Class 11

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 16 Probability Ex 16.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 16 Probability Ex 16.3

Question 1.
Which of the following can not be valid assignment of probabilities for outcomes of sample space
S = {ω₁, ω₂, ω₃, ω₄, ω₅, ω₆, ω₇}

Answer.
(a)

Here, each of the numbers pω₁, is positive and less than 1.
Sum of probabilities = p (ω₁ ) + p (ω₂) + p (ω₃) + p(ω₄) + p (ω₅) + p (ω₆) + p (ω₇)
= 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1
Thus, the assignment is valid.

(b)

Here, each of the numbers pea f is positive and less than 1.
Sum of probabilities = p (ω₁ ) + p (ω₂) + p (ω₃) + p(ω₄) + p (ω₅) + p (ω₆) + p (ω₇)
= \(\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}=7 \times \frac{1}{7}\) = 1
Thus, the assignment is valid.

(c)

Sum of probabilities = p (ω₁) + p (ω₂) + p (ω₃) + p(ω₄) + p (ω₅) + p (ω₆) + p (ω₇)
= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8 ≠ 1
Thus, the assignment is not valid.

(d)

Here, p(ω₁) and p(ω₂) are negative.
Hence, the assignment is not valid.

(e)

Hence, the assignment is not valid.
p(ω₇) = \(\frac{15}{14}\) > 1.

Question 2.
A coin is tossed twice, what is the probability that atleast one tail occurs?
Answer.
When a coin is tossed twice, the sample space is given by S = (HH, HT, TH, TT}
Let A be the event of the occurrence of atleast one tail.
Accordingly, A – {HT, TH, TT}
∴ P(A) = \(\frac{\text Number of outcomes favourable}{\text { Total number of possible outcomes }}=\frac{n(A)}{n(S)}\)

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

Question 3.
A die is thrown, find the probability of following events :
(i) A prime number will appear,
(ii) A number greater than or equal to 3 will appear,
(iii) A number less than or equal to one will appear,
(iv) A number more than 6 will appear,
(v) A number less than 6 will appear.
Answer.
The sample space of the given experiment is given by S = {1, 2, 3, 4, 5, 6}
(i) Let A be the event of the occurrence of a prime number. Accordingly, A = {2, 3, 5}
∴ P(A) = \(\frac{\text { Number of outcomes favourable to } A}{\text { Total number of possible outcomes }}=\frac{n(A)}{n(S)}\)

= \(\frac{3}{6}=\frac{1}{2}\)

Let B be the event of the occurrence of a number greater than or equal to 3. Accordingly, B = {3, 4, 5, 6}
∴ P(B) = \(\frac{\text { Number of outcomes favourable to } B}{\text { Total number of possible outcomes }}=\frac{n(A)}{n(S)}\)

= \(\frac{4}{6}=\frac{2}{3}\)

Let C be the event of the occurrence of a number les than or equal to one. Acordingly, C ={1}
∴ P(C) = \(\frac{\text { Number of outcomes favourable to } C}{\text { Total number of possible outcomes }}=\frac{n(C)}{n(S)}\)

= \(\frac{1}{6}\)

Let D be the event of the occurrence of a number greater than 6. Accordingly, D = {} = Φ
∴ P[D) = \(\frac{\text { Number of outcomes favourable to } D}{\text { Total number of possible outcomes }}=\frac{n(D)}{n(S)}\)

= \(\frac{0}{6}\) = 0

Let E be the event of the occurrence of a number less than 6. Accordingly, E = {1, 2, 3, 4, 5}
∴ P(E) = \(\frac{\text { Number of outcomes favourable to } E}{\text { Total number of possible outcomes }}=\frac{n(E)}{n(S)}\)

= \(\frac{5}{6}\).

Question 4.
A card is selected from a pack of 52 cards.
(a) How many points are there in the sample space?
(b) Calculate the probability that the card is an ace of spades.
(c) Calculate the probability that the card is
(i) an ace
(ii) black card.
Answer.
(a) when a card is selected from a pack of 52 cards, the number of possible outcomes is 52 i.e., the sample space contains 52 elements. Therefore there are 52 points in the sample space.

(b) Let A be the event in which the card drawn is an ace of spades. Accordingly, (n)A = 1
∴ P(A) = \(\frac{\text { Number of outcomes favourable to } A}{\text { Total number of possible outcomes }}=\frac{n(A)}{n(S)}\)

= \(\frac{1}{52}\)

(c) (i) Let E be the event in which the card drawn is an ace.
Since there are 4 aces in a pack of 52 cards,
Number of outcomes favourable to E n(E) = 4
∴ P(E) = \(\frac{\text { Number of outcomes favourable to } E}{\text { Total number of possible outcomes }}=\frac{n(E)}{n(S)}\)

= \(\frac{4}{52}=\frac{1}{13}\)

(ii) Let F be the event in which the card drawn is black.
Since there are 26 black cards in a pack of 52 cards, n(F) = 26

∴ P(F) = \(\frac{\text { Number of outcomes favourable to } F}{\text { Total number of possible outcomes }}=\frac{n(F)}{n(S)}\)

= \(\frac{26}{52}=\frac{1}{2}\).

Question 5.
A fair coin wIth 1 marked on one face and 6 on the other and a fair die are both tos,ed. Find the probability that the sum of numbers that turm up is
(i) 3
(ii) 12.
Answer.
Since the fair coin has 1 marked on one face and 6 on the other, and the die has six faces that are numbered 1,2, 3,4, 5, and 6, the sample space is given by
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Accordingly, n(S) = 12

(i) Let A be the event in which the sum of number that turn up is 3. Accordingly, A = {(1, 2)}
∴ P(A) = \(\frac{\text { Number of outcomes favourable to } A}{\text { Total number of possible outcomes }}=\frac{n(A)}{n(S)}\)

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

(ii) Let B be the event in which the sum of numbers that turn up is 12. Accordingly, B = {(6, 6)}
∴ P(B) = \(\frac{\text { Number of outcomes favourable to } B}{\text { Total number of possible outcomes }}=\frac{n(B)}{n(S)}\)

= \(\frac{1}{12}\).

Question 6.
There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?
Answer.
There are four men and six women on the city council.
As one council member is to be selected for a committee at random, the sample space contains 10 (4 + 6) elements.
Let A be the event in which the selected council member is a woman.
Accordingly, n(A) = 6
∴ P(A) = \(\frac{\text { Number of outcomes favourable to } A}{\text { Total number of possible outcomes }}=\frac{n(A)}{n(S)}\)

= \(\frac{6}{10}=\frac{3}{5}\)

Question 7.
A fair coin is tossed four times, and a person win Re 1 for each head and lose Rs 1.50 for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of have each of these amounts.
Answer.
There are five ways in which heads and tails appear.

(i) No head and 4 tail, appear
Money lost = Rs. 4 × 1.50 = Rs. 6.00
There is only 1 way when TTTT occurs.
No. of exhaustive cases = 2⁴ = 16
∴ Probaility of getting no head or 4 tails = \(\frac{1}{16}\).

(ii) When 1 head and 3 tail, appear
Money lost = Rs. (- 1 × 1 + 3 × 1.50) = Rs. 3.50
There are 4 ways when 1 head and 3 tails occur i.e., HTTT, THTT, TTHT, TTTH.
∴ Probaility of getting 1 head and 3 tails = \(\frac{4}{16}=\frac{1}{4}\).

(iii) When 2 head and 2 tail, appear
Money lost = Rs. (2 × 1.5 – 1 × 2) = Rs. (3 – 2) = Re. 1
2 heads and 2 tail may occur as HHTT, HTHT, HTTH, THHT, THTH, TTHH
Thus 2 head and 2 tails may appear in 6 ways
∴ Probability of getting 2 heads and 2 tails = \(\frac{6}{16}=\frac{3}{8}\)

(iv) When 3 head and 1 tail, appear
Money gained Rs. (3 × 1 – 1 × 1.5) = Rs. 1.50
3 heads and 1 tail may occurs as HHHT, HHTH, HTHH, THHH
∴ 3 heads and 1 tail appear in 4 ways
∴ Probability of getting 3 heads and 1 tail = \(\frac{4}{16}=\frac{1}{4}\).

