PSEB 12th Class Maths Solutions Chapter 13 Probability Miscellaneous Exercise

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).