ F) ∩ G = {3} ∩ {2, 3, 4, 5} = {3}

Question 12.
Assume that each horn child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that
(i) the youngest is a girl,
(ii) at least one is a girl?
Solution.
Let b and g represent the boy and the girl child respectively. If a family has two children, the sample space will be
S = {(b, b), (b, g), (g, b), (g, g)}
Let A be the event that both children are girls.
∴ A = {(g, g)}

(i) Let B be the event that the youngest child is a girl.
B = [(b, g), (g, g)] =$
⇒ A ∩ B = {(g, g)}
⇒ P(B) = \(\frac{2}{4}=\frac{1}{2}\) and P(A ∩ B) = \(\frac{1}{4}\)

The conditional probability that both are girls, given that the youngest child is a girl, is given by P(A/B).

P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}\)
Therefore, the required probability is \(\frac{1}{2}\).

(ii) Let C be the event that at least one child is a girl.
∴ C = {(b, g), (g, b)(g, g)}
⇒ A ∩ C = {g, g}
⇒ P(C) = \(\frac{3}{4}\) and P(A ∩ C) = \(\frac{1}{4}\).
The conditional probability that both are girls, given that at least one child is a girl, is given by P(A/C).

Therefore, P(A/C) = \(\frac{P(A \cap C)}{P(C)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}\).

Question 13.
An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question?
Solution.

Let us denote E = easy questions, M = multiple choice questions, D =
difficult questions, and T = True/False questions
Total number of questions = 1400
Total number of multiple choice questions = 900
Therefore, probability of selecting an easy multiple choice questions is
P(E ∩ M) = \(\frac{500}{1400}=\frac{5}{14}\)
Probability of selecting a multiple choice questions, P(M) is
\(\frac{900}{1400}=\frac{9}{14}\)
Therefore, the required probability is \(\frac{5}{9}\).

Question 14.
Given that the two numbers appearing on throwing the two dice are different. Find the probability of the event, “the sum of numbers on the dice is 4”.
Solution.
When dice is thrown, number of observations in the sample space = 6 × 6 = 36
Let A be the event that the sum of the numbers on the dice is 4 and B be the event that the two numbers appearing on throwing the two dice are different.

P(B) = \(\frac{30}{36}=\frac{5}{6}\) and

P(A ∩ B) = \(\frac{2}{36}=\frac{1}{18}\)

Let P(A/B) represent the probability that the sum of the numbers on the dice is 4, given that the two numbers appearing on throwing the two dice are different.

∴ P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{18}}{\frac{5}{6}}=\frac{1}{15}\)
Therefore, the required probability is \(\frac{1}{15}\).

 

Question 15.
Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.
Solution.
The outcomes of the given experiment can be represented by the following set.
The sample space of the experiment is,

Direction (16 – 17): Choose the correct answer.

Question 16.
If P(A) = \(\frac{1}{2}\), P(B) = 0, then P(A/B) is
(A) 0
(B) \(\frac{1}{2}\)
(C) not defined
(D) 1
Solution.
It is given that P(A) = \(\frac{1}{2}\) and P(B) = 0
P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{P(A \cap B)}{0}=\infty\)
Therefore, P(A/B) is not defined.
Thus, the correct answer is (C).

Question 17.
If A and B are events such that P(A/B) = P(B/A), then
(A) A ⊂ B but A±B
(B) A = B
(C) A ∩ B = Φ
(D) P(A) = P(B)
Solution.
It is given that, P(A/B) = P(B/A)
⇒ \(\frac{P(A \cap B)}{P(B)}=\frac{P(A \cap B)}{P(A)}\)
⇒ P(B) = P(A).
Thus, the correct answer is (D).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 12 Linear Programming Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 12 Miscellaneous Exercise

Question 1.
A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 240 units of calcium, atleast 460 units of iron and at most 300 units of cholesterol.

How many packets of each food should be used to maximise the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet?
Solution.
Let the diet contain x and y packets of foods P and Q respectively.
Therefore, x ≥ 0 and y ≥ 0.
The mathematical formulation of the given problem is as follows.
Maximise Z = 6x + 3y ……………..(i)
subject to the constraints, 4x + y ≥ 80 ………… (ii)
x + 5y ≥ 115 …………..(iii)
3x + 2y ≤ 150 ……………(iv)
x, y ≥ 0 …………….(v)
The feasible region determined by the system of constraints is as follows.

The comer points of the feasible region are A (15, 20), B (40, 15) and C (2, 72).
The values of Z at these corner points are as follows.

Thus, the maximum value of Z is 285 at (40, 15).

Therefore to maximise the amount of vitamin A in the diet, 40 packets of food P and 15 packets of food Q should be used. The maximum amount of vitamin A in the diet is 285 units.

Question 2.
A farmer mixes two brands P and Q of cattle feed. Brand P costing ₹ 250 per bag contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing ₹ 200 per bag contains 1.5 units of nutritional element A, 11.25 units of element B and 3 units of element C.

The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?
Solution.
Let the farmer mix x bags of brand P and y bags of brand Q.
The given information can be compiled in a table as follows

The given problem can be formulated as follows.
Minimise Z = 250x + 200y …………(i)
subject to the constraints, 3x + 1.5y ≥ 18 …………(ii)
2.5x + 11.25y ≥ 45 …………(iii)
2x + 3y ≥ 24 …………..(iv)
x, y ≥ 0 ………….(v)
The feasible region determined by the system of constraints is as follows.

The comer points of the feasible region are A (18, 0) B (9, 2) C (3, 6) and D (0, 12).
The values of Z at these corner points are as follows.

As the feasible region is unbounded, therefore, 1950 may or may not be minimum value of Z.
For this, we draw a graph of the inequality, 250x + 200y < 1950 or 5x + 4y < 39 and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 5x + 4y < 39.
Therefore, the minimum value of Z is 2000 at (3, 6).
Thus, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimise the cost to ₹ 1950.

Question 3.
A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:

One kg of food X costs ₹ 16 and one kg of food Y costs ₹ 20. Find the least cost of the mixture which will produce the required diet?
Solution.
Let the mixture contain x kg of food X and y kg of food Y.
The mathematical formulation of the given problem is as follows,
Minimise Z = 16x + 20y ………….(i)
subject to the constraints, x + 2y ≥ 10 ……… (ii)
x + y ≥ 6 …………….(iii)
3x + y ≥ 8 ………….(iv)
x, y ≥ 0 ………….(v)
The feasible region determined by the system of constraints is as follows.

The corner points of the feasible region are A (10, 0) B (2, 4), C (1, 5) and D (0, 8).
The values of Z at these corner points are as follows.

As the feasible region is unbounded, therefore, 112 may or may not be the minimum value of Z.
For this, we draw a graph of the.. inequality, 16x + 20y <112 or 4x + 5y < 28 and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 4x + 5y < 28.
Therefore, the minimum value of Z is 112 at (2, 4).
Thus, the mixture should contain 2 kg of food X and 4 kg of food Y.
The mixture is ₹ 112.