(v) When all the heads appear
Money gained Rs. 4 × 1 = Rs. 4
4 heads occur as HHHH i.e., in one way.
Probability of getting 3 heads = \(\frac{1}{16}\).

Question 8.
Three coins are tossed once. Find the probability of getting
(i) 3 heads
(ii) 2 heads
(iii) at least 2 heads
(iv) at most 2 heads
(v) no head
(vi) 3 tails
(vii) exactly two tails
(viii) no tail
(ix) at most two tails.
Answer.
When three coins are tossed once, the sample space is given by
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
∴ Accordingly, n(S) = 8
It is known that the probability of an event A is given by,
P(A) = \(\frac{\text { Number of outcomes favourable to } A}{\text { Total number of possible outcomes }}=\frac{n(A)}{n(S)}\)

(i) Let B be the event of the occurrence of 3 heads. Accordingly, B={HHH}
∴ P(B) = \(\frac{n(B)}{n(S)}\)

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

(ii) Let C be the event of the occurrence of 2 heads.
Accordingly, C = {HHT, HTH, THH}
∴ P(C) = \(\frac{n(C)}{n(S)}\)

= \(\frac{3}{8}\)

(iii) Let D be the event of the occurrence of at least 2 heads.
Accordingly, D = {HHH, HHT, HTH, THH}
∴ P(D) = \(\frac{n(D)}{n(S)}\)

= \(\frac{4}{8}=\frac{1}{2}\)

(iv) Let E be the event of the occurrence of almost 2 heads.
Accordingly, E = {HHT, HTH, THH, HTT, THT, TTH, TTT}
∴ P(E) = \(\frac{n(E)}{n(S)}\)

= \(\frac{7}{8}\)

(v) Let F be the event of the occurrence of no head.
Accordingly, F = {TTT}
∴ P(F) = \(\frac{n(F)}{n(S)}\)

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

(vi) Let G be the event of the occurrence of 3 tails.
Accordingly, G = {TTT}
∴ P(G) = \(\frac{n(G)}{n(S)}\)

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

(vii) Let H be the event of the occurrence of exacdy 2 tails.
Accordingly, H = {HTT, THT, TTH}
∴ P(H) = \(\frac{n(H)}{n(S)}\)

= \(\frac{3}{8}\)

(viii) Let I be the event of the occurrence of no tail.
Accordingly, I = {HHH}
∴ P(I) = \(\frac{n(I)}{n(S)}\)

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

(ix) Let J be the event of the occurrence of atmost 2 tails.
Accordingly, J = {HHH, HHT, HTH, THH, HTT, THT, TTH}
∴ P(G) = \(\frac{n(J)}{n(S)}\)

= \(\frac{7}{8}\)

Question 9.
If \(\frac{2}{11}\) is the probability of an event, what is the probability of the event ‘not A’.
Answer.
It is given that P(A) = \(\frac{2}{11}\)
Accordingly, P (not A) = 1 – P(A)
= 1 – \(\frac{2}{11}\) = \(\frac{9}{11}\).

Question 10.
A letter is chosen at random from the word ‘ASSASSINATION’.
Find the probability that letter is (i) a vowel (ii) a consonant.
Answer. There are 13 letters in the word ASSASSINATION.
∴ Hence, n(S) = 13

(i) There are 6 vowels in the given word.
∴ Probability (vowel) = \(\frac{6}{13}\).

(ii) There are 7 consonants in the given word.
∴ Probability (consonant) = \(\frac{7}{13}\).

Question 11.
In a lottery, person choses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers airedy fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game?
[Hint: order of the numbers is not important.]
Answer.
Total number of ways in which one can choose six different numbers from 1 to 20

= \({ }^{20} C_{6}=\frac{20 !}{6 !(20 !-6 !)}\)

= \(\frac{20 !}{6 ! 14 !}=\frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{1.2 .3 .4 .5 .6}\)

= 38760

Hence, there are 38760 combinations of 6 numbers.
Out of these combinations, one combination is already fixed by the lottery committee.
Required probability of winning the prize in the game = \(\frac{1}{38760}\).

Question 12.
Check whether the following probabilities P( A) and P(B) are consistently defined
(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6
(ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8 ;
Answer.
(i) Here, P(A ∩ B) > P(A)
So, given data is not consistent.

(ii) Here, P(A) = 05, P(B) = 0.4, P(A ∪ B) = 0.8
⇒ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ 0.8 = 0.5 + 0.4 – P(A ∩ B)
⇒ P(A ∩ B) = 0.9 – 0.8 = 0.1 < P(A) and P(B)
⇒ P(A) and P(B) are consistent.

Question 13.
Fill in the blanks in following table :

Answer.
(i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{1}{3}+\frac{1}{5}-\frac{1}{15}\)

= \(\frac{5+3-1}{15}=\frac{7}{15}\)

(ii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0.6 = 0.35 + P(B) – 0.25
P(B) = 0.6 – 0.35 + 0.25 = 0.5

(iii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0.7 = 0.5 + 0.35 – P(A ∩ B)
P(A ∩ B) = 0.5 + 0.35 – 0.7 = 0.15

Question 14.
Given P(A) = \(\frac{3}{5}\) and P(B) = \(\frac{1}{5}\). Find P(A or B), if A and B are mutually exclusive events.
Answer.
Here, P(A) = \(\frac{3}{5}\), P(B) = \(\frac{1}{5}\)
For mutually exclusive events A and B,
P(A or B) = P(A) + P(B)
∴ P(A or B) = \(\frac{3}{5}\) + \(\frac{1}{5}\) = \(\frac{4}{5}\).

Question 15.
If E and F are events such that P(E) = \(\frac{1}{4}\), P(F) = \(\frac{1}{2}\) and P(E and F) = \(\frac{1}{8}\) find:
(i) P(E or F),
(ii) P (not E and not F).
Answer.
(i) P(E or F) = P(E ∪ F) = P(E) + P(F) – P(E ∩ F)
= \(\frac{1}{4}+\frac{1}{2}-\frac{1}{8}\)

= \(\frac{2+4-1}{8}=\frac{5}{8}\)

(ii) not E and not F = E ∩ F’ = (E ∪ F)’ (De Morgan’s Law)
P(not E and not F) = P(E ∪ F)’
= 1 – P(E ∪ F)
= 1 – \(\frac{5}{8}\) = \(\frac{3}{8}\).

Question 16.
Events E and F are such that P(not E or not F) = 0.25. State whether E and F are mutually exclusive.
Answer.
It is given that P (not E or not F) = 0.25 i.e., P(E ∪ F) = 0.25
⇒ P(E ∪ F)’ = 0.25
[∵ E ∪ F = (E ∩ F)’, De Morgan’s law]
Now, P(E ∩ F) = 1 – P(E ∩ F)’
⇒ P(E ∩ F) = 1 – 0.25
⇒ P(E ∩ F) = 0.75 ≠ 0
⇒ E ∩ F ≠ 0
Thus, E and F are not mutually exclusive.

Question 17.
A and Bare events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine
(i) P(not A),
(ii) P (not B)
(iii) P(A or B).
Answer.
It is given that P(A) = 0.42, P(B) = 0.48, P(A and B) = 0.16
(i) P(not A) = 1 – P(A) = 1 – 0.42 = 0.58
(ii) P(not B) = 1 – P(B) = 1 – 0.48 = 0.52
(iii) We know that P(A or B) = P(A) + P(B) – P(A and B)
P(A or B) = 0.42 + 0.48 – 0.16 = 0.74

Question 18.
In Class XI of a school 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.
Answer.
Let A be the event in which the selected student studies Mathematics and B be the event in which the selected student studies Biology.
Accordingly, P(A) = 40% = \(\frac{40}{100}=\frac{2}{5}\);

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

P(A and B) = 10% = \(\frac{10}{100}=\frac{1}{10}\)

We know that P(A or B) = P(A) + P(B) – P(A and B)

∴ P(A or B) = \(\frac{2}{5}+\frac{3}{10}-\frac{1}{10}=\frac{6}{10}\) = 0.6

Thus, the probability that the selected student will be studying mathematics or Biology is 0.6.

Question 19.
In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both?
Answer.
Let A and B be the events of passing the first and second examinations respectively. .
P(A) = 0.8, P(B) = 0.7
Probability of passing at least one examination
= 1 – P(A’ ∩ B’) = 0.95 ……………….(i)
Now A’ ∩ B’ = (A ∪ B)’ (De Morgan’s law)
P(A’ ∩ B) = P(A ∪ B)’= 1 – (A ∪ B)
Putting the value in eQuestion (ii)
1 – [1 – P(A ∪ B)] = 0.95 or P(A ∪ B) = 0.95
Further P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= 0.8 + 0.7 – 0.95 = 1.5 – 0.95 = 0.55
Thus, probability that the student will pass in both the examinations = 0.55.

Question 20.
The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination?
Answer.
Let A and B be the events of passing English and Hindi examinations respectively.
Accordingly, P(A and B) = 0.5, P(not A and not B) = 0.1 i.e.,
P(A’ ∩ B’) = 0.1
P(A)= 0.75
Now, (A ∪ B)’= (A’ ∩ BO [De Morgan’s law]
∴ P(A ∪ B)’ = P(A’ ∩ B’) = 0.1
P(A ∪ B) = 1 – P(A ∪ B)’= 1 – 0.1 = 0.9
We know that P(A or B) = P(A) + P(B) – P(A and B)
∴ 0.9 = 0.75 + P(B) – 0.5
⇒ P(B) = 0.9 – 0.75 + 0.5
⇒ P(B) = 0.65
Thus, the probability of passing the Hindi examination is 0.65.

Question 21.
In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that
(i) The student opted for NCC or NSS.
(ii) The student has opted neither NCC nor NSS.
(iii) The student has opted NSS but not NCC.
Answer.
Let A and B denote the students in NCC and NSS, respectively.
Given, n(A) = 30,
n(B) = 32
n(A ∩ B) = 24,
n(S) = 60 [∵ 24 students opted for both NCC and NSS i.e., they are common in both]

P(A) = \(\frac{30}{60}\) P(B) = \(\frac{32}{60}\) and P(A ∩ B) = \(\frac{24}{60}\)

Then, P(A ∩ B) = \(\frac{2}{12}\)

Now, required probability,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{6}{12}+\frac{4}{12}-\frac{2}{12}\)

= \(\frac{8}{12}=\frac{2}{3}\)

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 16 Probability Ex 16.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 16 Probability Ex 16.2

Question 1.
A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive?
Answer.

If we roll a die, then all the possible outcomes will be
S= {1, 2, 3, 4, 5, 6}
E = Die shows 4 = {4}
F = Die shows an even number = (2, 4, 6}
⇒ E ∩ F = {4} ∩ {2, 4, 6} = {4}
⇒ E ∩ F ≠ Φ
Hence, E and F are not mutually exclusive.

Question 2.
A die is thrown. Describe the following events:
(i) A: a number less than 7
(ii) B: a number greater than 7
(iii) C: a multiple of 3
(iv) D: a number less than 4
(v) E: an even number greater than 4
(vi) F: a number not less than 3.
Also find A ∪ B, A ∩ B, B ∪ C, E ∩ F, D ∩ E, A – C, D – E, E ∩ F, F.
Answer.
When a die is throw, then sample space S = {1, 2, 3, 4, 5, 6}
(i) A : a number less than 7 = {1, 2, 3, 4, 5, 6}
(ii) B: a number greater than 7 = {} = {Φ}
(iii) C : a multiple of 3 = {3, 6}
(iv) D: a number less than 4 = {1, 2, 3}
(v) E : an even number greater than 4 = {6}
(vi) F : a number not less than 3 = {3, 4, 5, 6}
Now, A ∪ B = The elements which are in both A and B
{1, 2, 3, 4, 5, 6} ∪ – Φ = {1, 2, 3, 4, 5, 6}

A ∩ B – The elements which are common in both A and B
= {1, 2, 3, 4, 5, 6} ∩ Φ = Φ

B ∪ C = The elements which are in both B and C
= { } ∪ {3, 6} = {3, 6}

E ∩ F = The elements which are common in both E and F
= {6} ∩ {3, 4, 5, 6} = {6}

D ∩ E = The elements which are common in both D and E
= {1, 2, 3,} ∩ {6} = Φ

A – C = The elements which are in A but not in C
= {1, 2, 3, 4, 5, 6} – {3, 6} = {1, 2, 4, 5}

D – C = The elements which in D but not in E
= {1, 2, 3} – {6} = {1, 2, 3}

F’ = (S – F) = {1, 2, 3, 4, 5, 6} – {3, 4, 5, 6} = {1, 2}
and E ∩ F’ = E ∩ (S – F)
= {6} ∩ {1, 2} = Φ.

Question 3.
An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events:
A : the sum is greater than 8,
B : 2 occurs on either die
C : The sum is at least 7 and a multiple of 3.
Which pairs of these events are mutually exclusive?
Answer.
When a pair of dice is rolled, the sample space is given by S = {(x, y) : x, y = 1, 2, 3, 4, 5, 6}

Accordingly,
A = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}
B = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (1, 2), (3, 2), (4, 2), (5, 2), (6, 2)}
C = {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}
It is observed that
A ∩ B = Φ
B ∩ C = Φ
C ∩ A = {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}
Hence, events A and B and events B and C are mutually exclusive.

Question 4.
Three coins are tossed once. Let A denote the event “three heads show”, B denote the event “two heads and one tail show”, C denote the event “three tails show” and D denote the event “a head shows on the first coin”.
Which events are
(i) mutually exclusive?
(ii) simple?
(iii) compound?
Answer.
When three coins are tossed, the sample space is given by S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TIT}
Accordingly,
A = {HHH},
B = {HHT, HTH, HTT},
C = (TIT),
D = {HHH, HHT, HTH, HTT}
We now observe that
A ∩ B = Φ A ∩ C = Φ; B ∩ C = Φ; C ∩ D = Φ
A ∩ B ∩ C = Φ

(i) Event A and B; event A and C; event B and C; event C and D and event A, B,C are all mutually exclusive.
(ii) If an event has only one sample point of a sample space, it is called a simple event, Thus, A and C are simple events.
(iii) If an event has more than one sample point of a sample space, it is called a compound event. Thus, B and D are compound events.

Question 5.
Three coins are tossed. Describe
(i) Two events which are mutually exclusive.
(ii) Three events which are mutually exclusive and exhaustive.
(iii) Two events, which are not mutually exclusive. 1
(iv) Two events which are mutually exclusive but not exhaustive,
(v) Three events which are mutually exclusive but not exhaustive.
Answer.
When three coins are tossed, then the sample space S is S={HHH, HHT, HTH, HTT, THH, THT, TTH, TIT}
(i) Two events A and B which are mutually exclusive are
A : “getting at least two heads” and
B : “getting at least two tails”.

(ii) Three events A, B and C which are mutually exclusive and exhaustive are
A : “getting at most one head”
B : “getting exactly two heads”
C : “getting exactly three heads”.
Alternatively A : “getting no head”
B : “getting exactly on head”
C : “getting at least two heads”.

(iii) Two events A and B which are not mutually exclusive are
A : “getting at most two tails” and
B : “getting exactly two heads” or “getting exacdy two tails”.

(iv) Two events A and B which are mutually exclusive but not exhaustive are
A : “getting exactly one head” and
B : “getting exactly two heads”.

(v) Three events A, B and C which are mutually exclusive but not exhaustive are
A : “getting exacdy one tail”
B : “getting exacdy two tails” and
C : “getting exacdy three tails”.

Question 6.
Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die.
B: getting an odd number on the first die.
C: getting the sum of the numbers on the dice ≤ 5
Describe the events
Answer.
When two dice are thrown, the sample space is given by
S = {(x, y): x, y 1, 2, 3, 4, 5, 6}
= [(1, 1), (1, 2), (1, 3),(1, 4), (1, 5) ,(1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)]

Accordingly,
A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3) (4, 4), (4, 5), (4, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.

B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}.

C = {(1 ,1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

(i) A’ = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} = B

(ii) Not B = B’ = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
= A.

(iii) A or B = A ∪ B
= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2),(4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} = S

(iv) A and B = A ∩ B = Φ

(v) A but not C = A – C
= {(2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6,6)}

(vi) B or C = B ∪ C
= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

(vii) B and C = B ∩ C = {(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)}

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

∴ A ∩ B’ ∩ C’ = A ∩ A ∩ C’ = A ∩ C’
= {(2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.