Question 4.
A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below.

Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is ₹ 7.50 and that on each toy of type B is ₹ 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.
Solution.
Let x and y toys of type A and type B respectively be manufactured in a day.
The given problem can be formulated as follows.
Maximise Z = 7.5x + 5y ……………….(i)
subject to the constraints, 2x + y ≤ 60 …………….(ii)
x ≤ 20 ………………..(iii)
2x + 3y ≤ 120 ……………..(iv)
x, y ≥ 0 …………………(v)
The feasible region determined by the constraints is as follows.
The corner points of the feasible region are A (20, 0), B (20, 20) C (15, 30) and D (0, 40).

The values of Z at these comer points are as follows.

The values of Z is 262.5 at (15, 30).
Thus, the manufacturer should manufacture 15 toys of type A and 30 toys of type B to maximise the profit.

Question 5.
An aeroplane can carry a maximum of 200 passengers. A profit of ₹ 1000 is made on each executive class ticket and a profit of ₹ 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?
Solution.
Let the airline sell x tickets of executive class and y tickets of economy class.
The mathematical formulation of the given problem is as follows.
Maximise Z = 1000x + 600y ……………..(i)
subject to the constraints, x + y ≤ 200 …………(ii)
x ≥ 20 …………(iii)
y – 4x ≥ 0 ……….(iv)
x, y ≥ 0 ……………..(v)
The feasible region determined by the constraints is as follows.

The corner points of the feasible region are A (20, 80), B (40, 160) and C (20, 180).
The values of Z at these corner points are as follows.

The maximum value of Z is 136000 at (40, 160).
Thus, 40 tickets of executive class and 160 tickets of economy class should be sold to maximise the profit and the maximum profit is ₹ 136000.

Question 6.
Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table.

How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost? Solution.
Let godown A supply x and y quintals of grain to the shops D and E respectively. Then, (100 – x – y) will be supplied to shop F.

The requirement at shop D is 60 quintals since x quintals are transported from godown A.
Therefore, the remaining (60 – x) quintals will be transported from godown. B.
Similarly, (50 – y) quintals and 40 – (100 – x – y) = (x + y – 60) quintals will be transported from godown B to shop E and F respectively.
The given problems can be represented diagrammatically in above figure x ≥ 0, y ≥ 0 and 100 – x – y ≥ 0
=» x ≥ 0, y ≥ 0 and x + y ≤ 100 60 – x ≥ 0, 50 – y ≥ 0 and x + y – 60 ≥ 0
⇒ x ≤ 60, y ≤ 50 and x + y ≥ 60
Total transportation cost Z is given by,
Z = 6x + 3y + 2.5 (100 – x – y) + 4 (60 – x) + 2 (50 – y) + 3 (x + y – 60)
= 6x + 3y + 250 – 2.5x – 2.5y + 240 – 4x +100 – 2y + 3x + 3y -180
= 2.5x + 1.5y + 410
The given problem can be formulated as
Minimise Z = 2.5x + 1.5y + 410 …………..(i)
subject to the constraints, x + y ≤ 100 ………..(ii)
x ≤ 60 …………..(iii)
y ≤ 50 …………(iv)
x + y ≥ 60 ………….(v)
x, y ≥ 0 ……………(vi)
The feasible region determined by the constraints is as follows.

The values of Z at these corner points are as follows.

The minimum value of Z is 510 at (10, 50).
Thus, the amount of grain transported from A to D, E and F is 10 quintals, 50 quintals, and 40 quintals respectively and from B to D, E and F is 50 quintals, 0 quintals, and 0 quintals respectively.
The minimum cost is ₹ 510.

Question 7.
An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500 L, 3000 L and 3500 L respectively. The distance (in km) between the depots and the petrol pumps is given in the following table:

Assuming that the transportation cost of 10 litres of oil is ₹ 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?
Solution.
Let x and y litres of oil be supplied from A to the petrol pumps, D and E. Then, (7000 – x – y) will be supplied from A to petrol pump F.
The requirement at petrol pump D is 4500 L.
Since xL are transported from depot A, the remaining (4500 – x) L will be transported from petrol pump B.
Similarly, (3000 – y) L and 3500 – (7000 – x – y) = (x + y -3 500) L will be transported from depot B to petrol E and F respectively.
The given problems can be represented diagrammatically as follows.
x ≥ 0, y ≥ 0 and (7000 – x – y) ≥ 0
⇒ x ≥ 0, y ≥ 0 and x + y ≤ 7000
4500 – x ≥ 0, 3000 – y ≥ 0 and x + y – 3500 ≥ 0
⇒ x ≤ 4500, y ≤ 3000 and x + y ≥ 3500
Cost of transporting 10 L of petrol = ₹ 1
Cost of transporting 1 L of petrol = ₹ \(\frac{1}{10}\)

Therefore total transportation cost is given by,
Z = \(\frac{7}{10}\) x + \(\frac{6}{10}\) y + \(\frac{3}{10}\) (7000 – x – y) + \(\frac{3}{10}\) (4500 – x) + \(\frac{4}{10}\) (3000 – y) + \(\frac{2}{10}\) (x + y – 3500)
= 0.3x + 0.1 y + 3950
The problem can be formulated as follows
Minimise Z = 0.3x + 0.1y + 3950 …………….(i)
subjectto the constraints, x + y ≤ 7000 ………….(ii)
x ≤ 4500 ……………(iii)
y ≤ 3000 ………….(iv)
x + y ≥ 350 …………….(v)
x, y ≥ 0 ………….(vi)
The feasible region determined by the constraints is as follows.

The corner points of the feasible region are A (3500, 0), B (4500, 0) C (4500, 2500), D (1000, 3000) and E (500, 3000).
The values of Z at these corner points are as follows.

The minimum value of Z is 4400 at (500, 3000).
Thus, the oil supplied from depot A is 500 L, 3000 L and 3500 L, and from depot B is 4000 L, 0L and 0 L to petrol pumps D, E and F respectively. The minimum transportation cost is ₹ 4400.

Question 8.
A fruit grower can use two types of fertiliser in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid atleast 270 kg of potash and at most 310 kg of chlorine.

If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?

Solution.
Let the fruit grower use x bags of brand P and y bags of brand Q.
The problem can be formulated as follows.
Minimise Z = 3x + 3.5y
subject to the constraints, x + 2y ≥ 240 ………….(ii)
x + 0.5y ≥ 90 ……………….(iii)
1.5x + 2y ≤ 310 …………….(iv)
x, y ≥ 0
The feasible region determined by the system of constraints is as follows.

The corner points are A(140, 50), B(20, 140) and C(40, 100).