Question 7.
Refer to question 6 above state true or false (given reason for your answer).
(i) A and B are mutually exclusive.
(ii) A and B are mutually exclusive and exhaustive.
(iii) A = B’
(iv) A and C are mutually exclusive.
(v) A and B’ are mutually exclusive.
(vi) A’, B’ and C are mutually exclusive and exhaustive.
Answer.
(i) True
∵ A = getting an even number on the first die
B = getting an odd number on the first die
⇒ A ∩ B = Φ
∴ A and B are mutually exclusive events.

(ii) True
A ∪ B = S, i.e., exhaustive. Also, A ∩ B = Φ

(iii) True
∴ B = getting an odd number on the first die
⇒ B’ = getting an even number on the first die = A
∴ A = B’

(iv) False
∵ A ∩ C = {(2, 1), (2, 2), (2, 3), (4, 1)} ≠ Φ.
So A and C are not mutually exclusive.

(v) False
∵ B’ = A
A ∩ B’ = A ∩ A = A ≠ Φ (∵ B’ = A)
So, A and B’ are not mutually exclusive.

(vi) False
A’ ∩ B’ = Φ
A’ ∩ B’ ∩ C = Φ and A’ ∪ B’ ∪ C = S
But A’ ∩ C = B ∩ C = {(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)} ≠ Φ
(∵ A’ = B)
and B’ ∩ C = A ∩ C = {(2, 1), (2, 2), (2, 3), (4, 1)} ≠ Φ (∵ B’ = A)
∴ A’, B’ and C are not mutually exclusive.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 16 Probability Ex 16.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 16 Probability Ex 16.1

Question 1.
Describe the sample space for the indicated experiment: A coin is tossed three times.
Answer.
A coin has two faces : head (H) and tail (T).
When a coin is tossed three times, the total number of possible outcomes is 2³ = 8
Thus, when a coin is tossed three times, the sample space is given by:
S = {HHH, HHT, HTH, HIT, THH, THT, TTH, TTT}.

Question 2.
Describe the sample space for the indicated experiment. A die is thrown two times.
Answer.
When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6. When a die is thrown two times, the sample space is given by S={(x, y): x, y = 1, 2, 3, 4, 5, 6}.
The number of elements in this sample space is 6 × 6 = 36, while the sample space is given by :
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.

Question 3.
Descibe the sample space for the indicated experiment: A coin is tossed four times.
Answer.
When a coin is tossed once, there are two possible outcomes: head (H) and tail (T).
When a coin is tossed four times, the total number of possible outcomes is 2⁴ = 16
Thus, when a coin is tossed four times, the sample space is given by:
S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}.

Question 4.
Describe the sample space for the indicated experiment: A coin is tossed and a die is thrown.
Answer.
A coin has two faces: head (H) and tail (T).
A die has six faces that are numbered from 1 to 6, with one number on each face.
Thus, when a coin is tossed and a die is thrown, the sample space is given by:
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}.

Question 5.
Describe the sample space for the indicated experiment: A coin is tossed and then a die is rolled only in case a head is shown on the coin.
Answer.
A coin has two faces: head (H) and tail (T).
A die has six faces that are numbered from 1 to 6, with one number on each face.
Thus, when a coin is tossed and then a die is rolled only in case a head is shown on the coin, the sample space is given by:
S = {H1, H2, H3, H4, H5, H6,T}.

Question 6.
2 boys and 2 girls are in Room X, and 1 boy and 3 girls in Room Y. Specify the sample space for the experiment in which a room is selected and then a person.
Answer.
Let us denote 2 boys and 2 girls in room X as B1 B2 and G₁, G₂ respectively.
Let us denote 1 boy and 3 girls in room Y as B₃ and G₃, G₄, G₅ respectively. Accordingly, the required sample space is given by
S = {XB₁, XB₂, XG₁, XG₂, YB₃, YG₃, YG₄, YG₅}.

Question 7.
One die of red colour, one of white colour and one of blue colour are placed in a bag. One die is selected at random and rolled, its colour and the number on its uppermost face is noted. Describe the sample space,
Answer.
A die has six faces that are numbered from 1 to 6, with one number on each face.
Let us denote the red, white, and blue dices as R, W, and B respectively. Accordingly, when a die is selected and then rolled, the sample space is given by
S = {R1, R2, R3, R4, R5, R6, W1, W2, W3, W4, W5, W6, B1, B2, B3, B4, B5, B6}.

Question 8.
An experiment consists of recording boy-girl composition of families with 2 children.
(i) What is the sample space if we are interested in knowing whether it is a boy Or girl in the order of their births?
(ii) What is the sample space if we are interested in the number of girls in the family?
Answer.
(i) When the order of the birth of a girl or a boy is considered, the sample space is given by S = {GG, GB, BG, BB}.
(ii) Since the maximum number of children in each family is 2, a family can either have 2 girls or 1 girl or no girl.
Hence, the required sample space is S = {0, 1, 2}.

Question 9.
A box contains 1 red and 3 identical white balls. Two balls are drawn at random in succession without replacement. Write the sample space for this experiment.
Answer.
It is given that the box contains 1 red ball and 3 identical white balls. Let us denote the red ball with R and a white ball with W.
When two balls are drawn at random in succession without replacememt, the sample space is given by
S = {RW, WR, WW}.

Question 10.
An experiment consists of tossing a coin and then throwing it second time if a head occurs. If a tail occurs on the first toss, then a die is rolled once. Find the sample space.
Answer.
It is a two stage experiment. First stage results is either ‘head’ or ‘tail’ when the coin shows up head, then it is tossed again showing up either a head or a tail. If the first toss shows up a tail, then a die is rolled once which may show up any one of the six numbers 1, 2, 3, 4, 5, 6.

∴ S = {(H, H), (H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}.

Question 11.
Suppose 3 bulbs are selected at random from a lot. Each bulh is tested and classified as defective (D) or non-defective (N). Write the sample space of this experiment?
Answer.
Let we denote defective bulb by D and non-defective bulb by N.
There are 2³ = 8 possible outcomes.

Question 12.
A coin is tossed. If the out come is a head, a die is thrown. If the die shows up an even number, the die is thrown again. What is the sample space for the experiment?
Answer.
When a coin is tossed, the possible outcomes are head (H) and tail (T).
When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6.
Thus, the sample space of this experiment is given by:
S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66}.

Question 13.
The numbers 1, 2, 3 and 4 are written separately on four slips of paper. The slips are put in a box and mixed thoroughly. A person draws two slips from the box, one after the other, without replacement. Describe the sample space for the experiment.
Answer.
If 1 appears on the first drawn slip, then the possibilities that the number appears on the second drawn slip are 2, 3, or 4. Similarly, if 2 appears on the first drawn slip, then the possibilities that the number appears on the second drawn slip are 1, 3, or 4. The same holds true for the remaining numbers too.
Thus, the sample space of this experiment is given by:
S = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}.

Question 14.
An experiment consists of rolling a die and then tossing a coin once if the number on the die is even. If the number on the die is odd, the coin is tossed twice. Write the sample space for this experiment.
Answer.
A die has six faces that are numbered from 1 to 6, with one number on each face. Among these numbers, 2, 4, and 6 are even numbers, while 1, 3, and 5 are odd numbers. ‘
A coin has two faces: head (H) and tail (T).
Hence, the sample space of this experiment is given by:
S = {2H, 2T, 4H, 4T, 6H, 6T, 1HH, 1HT, 1TH, ITT, 3HH, 3HT, 3TH, 3TT, 5HH, 5HT, 5TH, 5TT}.

Question 15.
A coin is tossed. If it shows a tail, we draw a ball from a box which contains 2 red and 3 black balls. If it shows head, we throw a die. Find the sample space for this experiment.
Answer.
The box contains 2 red balls and 3 black balls. Let us denote the 2 red balls as Rj R2 and the 3 black balls as B1 B2 and B3.
The sample space of this experiment is given by
S = {TR₁, TR₂, TB₁, TB₂, TB₃, H1, H2, H3, H4, H5, H6}.

Question 16.
A die is thrown repeatedly until a six comes up. What is the sample space for this experiment?
Answer.
In this experiment, six may come up on the first throw, the second throw, the third throw and so on till six is obtained.
Hence, the sample space of this experiment is given by
S = {6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), (1, 5, 6), (2,1, 6,), (2, 2, 6), (2, 5, 6), (5, 1, 6), (5, 2, 6)}.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 15 Statistics Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 15 Statistics Miscellaneous Exercise

Question 1.
The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.
Answer.
Let the remaining two observations be a and b.
Given, \(\bar{x}\) = 9 and σ₂ = 925
∴ \(\bar{x}\) = 9
⇒ Sum of all observations / Number of all observations = 9
⇒ Sum of all observations = 9 × 8 = 72
∴ Number of all observations = 8
⇒ 6 + 7 + 10 + 12 + 12 + 13 + a + b = 72
⇒ 60 + a + b = 72
⇒ a + b = 72 – 60
⇒ a + b = 12
Again, σ² = \(\frac{\sum x_{i}^{2}}{n}-\left(\frac{\sum x_{i}}{n}\right)^{2}\)

⇒ σ² = \(\frac{\sum x_{i}^{2}}{n}-(\bar{x})^{2}\)

9.25 = \(\frac{36+49+100+144+144+169+a^{2}+b^{2}}{8}\) – (9)²

9.25 = \(\frac{642+a^{2}+b^{2}}{8}\) – 81

9.25 + 81 = \(\frac{642+a^{2}+b^{2}}{8}\)

90.25 × 8 = 642 + a² + b²
a² + b² = 722 – 642
a² + b² = 80 …………….(i)
Now, from eq. (1), putb = 12 – a in eq. (ii), we get
a² + (12 – a)² = 80
a² + 144 + a² – 24a = 80
2a² – 24a + 144 – 80 = 0
2a² – 24a + 64 = 0
a² – 12a + 32 = 0 [divide both sides by 2]
a² – 8a – 4a + 32 = 0
a (a – 8) – 4 (a – 8) = 0
(a – 4) (a – 8) = 0
a = 4 or a = 8
One putting a = 4 or a = 8 in eq. (i), we get
b = 8 or b = 4
Hence, observations are 4 and 8.

Question 2.
The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12 and 14,
Find the remaining two observations.
Answer.
Let the remaining two observations be x and y.
Given, \(\bar{x}\) = 8, x₁ = 2, x₂ = 4 x₃ = 10, x₄ = 12 and x₅ = 14.

⇒ \(\frac{x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x+y}{7}\) = 8

⇒ \(\frac{2+4+10+12+14+x+y}{7}\) = 8
⇒ 42 + x + y = 56
⇒ x + y = 14 ………………..(i)
Also, variance = 16
⇒ σ² = 16
⇒ \(\frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}+x^{2}+y^{2}}{7}-(\bar{x})^{2}\) = 16

⇒ \(\frac{2^{2}+4^{2}+10^{2}+12^{2}+14^{2}+x^{2}+y^{2}}{7}\) – (8)² = 16

⇒ \(\frac{4+16+100+144+196+x^{2}+y^{2}}{7}\) – 64 = 16

⇒ \(\frac{460+x^{2}+y^{2}}{7}\) = 80
⇒ x² + y² = 7 × 80 – 460
= 560 – 460
x² + y² = 100
From eq. (i), y = 14 – x
Put this value of y in eq.(ii), we get
x² + (14 – x)² = 100
⇒ x² +196 + x² – 28x = 100
⇒ 2x² – 28x + 96 = 0
⇒ x² – 14x + 48 = 0 [divide both sides by 2]
⇒ (x – 6) (x – 8) = 0
⇒ x = 6, 8
From eq. (i),
If x = 6, then y = 14 – 6 = 8
If x = 8, then y = 14 – 8 = 6
Hence, the remaining two observations are 6 and 8.

Question 3.
The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.
Answer.
Let the observations be x₁, x₂, x₃, x₄, x₅ and x₆.
Then, their mean, \(\bar{x}\) = \(\frac{\sum_{i=1}^{6} x_{i}}{6}\) = 8 [given]
\(\sum_{i=1}^{6} x_{i}\) = 8 × 6 = 48 …………….(i)

On multiplying each observation by 3, we get the new observation as 3x₁, 3x₂, 3x₃, 3x₄, 3x₅ and 3x₆.

Now, their new mean, \(\bar{x}\) = \(=\frac{\sum_{i=1}^{6} 3 x_{i}}{6}=\frac{3 \sum_{i=1}^{6} x_{i}}{6}\)

= \(\frac{3 \times 48}{6}\)

= 24 [from eq. (i)]

∴ Variance of new observation = \(\frac{\sum_{i=1}^{6}\left(3 x_{i}-24\right)^{2}}{6}\)

= \(\frac{3^{2} \sum_{i=1}^{6}\left(x_{i}-8\right)^{2}}{6}\)

= \(\frac{9}{1}\) × Variance of old observation

= \(\frac{9}{1}\) (4)² = 144
[∵ given S.D. of old observation is 4]
Thus, S.D. of new observation = √(Variance) = √(144) = 2.

Question 4.
Given that \(\bar{x}\) is the mean and σ² is the variance of observations x₁, x₂, ………….. x_n. Prove that the mean and variance of the observations ax₁, ax₂, ax₃, ……………., ax_n are a\(\bar{x}\) and a^{2 2} respectively,
Answer.
The given n observations are x₁, x₂, ………………., x_n.
Mean = \(\bar{x}\)
Variance = σ²
∴ σ² = \(\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}\) ………………(i)
If each observation is multiplied by a and the new observations are y_i then
y_i = ax_i i.e., x_i = \(\frac{1}{a}\) y_i
∴ \(\bar{y}\) = \(\frac{1}{n} \sum_{i=1}^{n} y_{i}=\frac{1}{n} \sum_{i=1}^{n} a x_{i}\)

= \(\frac{a}{n} \sum_{i=1}^{n} x_{i}\) = a\(\bar{x}\)

(∵ \(\bar{x}\) = \(\frac{1}{n} \sum_{i=1}^{n} x_{i}\))

Therefore, mean of the observations, ax₁, ax₂, ax₃, ……………., ax_n is a\(\bar{x}\).
Substituting the values of x_iand \(\bar{x}\) in eq. (i), we obtain

⇒ σ² = \(\frac{1}{n} \sum_{i=1}^{n}\left(\frac{1}{a} y_{i}-\frac{1}{a} \bar{y}\right)^{2}\)

a² σ² = \(\frac{1}{n} \sum_{i=1}^{n}\left(y_{i}-\bar{y}\right)^{2}\)
Thus, the variance of the observations, ax₁, ax₂, ax₃, ……………., ax_n is a² σ².

Question 5.
The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases :
(i) If wrong item is omitted.
(ii) If it is replaced by 12.
Answer.
(i) Number of observations (n) = 20
Incorrect mean = 10
Mean = \(\frac{\sum x_{i}}{n}\)

\(\frac{\sum_{i=1}^{20} x_{i}}{20}\) = 10

\(\sum_{i=1}^{20} x_{i}\) = 20 × 10 = 200

(i) Since the wrong item which was 8 has been omitted, the correct mean = \(\frac{\sum_{i=1}^{20} x_{i}-8}{19}\) because, on omitting the wrong item, the total number of observations left are 19.

∴ Correct mean = \(\frac{200-8}{19}=\frac{192}{19}\) = 10.11

Question 6.
The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:

Which of the three subjects shows the highest variability in marks and which shows the lowest?
Answer.
Here, n = 50

For Mathematics
Given, \(\bar{x}\) = 42 and σ = 12
Coefficient of variation, (CV) = \(\frac{\sigma}{\bar{x}}\) × 100
= \(\frac{12}{42}\) × 100
= \(\frac{2}{7}\) × 100
= \(\frac{200}{7}\)
= 2857 ……………(i)

For Physics:
Given, \(\bar{x}\) = 32 and σ = 15
Coefficient of variation, (CV) = \(\frac{\sigma}{\bar{x}}\) × 100
= \(\frac{15}{32}\) × 100
= \(\frac{1500}{32}\)
= 46.87 ………………(ii)

For Chemistry:
Given, \(\bar{x}\) = 40.9 and σ = 20
Coefficient of variation, (CV) = \(\frac{\sigma}{\bar{x}}\) × 100
= \(\frac{20}{40.9}\) × 100
= \(\frac{2000}{40.9}\)
= 2000 = 48.89 ………………(iii)

From eqs. (i), (ii) and (iii), we have
CV of Chemistry > CV of Physics > CV of Mathematics
Hence, Chemistry shows the highest variability and Mathematics shows the least variability.