The values of Z at these corner points are as follows.
The minimum value of Z is 470 at (40, 100).
Thus, 40 bags of brand P and 100 bags of brand Q should be added to the garden to minimise the amount of nitrogen.
The minimum amount of nitrogen added to the garden is 470 kg.

Question 9.
Refer to question 8. If the grower wants to maximise the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?
Solution.
Let the fruit grower use x bags of brand P and y bags of brand Q.
The problem can be formulated as follows.
Maximise Z = 3x + 3.5y ……………(i)
subject to the constraints, x + 2y ≥ 240 …………..(ii)
x + 0.5y ≥ 90 …………(iii)
1.5x + 2y ≤ 310 …………..(iv)
x, y ≥ 0 ……….(v)
The feasile region determined y the system of constraints is as follows.

The corner points are A (1. 40, 50), B(20, 140) and C(40, 100).
The values of Z at these corner points are as follows.

The maximum value of Z is 595 at (140, 50).
Thus, 140 bags of brand P and 50 bags of brand Q should be used to maximise the amount of nitrogen.
The maximum amount of nitrogen added to the garden is 595 kg.

Question 10.
A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A.

Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of ₹ 12 and 116 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit?
Solution.
Let x and y be the number of dolls of type A and B respectively that are produced per week.
The given problem can be formulated as follows.
Maximise Z = 12x + 16y ……………(i)
subject to the constraints, x + y ≤ 1200……….(ii)
y ≤ \(\frac{x}{2}\)
⇒x ≥ 2y ………..(iii)
x – 3y ≤ 600 ……………(iv)
x, y ≥ 0 …………….(v)
The feasible region determined by the system of constraints is as follows.

The corner points are A(600, 0), B(1050, 150) and C(800, 400).
The Y’ values of Z at these corner points are as follows.

The maximum value of Z is 16000 at (800, 400).
Thus, 800 and 400 dolls of type A and type B should be produced respectively to get the maximum profit of ₹ 16000.

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 12 Linear Programming Ex 12.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.2

Question 1.
Reshma wishes to mix two types of food P and Q in such a way that vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs ₹ 60/kg and food Q costs ₹ 80/kg. Food P contains 3 units/kg of vitamin A and 5 units/kg of vitamin B while food Q contains 4 units/kg of vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture?
Solution.
Let the mixture contains x kg of food P and y kg of food Q.
Therefore, x ≥ 0 and y ≥ 0
The given information can be compiled in a table as follows.

The mixture must contain at least 8 units of vitamin A and 11 units of vitamin B.
Therefore, the constraints are 3x + 4y ≥ 8; 5x + 2y ≥ 11.
Total cost, Z of purchasing food is Z = 60x + 80 y
The mathematical formulation of the given prolem is

Minimise Z = 60x + 80y …………(i)
subject to the constraints,
3x + 4y ≥ 8 ……………(ii)
5x + 2y ≥ 1 …………….(iii)
x,y ≥ 0 …………..(iv)
The feasible region determined by the system of constraints is as follows.
It can be seen that the feasible region is unbounded.
The corner points of the feasible region are A (\(\frac{8}{3}\), 0), (2, \(\frac{1}{2}\)) and C (0, \(\frac{11}{2}\)).
The value of Z at these corner points are as follows.

As the feasible region is unbounded, therefore, 160 may or may not be the minimum value of Z.
For this, we graph the inequality, 60x + 80 y < 160 or 3x + 4y < 8 and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 3x + 4y< 8.
Therefore, the minimum cost of the mixture will be ₹ 160 at the line segment joining the points (\(\frac{8}{3}\), 0) and (2, \(\frac{1}{2}\)).

Question 2.
One kind of cake requires 200g of flour and 25g of fat and another kind of cake requireds 100g of flour and 50g of fat. Find the maximum number of cakes which can he made from 5kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.
Solution.
Let there be x cakes of first kind and y cakes of second kind.
Therefore, x ≥ 0 and y ≥ 0
The given information can be compiled in a table as follows

∴ 200x +100y ≤ 5000
⇒ 2x + y ≤ 50
25x + 50 y ≤ 1000
⇒ x + 2y ≤ 40 Total number of cakes, Z, that can be made are, Z = x + y
The mathematical formulation of the given problem is
Maximise Z = x + y ……………(i)
subject to the constraints,
2x + y ≤ 50 …………..(ii)
x + 2y < 40 ……………(iii)
x , y ≥ 0 ………………(iv)

The feasible region determined by the system of constraints is as follows

The corner points are A(25, 0), B(20, 10), C(0, 20) and 0(0, 0). The values of Z at these comer points are as follows

Thus, the maximum number of cakes that can be made are 30 (20 of one kind and 10 of the other kind).

Question 3.
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsman’s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftsman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time.
(i) What number of rackets and bats must be made if the factory is to work at full capacity?
(ii) If the profit on a racket and on a bat is ₹ 20 and ₹ 10 respectively, find the maximum profit of the factory when it works at full capacity.
Solution.
(i) Let the number of rackets and the number of bats to be made by x and y respectively.
The machine time is not available for more than 42 hours.
∴ 1.5x + 3y ≤ 42 …………(i)
The craftsman’s time is not available for more than 42 hours.
∴ 3x + y ≤ 24 ……………(ii)
The factory is to work at full capacity.
Therefore, 1.5x + 3y = 42;
3x + y = 24
On solving these equations, we get x = 4 and y = 12
Thus, 4 rackets and 12 bats must be made.

(ii) The given information can be compiled in a tables as follows

∴ 1.5x + 3y ≤ 42;
3x + y ≤ 24;
x, y ≥ 0
The profit on a racket is ₹ 20 and on a bat is ₹ 10.
∴ Z = 20x +10y
The mathematical formulation of the given problem is
Maximise Z = 20x+10y …………….(i)
subject to the constraints,
1.5x + 3y ≤ 42 …………(ii)
3x + y ≤ 24 …………..(iii)
x, y ≥ 0 ……………(iv)
The feasible region determined by the system of constraints is as follows

The corner points are A (8, 0) B (4, 12), C (0, 14) and O (0, 0).
The values of Z at these comer points are as follows

Thus, the maximum profit of the factory when it works to its full capacity is ₹ 200.

Question 4.
A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of holts. He earns a profit, of ₹ 17.50 per package on nuts and ₹ 7.00 per package on bolts. How many packages of each should be produced each day so as to maximise his profit, if he operates his machines for at the most 12 hours a day?
Solution.
Let the manufacturer produce x packages of nuts and y packages of bolts.
Therefore, x ≥ 0 and y ≥ 0
The given information can be compiled in a table as follows.

The profit on a package of nuts if ₹ 17.50 and on a package of bolts is ₹ 7.
Therefore, the constraints are x + 3y ≤ 12; 3x + y ≤ 12.
Total profit Z = 17.5x + 7y
The mathematical formulation of the given problem is
Maximise Z = 17.5x + 7y ……………(i)
subject to the constraints, x + 3y ≤12 ……….(ii)
3x + y ≤ 12………….(iii)
x, y ≥ 0 …………(iv)
The feasible region determined by the system of constraints is as follows.