Question 7.
The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.
Answer.
Given, n = 100, \(\bar{x}\) = 20, σ = 3
∵ Mean, \(\bar{x}\) = 20
∴ \(\frac{\sum x_{i}}{100}\) = 20
⇒ Σx_i = 100 × 20
⇒ Σx_i = 2000
Now, incorrect observations 21, 21 and 18 are omitted, then correct sum is
Σx_i = 2000 – 21 – 21 – 18 = 2000 – 60 = 1940
Now, correct mean of remaining 97 observations is \(\bar{x}\) = \(\frac{1940}{97}\) = 20
Again, σ = 3
⇒ \(\sqrt{\frac{\sum x_{i}^{2}}{n}-(\bar{x})^{2}}\) = 3
On squaring both sides, we get \(\frac{\sum x_{i}^{2}}{100}\) – (20)² = 9
\(\frac{\sum x_{i}^{2}}{100}\) = 9 + 400 = 409
Σx_i² = 409 × 100 = 40900

Now, correct, Σx_i² = 40900 – (21)² – (21)² – (18)²
= 40900 – 441 – 441 – 324
= 40900 – 1206 = 39694
Now, correct standard deviation for remaining 97 observations is σ = \(\sqrt{\frac{39694}{97}-(20)^{2}}\)

= \(\sqrt{409.2-(20)^{2}}\)

= \(\sqrt{409.2-400}=\sqrt{9.2}\) = 3.03.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 15 Statistics Ex 15.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.3

Question 1.
From the data given below state which group is more variable, A or B?

Answer.

For Group A:
Mean, \(\bar{x}\) = A + \(\frac{\sum f_{i} y_{i}}{\sum f_{i}}\) × h

= 45 + \(\frac{-6}{150}\) × 10

= 45 – 0.4 = 44.6

Variance (σ²) = \(\frac{\sum f_{i} y_{i}}{\sum f_{i}}\) [N Σ f_i y_i² – (Σ f_i y_i)²]
= \(\frac{100}{22500}\) [150 × 342 – (- 6)²]
= \(\frac{1}{225}\) [51300 – 36]
= \(\frac{51264}{225}\)
= 227.84

Coefficient of variation (C.V.) = \(\frac{\sigma}{\bar{x}}\) × 100
= \(\frac{15.09}{44.6}\) × 100
= 33.83

For Group B :
Mean, \(\bar{x}\) = A + \(\frac{\sum f_{i} y_{i}}{\sum f_{i}}\) × h

= 45 + \(\frac{-6}{150}\) × 10

= 45 – 0.4 = 44.6

Variance (σ²) = \(\frac{\sum f_{i} y_{i}}{\sum f_{i}}\) [N Σ f_i y_i² – (Σ f_i y_i)²]

= \(\frac{100}{22500}\) [150 × 366 – (- 6)²]

= \(\frac{1}{225}\) (54900 – 36)
= \(\frac{54864}{225}\) = 243.84
∴ σ = 15.61

Coefficient of variation (C.V.) = \(\frac{\sigma}{\bar{x}}\) × 100
= \(\frac{15.61}{44.6}\) × 100 = 35

Coefficient of variation of group B is greater than the coefficient of variation of group A.
Therefore, group B is more variable than group A.

Question 2.
From the prices of shares X and Y below, find out which is more stable in value.

Answer.

For shares X:

For shares Y:
Mean, \(\bar{x}\) = A + \(\frac{\sum y_{i}}{10}\)
= 105 + 0 = 105
S.D., σ = \(\frac{1}{N} \sqrt{N \sum y_{i}^{2}-\left(\sum y_{i}\right)^{2}}\)

= \(\frac{1}{10} \sqrt{10 \times 40-0}=\frac{20}{10}\)

= 0.2
Coefficient of variation (C.V.) = \(\frac{\sigma}{\bar{x}}\) × 100
= \(\frac{2}{105}\) × 100 = 1.9

Coefficient of variation in shares Y is less than the coefficient of variation in shares X.
Therefore, the share Y is more stable than the share X.

Question 3.
An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

(i) Which firm A or B pays larger amount as monthly wages?
(ii) Which firm, A or B, shows greater variability in individual wages?
Answer.
(i) Monthly wages of firm A = Rs. 5253
Number of wage earners in firm A = 586
∴ Total amount paid = Rs. 5253 × 586
Monthly wages of firm B = Rs. 5253
Number of wage earners in firm B = 648
∴ Total amount paid = Rs. 5253 × 648
Thus, firm B pays the larger amount as monthly wages as the number of wage earners in firm B are more than the number of wage earners in firm A.

(ii) Variance of the distribution of wages in firm A (σ₁²) = 100
∴ Standard deviation of the distribution of wages in firm
A(σ₁) = √100 = 10
Variance of the distibution of wages in firm B(σ₂²) = 121
∴ Standard deviation of the distribution of wages in firm
B(σ₂) = √121 = 11
The mean of monthly wages of both the firms is same i.e., 5253.
Therefore, the firm with greater standard deviation will have more varibility.
Thus, firm B has greater variability in the individual wages.

Question 4.
The following is the record of goals scored by team A in a football session:

For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?
Answer.
Make a table from the given data.

Here, Σf_i = 25, Σf_ix_i = 50 and Σ f_i x_i² = 130

For team A:
Mean, \(\bar{x}\) = \(\frac{\sum f_{i} x_{i}}{\sum f_{i}}=\frac{50}{25}\) = 2

Standard deviation, σ = \(\frac{1}{N} \sqrt{N \sum f_{i} x_{i}^{2}-\left(\sum f_{i} x_{i}\right)^{2}}\)

= \(\frac{1}{25} \sqrt{25 \times 130-(50)^{2}}\)

= \(\frac{5}{25} \sqrt{130-100}=\frac{\sqrt{30}}{5}\)

= \(\frac{5.477}{5}\)

= 1.095
∴ Coefficient of variation = \(\frac{\sigma}{\bar{x}}\) × 100
= \(\frac{1.095}{2}\) × 100 = 54.75

For team B:
Given, Mean, \(\bar{x}\) = 2 and SD = σ = 1.25
= \(\frac{1.25}{2}\) × 100 = 625

Since, the coefficient of variation of goals of team A is less than that of B.
Therefore, team A is more consistent than team B.

Question 5.
The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:
\(\sum_{i=1}^{50} x_{i}\) = 212, \(\sum_{i=1}^{50} x_{i}^{2}\) = 902.8, \(\sum_{i=1}^{50} \mathbf{y}_{i}\) = 261 \(\sum_{i=1}^{50} y_{i}^{2}\) = 1457.6
which is more varying the length or weight?
Answer.
Given, \(\sum_{i=1}^{50} x_{i}\) = 212, \(\sum_{i=1}^{50} x_{i}^{2}\) = 902.8
Here, N = 50
∴ Mean, \(\bar{x}=\frac{\sum_{i=1}^{50} x_{i}}{N}=\frac{212}{50}\) = 4.24

∴ Standard deviation, σ₂ (Weight) = √1.89 = 1.37
∴ C.V. (Weight) = Standard deviation / Mean × 100
= \(\frac{1.37}{5.22}\) × 100 = 26.24
Thus, C. V. of weights is greater than the CV. of lengths.
Therefore, weights vary more than the lengths.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 15 Statistics Ex 15.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.2

Question 1.
Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12.
Answer.
The given data is 6, 7, 10, 12, 13, 4, 8, 12

Mean \(\bar{x}=\frac{\sum_{i=1}^{8} x_{i}}{n}\)

= \(\frac{6+7+10+12+13+4+8+12}{8}=\frac{72}{8}\) = 9.

The following table is obtained.

Variance (σ²) = \(\frac{1}{n} \sum_{i=1}^{8}\left(x_{1}-\bar{x}\right)^{2}\)

= \(\frac{1}{8}\) × 74 = 9.25.

Question 2.
Find the mean and variance for the first n natural numbers.
Answer.
The mean of first n natural numbers is calculated as follows.

Mean = \(\frac{\text { Sum of all observations }}{\text { Number of observations }}\)

Mean = \(\frac{\frac{n(n+1)}{2}}{n}=\frac{n+1}{2}\)

Question 3.
Find the mean and variance for the first 10 multiples of 3.
Answer.
The first 10 multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Here, number of observations, n = 10
Mean, \(\bar{x}=\frac{\sum_{i=1}^{10} x_{i}}{10}=\frac{165}{10}\) = 16.5
The following table is obtained.

Variance (σ²) = \(\frac{1}{n} \sum_{i=1}^{10}\left(x_{1}-\bar{x}\right)^{2}\)
= \(\frac{1}{10}\) × 742.5 = 74.25.

Question 4.
Find the mean and variance for the data

Answer.
The data is obtained in tabular form as follows.

Question 5.
Find the mean and variance for the data

Answer.

Question 6.
Find the mean and standerd deviation using short-cut method.

Answer.
The data is obtained in tabular form as follows.

Question 7.
Find the mean and variance for the following frequency distribution.

Answer.

Mean, \(\bar{x}\) = A + \(\frac{\sum_{i=1}^{7} f_{i} y_{i}}{N}\) × h
= 105 + \(\frac{2}{30}\) × 30
= 105 + 2 = 107

Variance (σ²) = \(\frac{h^{2}}{N^{2}}\left[N \sum_{i=1}^{7} f_{i} y_{i}^{2}-\left(\sum_{i=1}^{7} f_{i} y_{i}\right)^{2}\right]\)

= \(\frac{(30)^{2}}{(30)^{2}}\) [30 × 76 – (2)²]
= 2280 – 4 = 2276.

Question 8.
Find the mean and variance for the following frequency distribution.

Answer.

Mean, \(\bar{x}\) = A + \(\frac{\sum_{i=1}^{5} f_{i} y_{i}}{N}\) × h
= 25 + \(\frac{10}{50}\) × 10
= 25 + 2 = 27

Variance (σ²) = \(\frac{h^{2}}{N^{2}}\left[N \sum_{i=1}^{5} f_{i} y_{i}^{2}-\left(\sum_{i=1}^{5} f_{i} y_{i}\right)^{2}\right]\)

= \(\frac{(10)^{2}}{(50)^{2}}\) [50 × 68 – (10)²]

= \(\frac{1}{25}\) [3400 – 100]

= \(\frac{3300}{25}\) = 132.

Question 9.
Find the mean, variance and standard deviation using short-cut method.

Answer.

Mean, \(\bar{x}\) = A + \(\frac{\sum_{i=1}^{5} f_{i} y_{i}}{N}\) × h
= 92.5 + \(\frac{6}{60}\) × 5
= 92.5 + 0.5 = 93

Variance (σ²) = \(\frac{h^{2}}{N^{2}}\left[N \sum_{i=1}^{9} f_{i} y_{i}^{2}-\left(\sum_{i=1}^{9} f_{i} y_{i}\right)^{2}\right]\)

= \(\frac{(5)^{2}}{(60)^{2}}\) [60 × 254 – (6)²]

= \(\frac{25}{3600}\) (15204) = 105.58

∴ Standard deviation (σ) = \(\sqrt{105.58}\) = 10.27.

Question 10.
The diameters of circles (in mm) drawn in a design are given below:

Answer.

Here, N = 100, h = 4
Let the assumed mean, A, be 42.5

Mean, \(\) = A + \(\frac{\sum_{i=1}^{5} f_{i} y_{i}}{N}\) × h
= 42.5 + \(\frac{25}{100}\) × 4 = 43.5

Variance (σ²) = \(\frac{h^{2}}{N^{2}}\left[N \sum_{i=1}^{5} f_{i} y_{i}^{2}-\left(\sum_{i=1}^{5} f_{i} y_{i}\right)^{2}\right]\)

= \(\frac{16}{10000}\) [100 × 199 – (25)²]
= \(\frac{16}{10000}\) [19900 – 625]
= \(\frac{16}{10000}\) × 19275 = 30.84
∴ Standard deviation (σ) = 5.55.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 15 Statistics Ex 15.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1

Question 1.
Find the mean deviation about the mean for the data 4, 7, 8, 9, 10, 12, 13, 17.
Answer.
The given data is 4, 7, 8, 9, 10, 12, 13, 17
Mean of the data, \(\bar{x}=\frac{4+7+8+9+10+12+13+17}{8}=\frac{80}{8}\) = 10
The deviations of the respective observations from the mean \(\bar{x}\), i.e., x_i – \(\bar{x}\), are – 6, – 3, – 2, – 1, 0, 2, 3, 7.
The absolute values of the deviations, i.e., |x_i – \(\bar{x}\)|, are 6, 3, 2, 1, 0, 2, 3, 7
The required mean deviation about the mean is M.D. (\(\bar{x}\)) = \(\frac{\sum_{i=1}^{8}\left|x_{i}-\bar{x}\right|}{8}\)

= \(\frac{6+3+2+1+0+2+3+7}{8}=\frac{24}{8}\) = 3.

Question 2.
Find the mean deviation about the mean for the data 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.
Answer.
The given data is 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Mean of the data \(\bar{x}\) = \(\frac{38+70+48+40+42+55+63+46+54+44}{10}=\frac{500}{10}\) = 50
The deviations of the respective observations from the mean \(\bar{x}\), i.e., x_i – \(\bar{x}\), are – 12, 20, – 2, – 10, – 8, 5, 13, – 4, 4, – 6.
The absolute values of the deviations, i.e., |x_i – \(\bar{x}\)|, are 12, 20, 2, 10, 8, 5, 13, 4, 4, 6.
The required mean deviation about the mean is
M.D (\(\bar{x}\)) = \(\frac{\sum_{i=1}^{10}\left|x_{i}-\bar{x}\right|}{10}\)

= \(\frac{12+20+2+10+8+5+13+4+4+6}{10}\)

= \(\frac{84}{10}\) = 8.4.

Question 3.
Find the mean deviation about the median for the data 13, 17, 16,14, 11, 13,10,16,11,18,12,17.
Answer.
The given data is 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.
Here, the numbers of observations are 12, which is even.
Arranging the data in ascending order, we obtain 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

The deviations of the respective observations from the median, i.e., x_i – M, are – 3.5, – 2.5, – 2.5, – 1.5, – 0.5,- 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5.
The absolute values of the deviations, |x_i – M|, are 3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5
The required mean deviation about the median is
M.D.(M) = \(\frac{\sum_{i=1}^{12}\left|x_{i}-M\right|}{12}\)

= \(\frac{3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.5}{12}\)

= \(\frac{28}{12}\) = 2.33.

Question 4.
Find the mean deviation about the median for the data 36, 72, 46,42,60,45,53,46,51,49
Answer.
The given data is 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.
Here, the numbers of observations are 10, which is even.
Arranging the data in ascending order, we obtain 36, 42, 45, 46, 46, 49, 51, 53, 60, 72

The deviations of the respective observations from the median i.e.,x_i – M, are – 11.5, – 5.5, – 2.5, – 1.5, – 1.5, 1.5, 3.5, 5.5, 12.5, 24.5
The absolute values of the deviations, |x_i – M|, are 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5.
Thus, the required mean deviation about the median is

M.D.(M) = \(\frac{\sum_{i=1}^{10}\left|x_{i}-M\right|}{10}\)

= \(\frac{11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5}{10}\)

= \(\frac{70}{10}\) = 7.

Question 5.
Find the mean deviation about the median for the data.

Answer.

Question 6.
Find the mena deviation about the mean for the data.

Answer.

Question 7.
Find the men and deviation about the mean for the data.

Answer.
The given observations are already in ascending order. Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

Here, N = 26, which is even.
Median is the mean of 13th and 14th observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.
∴ Median, (M) = \(\frac{13^{\text {th }} \text { observation }+14^{\text {th }} \text { observation }}{2}=\frac{7+7}{2}\) = 7

The absolute values of the deviations, i.e., |x_i – M|, are

Question 8.
Find the mean deviation about the median for the data.

Answer.
Here, N = Σf_i = 29, which is odd. So, the median is the \(\left(\frac{n+1}{2}\right)^{\text {th }}\) observation i.e., \(\frac{29+1}{2}\) = 15th observation, which is equal to 30. Thus, median is 30.