The corner points are A(4, 0), B(3, 3) and C(0, 4).
The values of Z at these corner points are as follows.

The maximum value of Z is ₹ 73.50 at (3, 3).
Thus, 3 packages of nuts and 3 packages of bolts should be produced each day to get the maximum profit ₹ 73.50.

Question 5.
A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B.

Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of ₹ 7 and screws B at a profit of ₹ 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit? Determine the maximum profit.
Solution.
Let the factory manufacture x screws of type A and y screws of type B on each day.
Therefore, x ≥ 0 and y ≥ 0.
The given information can be compiled in a table as follows.

The profit on a package of screws A is ₹ 7 and on the package of screws B is ₹ 10.
Therefore the constraints are 4x + 6y ≤ 240; 6x + 3y ≤ 240
Total profit, Z = 7x + 10y
The mathematical formulation of the given problem is
Maximise Z = 7x+10y …………..(i)
subjectto the constraints, 4x + 6y ≤ 240 …………..(ii)
6x + 3y ≤ 240 …………..(iii)
x, y ≥ 0 …………….(iv)
The feasible region determined by the system of constraints is shown in previous page graph.

The comer points are A (40, 0), B (30, 20) and C (0, 40).
The values of Z at these comer points are as follows.

The maximum value of Z is 410 at (30, 20).
Thus, the factory should produce 30 packages of screws A and 20 packages of screws B to get the maximum profit ₹ 410.

Question 6.
A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade.

On any day, the sprayer is available for the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is ₹ 5 and that from a shade is ₹ 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit?
Solution.
Let the cottage industry manufacture x pedestal lamps and y wooden shades.
Therefore, x ≥ 0 and y ≥ 0.
The given information can be compiled in a table as follows.

The profit on a lamp is ₹ 5 and on the shades is ₹ 3.
Therefore, the constraints are 2x + y ≤ 12; 3x + 2y ≤ 20
Total profit, Z = 5x + 3y ,
The mathematical formulation of the given problem is
Maximise Z = 5x + 3y …………(i)
subject to the constraints,
2x + y ≤ 12 ……………(ii)
3x + 2y ≤ 20 ………(iii)
x, y ≥ 0 …………..(iv)
The feasible region determined by the system of constraints is as follows.

The maximum value of Z at these comer points are as follows.

The maximum value of Z is 32 at (4, 4).
Thus, the manufacturer should produce 4 pedestal lamps and 4 wooden shades to maximise his profits.

Question 7.
A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is ₹ 5 each for type A and ₹ 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximise the profit.
Solution.
Let the company manufacture x souvenirs of type A and y souvenirs of type B.
Therefore, x ≥ 0 and y ≥ 0.
The given information can be compiled in the table as follows.

The profit on type A souvenirs is ₹ 5 and on type B souvenirs is ₹ 6.
Therefore, the constraints are
5x + 8y ≤ 200;
10x +8y ≤ 200 i.e., 5x +4y ≤ 120

The mathematical formulation of the given problem is
Maximise Z = 5x + 6y ………….(i)
subject to the constraints,
5x + 8y ≤ 200 ……….(ii)
5x + 4y ≤ 120 ……….(iii)
x,y ≥ 0 …………..(iv)
The feasible region determined by the system of constraints is as follows.
The corner points are A(24, 0), B(8, 20) and C(0, 25).
The values of Z and these comer points are as follows.

The maximum value of Z is 200 at (8, 20).
Thus, 8 souvenirs of type A and 20 souvenirs of type B should be produced each day to get the maximum profit of ₹ 160.

Question 8.
A merchant plans to sell two types of personal computers — a desktop model and a portable model that will cost ₹ 25000 and ₹ 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than ₹ 70 lakhs and if his profit on the desktop model is₹ 4500 and on portable model is ₹ 5000.
Solution.
Let the merchant stock x desktop models and y portable models.
Therefore, x ≥ 0 and y ≥ 0
The cost of a desktop model is ₹ 25000 and of a portable model ₹ 40000.
However, the merchant can invest a maximum of ₹ 70 lakhs.
∴ 25000x + 40000y ≤ 7000000
⇒ 5x + 8y ≤ 1400
The monthly demand of computers will not exceed 250 units.
∴ x + y ≤ 250
The profit on a desktop model is ₹ 4500 and the profit on a portable model is ₹ 5000.
Total profit, Z = 4500x + 5000y
Thus, the mathematical formulation of the given problem is
Maximise Z = 4500x + 5000y ………………..(i)
subject to the constraints, 5x + 8y < 1400 ……………(ii)
x + y ≤ 250 ……………..(iii)
x, y ≥ 0 ……………(iv)
The feasible region determined by the system of constraints is as follows.

The corner points are A (250, 0), B (200, 50) and C (0, 175).
The values of Z at these corner points are as follows.

The maximum value of Z is 1150000 at (200, 50).
Thus, the merchant should stock 200 desktop models and 50 portable models to get the maximum profit of ₹ 1150000.

Question 9.
A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F₁ and F₂ are available. Food F₁ costs ₹ 4 per unit food and F₂ costs ₹ 6 per unit. One unit of food F₁ contains 3 units of vitamin A and 4 units of minerals. One unit of food F₂ contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements?
Solution.
Let the diet contain x units of food F₁ and y units of food F₂.
Therefore, x ≥ 0 and y ≥ 0.
The given information can be compiled in a table as follows.

The values of Z at these comer points are as follows.
The cost of food F₁is 4 per unit and of food F₂ is 6 per unit.
Therefore, the constraints are 3x + 6y ≥80; 4x + 3y ≥100; x, y ≥ 0
Total cost of the diet, Z = 4x + 6y
The mathematical formulation of the given problem is
Minimise Z = 4x + 6y ………… …(i)
subject to the constraints, 3x + 6y ≥80 …………… (ii)
4x + 3y ≥ 100 …………..(iii)
x, y ≥ 0 ……………(iv)
The feasible region determined by the constraints is as follows.

It can be seen that the feasible region is unbounded.
The corner points of the feasible region are A (\(\frac{8}{3}\), o) B (2, \(\frac{1}{2}\)) and C(o, \(\frac{11}{2}\))
The comer points are A (\(\frac{80}{3}\), B (24, \(\frac{4}{3}\)) and C (o, \(\frac{100}{3}\))

The value of Z at these corner points are as follows.

As the feasible region is unbounded, therefore, 104 may or may not be the minimum value of Z.
For this, we draw a graph of the inequality, 4x + 6y < 104 or 2x + 3y < 52, and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 2x +3y < 52.
Therefore, the minimum cost of the mixture will be ₹ 104.

Question 10.
There are two types of fertilisers F₁ and F₂ F₁ consists of 10% nitrogen and 6% phosphoric acid and F₂ consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F₁ costs ₹ 6per kg and F₂ costs ₹ 5 kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?
Solution.
Let the farmer buy x kg of fertiliser F₁ and y kg of fertiliser F₂.
Therefore, x ≥ 0 and y ≥ 0.
The given information can be compiled in a table as follows.

F₁ consists of 10% nitrogen and F₂ consists of 5% nitrogen.
However, the farmer requires adeast 14 kg of nitrogen.
∴ 10% of x + 5% of y ≥ 14
⇒ \(\frac{x}{10}+\frac{y}{20}\) ≥ 14
⇒ 2x + y ≥ 280
F₁ consists of 6% phosphoric acid and F₂ consists of 10% phosphoric acid. However, the farmer requires at least 14 kg of phosphoric acid.
∴ 6 % of x +10 % of y ≥ 14
⇒ \(\frac{6 x}{100}+\frac{10 y}{100}\) ≥ 14
⇒ 3x + 56y ≥ 700
Total cost of fertilisers, Z = 6x + 5y
The mathematical formulation of the given problem is
Minimise Z = 6x + 5y ………….(i)
subject to the constraints,
2x + y ≥ 280 ………..(ii)
3x + 5y > 700 ……….(iii)
x,y ≥ 0 ………….(iv)
The feasible region determined by the system of constraints is as follows.

It can be seen that the feasible region is unbounded.
The comer points are A (\(\frac{700}{3}\), o) B (100, 80) and C (0, 280).
The values of Z of these points are as follows.

As the feasible region is unbounded, therefore, 1000 may or may not be the minimum value of Z.
For this, we draw a graph of the inequality, 6x + 5y < 1000 and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 6x + 5y < 1000
Therefore, 100 kg of fertiliser F₁ and 80 kg of fertlizer F₂ should be used to minimise the cost.
The minimum cost is ₹ 1000.

Question 11.
The comer points of the feasible region determined by the following system of liner inequalities.
2x + y ≤ 10, x + 3y ≤ 15, x, y > 0 are (0, 0), (5, 0), (3, 4) and (0, 5).
Let Z = px + qy, where p, q> 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is
(A) p = q
(B) p = 2q
(C) p = 3q
(D) q = 3p
Solution.
The maximum value of Z is unique.
It is given that the maximum value of Z occurs at two points, (3, 4) and (0, 5).
∴ Value of Z at (3, 4) = Value of Z at (0, 5)
⇒ p(3) + q(4) = p(0) + q(5)
⇒ 3p + 4q = 5q
⇒ q = 3P
Hence, the correct answer is (D).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 12 Linear Programming Ex 12.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1

Direction (1 – 6): Solve the following linear programming problems graphically.

Question 1.
Maximise Z = 3x + 4y
subject to the constraints : x + y ≤ 4, x ≥ 0, y ≥ 0.
Solution.
The feasible region determined by the constraints, x + y ≤ 4, x ≥ 0, y ≥ 0, is as follows.

The corner points of the feasible region are 0 (0, 0), A (4, 0) and B (0, 4).

Therefore, the maximum value of Z is 16 at the point B(0, 4).

Question 2.
Minimise Z = – 3x + 4y
Subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0
Solution.
The feasible region determined by the system of constraints, x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0 and y ≥ 0 is as follows.

The comer points of the feasible region are O (0, 0), A (4, 0), B (2, 3) and C (0, 4).
The values of Z at these comer points are as follows

Therefore, the minimum value of Z is – 12 at the point (4, 0).

Question 3.
Maximise Z = 5x + 3y
subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.
Solution.
The feasible region determined by the system of constraints, 3 + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0 and y ≥ 0 are as follows.
The corner points of the feasible region are O (0, 0), A (2, 0), B (0, 3) and C (\(\frac{20}{19}\), \(\frac{45}{19}\))
The values of Z at these comer points are as follows.

Therefore, the maximum value of Z is \(\frac{235}{19}\) at the point (\(\frac{20}{19}\), \(\frac{45}{19}\)).

Question 4.
Minimise Z = 3x + 5y
such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.
Solution.
The feasible region determined by the system of constraints, x + 3y ≥ 3, x + y ≥ 2 and x, y ≥ 0 is as follows

It can be seen that the feasible region is unbounded.
The corner points of the feasile region are A (3, 0), (\(\frac{3}{2}\), \(\frac{1}{2}\)) and C (0, 2)
The values of Z at these comer points are as follows

As the feasible region is unbounded, therefore, 7 may or may not be the minimum value of Z.
For this, we draw the graph of the inequality, 3x + 5y < 7 and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 3x + 5y < 7.
Therefore, the minimum value of Z is 7 at (\(\frac{3}{2}\), \(\frac{1}{2}\)).

Question 5.
Maximise Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.
Solution. T
he feasible region determined by the constraints, x + 2y ≤ 10, 3x + y ≤ 15, x ≥ 0 and y ≥ 0 is as follows.
The corner points of the feasible region are A(5, 0) B(4, 3) and C(0, 5).

The values of Z at these comer points are as follows.

Therefore, the maximum value of Z is 18 at the point B(4, 3).

Question 6.
Minimise Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.
Solution.
The feasible region determined by the constraints, 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0 and y ≥ 0 is as follows.

The corner points of the feasible region are A(6, 0) and B(0, 3).
The values of Z at these comer points are as follows

Comer point Z = x + 2y.
It can be seen that the value of Z at points A and B is same.
If we take any other point such as (2, 2) on line x + 2y = 6, then Z = 6.
Thus, the minimum value of Z occurs for more than 2 points.
Therefore, the value of Z is minimum at every point on the line, x + 2y = 6.

Direction (7 – 10): Show that the minimum of Z occurs at more than two points.

Question 7.
Minimise and Maximise Z = 5x + 110y subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0.
Solution.
The feasible region determined by the constraints, x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x ≥ 0, and y ≥ 0 is as follows

The corner points of the feasible region are A (60, 0), B (120, 0), C (60, 30) and D (40, 20).
The values of Z at the corner points are as follows

The minimum value of Z is 300 at (60, 0) and the maximum value of Z is 600 at all the points on the line segment joining (120, 0) and (60, 30).

Question 8.
Minimise and maximise Z = x + 2y subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0. Solution. The feasible region determined by the constraints, x + 2y ≥ 100, 2x – y ≤ 0,2x + y ≤ 200, x > 0 and y ≥ 0 is as follows

The comer points of the feasible region are A (0, 50), B (20, 40), C (50, 100) and D (0, 200).
The values of Z at the corner points are as follows

The minimum value of Z is 400 at (0, 200) and the maximum value of Z is 100 at all the points on the line segment joining (0, 50) and (20, 40).

Question 9.
Maximise Z = – x + 2y,
subject to the constraints : x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.
Solution.
The feasible region determined by the constraints, x ≥ 3, x + y ≥ 5, x + 2y ≥ 6 and y ≥ 0 is as given below

It can be seen that the feasible region is unbounded
The values of Z at corner points A(6,0), B(4,1) and C(3,2) are as follows.

As the feasible region is unbounded, therefore, Z = 1 may or may not be the maximum value.
For this, we graph the inequality, – x + 2y > 1 and check whether the resulting half plane has points in common with the feasible region or not. The resulting feasible reason has points in common with the feasible region.
Therefore, Z = 1 is not the maximum value. Hence, Z has no maximum value.

Question 10.
Maximise Z = x + y,
subject to x – y ≤ – 1, – x + y ≤ 0, x, y ≥ 0.
Solution.
The region determined by the constraints, x – y ≤ – 1, – x + y ≤ 0, x, y ≥ 0 is as follows.

There is no feasible region and thus, Z has no maximum value.

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 11 Three Dimensional Geometry Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Miscellaneous Exercise

Question 1.
Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5,-1), (4, 3,-1).
Solution.
Let OA be the line joining the origin, 0(0, 0, 0), and the point, A(2, 1, 1).
Also, let BC be the line joining the points, B (3, 5, – 1) and C(4, 3, – 1).
The direction ratios of OA are 2, 1 and 1 and of BC are (4 – 3) = 1, (3 – 5) = – 2 and (- 1 + 1) = 0
OA is perpendicular to BC, if a<sub1a₂ + b₁b₂ + c₁c₂ = 0
∴ a₁a₂ + b₁b₂ + c₁c₂ = 2 × 1 + 1 (- 2) + 1 × 0
= 2 – 2 = 0
Thus, OA is perpendicular to BC.

Question 2.
If l₁, m₁ and n₁ and l₂, m₂, n₂ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are m₁n₂ – m₂n₁, n₁l₂ – n₂l₁, l₁m₂ – l₂m₁.
Solution.
Given, lines are respectively parallel to unit vector

Question 3.
Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.
Solution.
The angle θ between the lines with direction cosines, a, b, c and b – c, c – a, a – b is given by
cos θ = \(\frac{a(b-c)+b(c-a)+c(a-b)}{\sqrt{a^{2}+b^{2}+c^{2}}+\sqrt{(b-c)^{2}+(c-a)^{2}+(a-b)^{2}}}\)
⇒ cos θ = 0
⇒ θ = cos^-1 0
⇒ θ = 90°
Thus, the angle between the lines is 90°.

Question 4.
Find the equation of a line parallel to x-axis and passing through the origin.
Solution.
The line parallel to x-axis and passing through the origin is x-axis itself.
Let A be a point on x-axis.
Therefore, the coordinates of A are given by (a, 0,0), where a ∈ R.
Direction ratios of OA are (a – 0), 0, 0 = a, 0,0
The equation of OA is given by,
\(\frac{x-0}{a}=\frac{y-0}{0}=\frac{z-0}{0}\)

⇒ \(\frac{x}{1}=\frac{y}{0}=\frac{z}{0}\) = a
Thus, the equation of line parallel to x-axis and passing through origin is \(\frac{x}{1}=\frac{y}{0}=\frac{z}{0}\) = a

Question 5.
If the coordinates of the points A, B, C,D be (1, 2, 3), (4, 5, 7), (- 4, 3, – 6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.
Solution.
The coordinates of A, B, C and D are (1, 2, 3),(4, 5, 7), (- 4, 3, – 6)and (2, 9, 2) respectively.
The direction ratios of AB are (4 – 1) = 3, (5 – 2) = 3 and (7 – 3) = 4
The direction ratios of CD are (2 – (- 4)) = 6, (9 – 3) = 6 and (2 – (- 6)) = 8
It can be seen that,

\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{2}\)

Therefore, AB is parallel to CD.
Thus, the angle between AB and CD is either 0° or 180°.

Question 6.
If the lines \(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\) and \(\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}\) are perpendicular, find the value of k.
Solution.
The direction of ratios of the lines, \(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\) and \(\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}\) are – 3, 2k, 2 and 3k, 1, – 5 respectively.
Itis known that two lines with direction ratios, a₁, b₁, c₁ and a₂, b₂, c₂ are perpendicular, if
a₁a₂ + b₁b₂ + c₁c₂ = 0
∴ – 3 (3k) + 2k × 1 + 2 (- 5) = 0
⇒ – 9k + 2k – 10 = 0
⇒ 7k = – \(\frac{10}{7}\)
⇒ k = – \(\frac{10}{7}\)
Therefore, for k = – \(\frac{10}{7}\), the given lines are perpendicular to each other.

Question 7.
Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane \(\vec{r}\) . (î + 2ĵ – 5k̂) + 9 = 0.
Solution.
Direction ratios of the normal of the plane \(\vec{r}\) . (î + 2ĵ – 5k̂) + 9 = 0 are 1, 2, – 5.
The equation of line passing through \(\overrightarrow{r_{1}}\) and with direction ratios b₁, b₂, b₃ is
\(\vec{r}=\vec{r}_{1}+\lambda\left(b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\right)\)
Hence, the line passing through (1, 2, 3) and having the direction ratios 1, 2, – 5 is \(\vec{r}\) = î + 2ĵ + 3k̂ + λ (î + 2ĵ – 5k̂)

Question 8.
Find the equation of the plane passing through (a, b, c) and parallel to the plane \(\vec{r}\) . (î + ĵ + k̂) = 2.
Solution.
Any plane parallel to the plane, \(\overrightarrow{r_{1}}\) . [(î + ĵ + k̂) = 2, is of the form
\(\vec{r}\) . (î + ĵ + k̂) = λ …………….(i)
The plane passes through the point (a, b, c).
Therefore, the position vector \(\vec{r}\) of this point is \(\vec{r}\) = aî + bĵ + ck̂
Therefore, equation (i) becomes
(aî + bĵ + ck̂) . (î + ĵ + k̂) = λ
⇒ a + b + c = λ
Substituting λ = a + b + c in equation (i), we get
\(\vec{r}\) . (î + ĵ + k̂) = a + b + c …(ii)
This is the vector equation of the required plane.
Substituting \(\vec{r}\) = xî + yĵ + zk̂ in equation (ii), we get
(xî + yĵ + zk̂) . (î + ĵ + k̂) = a + b + c
⇒ x + y + z = a + b + c.

Question 9.
Find the Shortest distance between lines \(\vec{r}\) = 6î + 2ĵ + 2k̂ + λ (î – 2ĵ + 2k̂) and r = – 4î – k̂ + μ (3î – 2ĵ – 2k̂).
Solution.

Question 10.
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane.
Solution.
It is known that the equation of the line passing through the points, (x₁, y₁, z₁) and (x₂, y₂, z₂) is
The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by \(\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}\)

⇒ \(\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}\)= k (say)

⇒ x = 5 – 2k, y = 3k + 1, z = 6 – 5k
Any point on the line is of the form (5 – 2k, 3k + 1, 6 – 5k).
The equation of YZ-plane is x = 0
Since the line passes through YZ-plane.
5 – 2k = 0
⇒ k = \(\frac{5}{2}\)
⇒ 3k + 1 = 3 × \(\frac{5}{2}\) + 1 = \(\frac{17}{2}\);

6 – 5k = 6 – 5 × \(\frac{5}{2}\) = – \(\frac{13}{2}\)

Therefore, the required point is (0, \(\frac{17}{2}\), \(-\frac{13}{2}\)).

Question 11.
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.
Solution.
It is known that the equation of the ime passing through the points, (x₁, y₁, z₁) and (x₂, y₂, z₂) is \(\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}\)

The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by \(\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}\)

⇒ \(\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}\) = k(say)
⇒ x = 5 – 2k, y = 3k + 1, z = 6 – 5k
Any point on the line is of the form (5 – 2k, 3k + 1, 6 – 5k).
Since the line passes through ZX-plane.
3k + 1 = 0
⇒ k = – \(\frac{1}{3}\)
⇒ 5 – 2k = 5 – 2(- \(\frac{1}{3}\)) = \(\frac{17}{3}\);

6 – 5k = 6 – 5(- \(\frac{1}{3}\)) = \(\frac{23}{3}\)

Therefore, the required point is (\(\frac{17}{3}\), 0, \(\frac{23}{3}\)).

Question 12.
Find the coordinates of the point where the line through (3, – 4, – 5) and (2, – 3, 1) crosses the plane 2x + y + z = 7.
Solution.
It is known that the equation of the line through the points, (x₁, y₁, z₁) and (x₂, y₂, z₂)
Since the line passing through the points, (3, – 4,- 5) and (2, – 3, 1), its equation is given by \(\frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}\)

⇒ \(\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}\) = k (say)
⇒ x = 3 – k, y = k – 4, z = 6k – 5
Therefore, any point on the line is of the form (3 – k, k – 4, 6 k – 5).
This point lies on the plane, 2x + y + z = 7
∴ 2 (3 – k) + (k – 4) + (6k – 5) = 7
⇒ 5k – 3 = 7
⇒ k = 2
Hence, the coordinates of the required point are (3 – 2, 2 – 4, 6 × 2 – 5), i.e., (1, – 2, 7).

Question 13.
Find the equation of the plane passing through the point (- 1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
Solution.
The equation of the plane passing through the point (- 1, 3, 2) is
a (x + 1) + b (y – 3) + c (z- 2) = 0 ……………(i)
where, a, b, c are the direction ratios of normal to the plane.
It is known that two planes, a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0, are perpendicular, if
a₁a₂ + b₁b₂ + c₁c₂ = 0
Plane (i) is perpendicular to the plane, x + 2y + 3z = 5
∴ a . 1 + b . 2 + c . 3 = 0
⇒ a + 2b + 3c = 0 …………..(ii)
Also, plane (i) is perpendicular to the plane, 3x + 3y +z = 0
∴ a . 3 + b . 3 + c . 1 = 0
⇒ 3a + 3b + c = 0 …………(iii)
From equations (ii) and (iii), we get
\(\frac{a}{2 \times 1-3 \times 3}=\frac{b}{3 \times 3-1 \times 1}=\frac{c}{1 \times 3-2 \times 3}\)
⇒ \(\frac{a}{-7}=\frac{b}{8}=\frac{c}{-3}\) = k (say)
⇒ a = – 7k, b = 8k, c = – 3k -7
Substituting the value of a, b and c in equation (i), we get
– 7k (x + 1) + 8k (y – 3) – 3k(z – 2) = 0
⇒ (- 7x – 7) + (8y – 24) – 3z + 6 = 0
⇒ – 7x + 8y – 3z – 25 = 0
⇒ 7x – 8y + 3z + 25 = 0
This is the required equation of the plane.

Question 14.
If the points (1, 1, p) and (- 3, 0, 1) be equidistant from the plane \(\vec{r}\) = (3î + 4ĵ – 12k̂) + 13 = 0, then find the value of p.
Solution.
The position vector through the point (1, 1, p) is \(\overrightarrow{a_{1}}\) = î + ĵ + pk̂
Similarly, the position vector through the point (- 3, 0, 1) is \(\overrightarrow{a_{2}}\) = – 4î + k̂
The equation of the given plane is \(\vec{r}\) . (3î + 4ĵ – 12k̂) + 13 = 0

Question 15.
Find the equation of the plane passing through the line of intersection of the planes \(\vec{r}\) . (î + ĵ + k̂) = 1 and \(\vec{r}\) . (2î + 3ĵ – k̂) + 4 = 0 and parallel to x-axis.
Solution.
The given planes are \(\vec{r}\) . (î + ĵ + k̂) = 1
⇒ \(\vec{r}\) . (î + ĵ + k̂) – 1 = 0
\(\vec{r}\) . (2î + 3ĵ – k̂) + 4 = 0
The equation of any plane passing through the line of intersection of these planes is
[\(\vec{r}\) . (î + ĵ + k̂) – 1] + λ [\(\vec{r}\) . (2î + 3ĵ – k̂) + 4] = 0
\(\vec{r}\) . [(2λ + 1)î + (3λ + 1)ĵ + (1 – λ) k̂] + (4λ +1) = 0 ………….(i)
Its direction ratios are (2λ + 1), (3λ + 1) and (1 – λ)
The required plane is parallel to x-axis.
Therefore, its normal is perpendicular to x-axis.
The direction ratios of x-axis are 1, 0 and 0.
∴ 1 (2λ + 1) + 0 (3λ + 1) + 0 (1 – λ) = 0
⇒ 2λ + 1 = 0
⇒ λ = – \(\frac{1}{2}\)
Substituting λ = – \(\frac{1}{2}\) in equation (i), we get
⇒ \(\vec{r} \cdot\left[-\frac{1}{2} \hat{j}+\frac{3}{2} \hat{k}\right]\) + (- 3) = 0
⇒ \(\vec{r}\) (ĵ – 3k̂) + 6 = 0
Therefore, its Cartesian equation is y – 3z + 6 = 0.
This is the equation of the required plane.

Question 16.
If O be the origin and the coordinates of P be (1, 2,- 3), then find the equation of the plane passing through P and perpendicular to OP.
Solution.
The coordinates of the points, O and P are (0, 0, 0) and (1, 2, – 3) respectively.
Therefore, the direction ratios of OP are (1 – 0) = 1, (2 – 0) = 2 and (- 3 – 0) = – 3
Here, the direction ratios of normal are 1, 2 and – 3 and the point P is (1, 2, – 3).
Thus, the equation of the required plane is 1 (x – 1) + 2 (y – 2) – 3 (z + 3) = 0
⇒ x – 1 + 2y – 4 – 3z – 9 = 0
⇒ x + 2y – 3z – 14 = 0.

Question 17.
Find the equation of the plane which contains the line of intersec¬tion of the planes \(\vec{r}\) . (î + 2ĵ + 3k̂) – 4 = 0, \(\vec{r}\) . (2î + ĵ – k̂) + 5 = 0 and which is perpendicular to the plane \(\vec{r}\) . (5î + 3ĵ – 6k̂) + 8 = 0.
Solution.
The equations of the given planes are
\(\vec{r}\) . (î + 2ĵ + 3k̂) – 4 = 0 …………..(i)
\(\vec{r}\) . (2î + ĵ – k̂) + 5 = 0 ………….(ii)
The equation of the plane passing through the line intersection of the plane given in equation (i) and equation (ii) is
[\(\vec{r}\) . (î + 2ĵ + 3k̂) – 4 ] + λ [\(\vec{r}\) . (2î + ĵ – k̂) + 5] = 0
\(\vec{r}\) . [(2λ + 1)î + (λ + 2)ĵ + (3 – λ)k̂] + (5λ – 4) = 0 ……………(iii)
The plane in equation (iii) is perpendicular to the plane,
\(\vec{r}\) . (5î + 3ĵ – 6k̂) + 8 = 0
∴ 5(2λ + 1) + 3 (λ + 2) – 6 (3 – λ) = 0
⇒ 19λ – 7 = 0
⇒ λ = \(\frac{7}{19}\)
Sustituting λ = \(\frac{7}{19}\) in equation (iii), we get
⇒ \(\vec{r} \cdot\left[\frac{13}{19} \hat{i}+\frac{45}{19} \hat{j}+\frac{50}{19} \hat{k}\right] \frac{-41}{19}\) = 0
⇒ \(\vec{r}\) . (33î + 45ĵ + 50k̂) – 41 = 0
This is the vector equation of the required plane.
The Cartesian equation of this plane can be geted by substituting \(\vec{r}\) = xî + yĵ + zk̂ in equation (iii).
(xî + yĵ + zk̂) . (33î + 45ĵ + 50k̂) – 41 = 0
⇒ 33x + 45y + 50z – 41 = 0.

Question 18.
Find the distance of the point (- 1, – 5, 10) from the point of intersection of the line \(\vec{r}\) = 2î – ĵ + 2k̂ + λ (3î + 4ĵ + 2k̂) and the plane \(\vec{r}\) . (î – ĵ + k̂) = 5.
Solution.
The equation of the given line is
\(\vec{r}\) = 2î – ĵ + 2k̂ + λ (3î + 4ĵ + 2k̂) ………….(i)
The equation of the given plane is
\(\vec{r}\) . (î – ĵ + k̂) = 5 ………….(ii)
Substituting the value of r from equation (i) in equation (ii), we get
[2î – ĵ + 2k̂ + λ (3î + 4ĵ + 2k̂)] . (î – ĵ + k̂) = 5
⇒ (3λ + 2)î + (4λ – 1)ĵ + (2λ + 2)k̂] . (î – ĵ + k̂) = 5
⇒ (3λ + 2) – (4λ – 1) + (2λ + 2) = 5
⇒ λ = 0
Substituting this value in equation (i), we get the equation of the line as
\(\vec{r}\) = 2î – ĵ + 2k̂
And r is position vector of the point (2, – 1, 2)
The distance d between the points (2, – 1, 2) and (- 1, – 5, – 10) is
d = \(\sqrt{(-1-2)^{2}+(-5+1)^{2}+(-10-2)^{2}}\)
= \(\sqrt{9+16+144}=\sqrt{169}\)
= 13 units.

Question 19.
Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes \(\vec{r}\) . (î – ĵ + 2k̂) = 5 and \(\vec{r}\) . (3î + ĵ + k̂) = 6.
Solution.
Let the required line be parallel to vector \(\vec{b}\) given by
\(\vec{b}\) = b₁ î + b₂ ĵ + b₃ k̂
The position vector of the point (1, 2, 3) is \(\vec{a}\) = î + 2 ĵ + 3 k̂
The equation of line passing through (1, 2, 3) and parallel to \(\vec{b}\) is given by
\(\vec{r}=\vec{a}+\lambda \vec{b}\)
⇒ \(\vec{r}\) (î + 2 ĵ + 3 k̂) + λ (b₁ î + b₂ ĵ + b₃ k̂) ……………(i)
The equations of the given planes are
\(\vec{r}\) . (î – ĵ + 2k̂) = 5 …………….(ii)
\(\vec{r}\) . (3î + ĵ + k̂) = 6 …………..(iii)
The line in equation (i) and plane in equation (ii) are parallel. Therefore, the normal to the plane of equation (ii) and the given line are perpendicular.
⇒ (î – ĵ + 2k̂) . λ (b₁ î + b₂ ĵ + b₃ k̂) = 0

Question 20.
Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines \(\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}\) and \(\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}\).
Solution.
Let the required line be parallel to the vector \(\vec{b}\) given by,

Question 21.
Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then \(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{p^{2}}\).
Solution.
The equation of a plane having Intercepts a, b, c with x, y and z axes respetively is given by \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1 …………….(i)

The distance (p) of the plane from the origin is given by

Direction (22 – 23): Choose the correct answer.
Question 22.
Distance between the two planes 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is
(A) 2 units
(B) 4 units
(C) 8 units
(D) units
Solution.
The equations of the planes are
2x + 3y + 4z = 4 ……………(i)
4x + 6y + 8z = 112
⇒ 2x + 3y + 4z = 6 ……………(ii)
It can be seen that the given planes are parallel.
k is known that the distance between two parallel planes, ax + by + cz = d₁ and ax + by + cz = d₂ given by
D = \(\frac{d_{2}-d_{1}}{\sqrt{a^{2}+b^{2}+c^{2}}}\)

D = \(\left|\frac{6-4}{\sqrt{(2)^{2}+(3)^{2}+(4)^{2}}}\right|\)

D = \(\frac{2}{\sqrt{29}}\)
Thus, the distance between the lines is \(\frac{2}{\sqrt{29}}\) units.
Hence, the correct answer is (D).

Question 23.
The planes 2x – y + 4z = 5 and 5x – 25y + 10z = 6are
(A) perpendicular
(B) parallel
(C) intersect y-axis
(D) passes through (o, o, \(\frac{5}{4}\))
Solution.
The equations of the planes are
2x – y + 4z = 5 ………….(i)
5x – 25y + 10z = 6 …………(ii)
It can be seen that
\(\frac{a_{1}}{a_{2}}=\frac{2}{5}\);

\(\frac{b_{1}}{b_{2}}=\frac{-1}{-2.5}=\frac{2}{5}\);

\(\frac{c_{1}}{c_{2}}=\frac{4}{10}=\frac{2}{5}\)

⇒ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\).

Therefore, the given planes are parallel.
Hence, the correct answer is (B).