∴ Mean deviation from the median = \(\frac{\sum f_{i}\left|x_{i}-30\right|}{\sum f_{i}}\)

= \(\frac{148}{29}\) = 5.1.

Question 9.
Find the mean deviation about the mean for the data.

Answer.
The following table is formed.

Question 10.
Find the mean deviation about the mean for the data.

Answers.
The following table is formed.

Question 11.
Find the mean deviation about median for the following data.

Answer.
Table to find cumulative frequencies and Σf_i |x_i – M| is given below :

Now C = 14, class corresponding to cumulative frequency 28 is 20 – 30.
∴ l = 20, f = 14, h = 10

Median = l + \(\frac{\frac{N}{2}-C}{f}\) × h
= 20 + \(\frac{25-14}{14}\) × 10
= 20 + \(\frac{11}{14}\) × 10
= 20 + 7.86 = 27.86 14
Σf_i |x_i – M| = 517.16
∴ Mean devintion about median = \(\frac{517.16}{50}\) = 10.34.

Question 12.
Calculate the mean deviation about median age for the age distribution of 100 persons given below:

Answer.
We make the table from the given data.

Here, N = 100
∴ \(\frac{N}{2}\) = 50
So, median class is 35.5 – 40.5
∴ l = 35.5, cf = 37, f = 26,h = 5
∵ M = l + \(\frac{\frac{N}{2}-c f}{f}\) × h

M = 35.5 + \(\frac{50-37}{26}\) × 5

= 35.5 + \(\frac{13}{26}\) × 5

= 35.5 + 2.5 = 38

∴ Mean deviation about median = \(\frac{\sum f_{i}\left|\boldsymbol{x}_{i}-M\right|}{\sum f_{i}}=\frac{735}{100}\) = 7.35
Hence, the mean deviation about median is 7.35.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 14 Mathematical Reasoning Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Miscellaneous Exercise

Question 1.
Write the negation of the following statements :
(i) p : For every positive real number x, the number #-l is also positive.
(ii) q : All cats scratch.
(iii) r : For every real number x, either x > 1 or x < 1.
(iv) s : There exists a number# such that 0 < x < 1.
Answer.
(i) ~ p : There exists a positive real number x such that x -1 is not positive.
(ii) ~ q : There exist a cat which does not scratch.
(iii) ~ r : There exists a real number x such that neither x > 1 nor x < 1.
(iv) ~ s: There does not exist a number x such that 0 < x < 1.

Question 2.
State the converse and contrapositive of each of the following statements:
(i) p : A positive integer is prime only if it has no divisors other than 1 and itself.
(ii) q : I go to a beach whenever it is a sunny day.
(iii) r : If it is hot outside, then you feel thirsty.
Answer.
(i) Converse : If a positive integer has no divisor other than 1 and itself then it is a prime. Contrapositive : If a positive integer has no divisor other than 1 and itself then it is not prime.
(ii) Converse : If it is a sunny day, then I go to beach.
Contrapositive : If it is not sunny day, then I do not go to beach.
(iii) Converse : If you feel thirsty then it is hot outside.
Contrapositive : If you do not feel thirsty then it is not hot outside.

Question 3.
Write each of the statements in the form “if p, then q
(i) p : It is necessary to have a password to log on to the server,
(ii) q : There is traffic jam whenever it rains.
(iii) r : You can access the website only if you pay a subscription fee.
Answer.
(i) Statement p can be written as follows. If you log on to the server, then you have a password.
(ii) Statement q can be written as follows. If if rains, then there is a traffice jam.
(iii) Statement r can be written as follows. If you can access the website, then you pay a subscription fee.

Question 4.
Rewrite each of the following statements in the form “p if and only if q”.
(i) p : If you watch television, then your mind is free and if your mind is free, then you watch television.
(ii) q : For you to get an A grade, it is necessary and sufficent that you do all the homework regularly.
(iii) r : If a quadrilateral is equiangular, then it is a rectangle and if a quadrilateral is a rectangle, then it is equiangular.
Answer.
(i) You watch television if and only if your mind is free.
(ii) You get an A grade if and only if you do all the homework regularly.
(iii) A quadrilateral is equiangular if and only if it is a rectangle.

Question 5.
Given below are two statements p : 25 is a multiple of 5. q : 25 is a multiple of 8. Write the compound statements connecting these two statements with “And” and “Or”. In both cases check the validity of the compound statement.
Answer.
The compound statement with ‘And’ is “25 is a multiple of 5 and 8”. This is a false statement, since 25 is not a multiple of 8. The compound statement with ‘Or’ is “25 is a multiple of 5 or 8”. This is a true statement, since 25 is not a multiple of 8 but it is a multiple of 5.

Question 6.
Check the validity of the statements given below by the method given against it.
(i) p : The sum of an irrational number and a rational number is irrational (by contradiction method).
(ii) q : If n is a real number with n > 3, then n² > 9 (by contradiction method).
Answer.
(i) Let √a be irrational number and b be a rational number.
Their sum = b + √a
Let it is not irrational. Therefore, it is a rational number. …(i)
b + √a = \(\frac{p}{q}\), where p, q are co-prime
√a = \(\frac{p}{q}\) – b ……………(i)
L.H.S. = √a = An irrational number
R.HS. = \(\frac{p}{q}\) – b = A rational number
It is a contradiction. Therefore, the sum of a rational and irrational number is irrational, which is a valid statement.

(ii) Let n > 3 and n² < 9
Put n = 3 + a, we have
n² = 9 + 6a + a² = 9 + a (6 + a), which is a contradiction.
⇒ If n > 3, then n² > 9, which is a valid statement.

Question 7.
Write the following statement in five different ways, conveying the same meaning.
p: If triangle is equiangular, then it is an obtuse angled triangle.
Answer.
(i) A triangle is equiangular implies that it is an obtuse angled triangle.
(ii) A triangle is equiangular only if it is an obtuse angled triangle.
(iii) For a triangle to be equiangular it is necessary that it is an obtuse angled triangle.
(iv) For a triangle to be obtuse angled triangle, it is sufficient that it is equiangular.
(v) If a triangle is not obtuse angled triangle, then it is not an equiangular triangle.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 14 Mathematical Reasoning Ex 14.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.5

Question 1.
Show that the statement
p: “If x is a real number such that x³ + 4x = 0, then x is 0″ is true by
(i) direct method
(ii) method of contradiction
(iii) method of contrapositive
Answer.
Let q and r be the statements given by q: x is a real number such that x³ + 4x =0.
r : x = 0
Then p : if q, there r.

(i) Direct Method:
Let q be true, then q is true.
⇒ x is real number such that x³ + 4x = 0.
⇒ x is a real number such that x (x² + 4) = 0.
⇒ x = 0, [∵ n ∈ R, ∴ x² + 4 ≠ 0]
⇒ r is true.
Thus, q is true
⇒ r is true.
Hence, p is true.

(ii) Method of contradiction :
If possible, let p be not true, then p is not true.
⇒ ~ p is true.
⇒ ~ (q ⇒ r) is true, [∵ p = q ⇒ r]
⇒ q and ~ r is true, [∵ ~ (q ⇒ r) ~ q and ~ r]
⇒ x is a real number such that x³ + 4x = 0 and x ≠ 0.
⇒ x = 0 and x ≠ 0.
This is a contradiction.
Hence, p is true.

(iii) Method of contrapositive :
Let r be not true, then r is not true.
⇒ x ≠ 0, x ∈ R.
⇒ x (x² + 4) ≠ 0, n ∈ R ⇒ q is not true.
Thus, ~ x ⇒ ~ q.
Hence, p : q ⇒ r is true.

Question 2.
Show that the statement “For any real numbers a and b, a² = b² implies that a = b” is not true by giving a counter-example.
Answer.
The given compound statement is of the form “if p then q”.
We assume that p is true then a, b efi such that a² = b²
Let us take a = – 3 and b = 3
Now a² = b² but a ≠ b
So when p is true, q is false.
Thus the given compound statement is not true.

Question 3.
Show that the following statement is true by the method of contrapositive.
p : If x is an integer and x² is even, then x is also even.
Answer.
Let q and r be the statements given by
q : if x is an integer and x2 is even,
r : x is an even integer.
Thus, p : “If q, then r”.
If possible, let r be false. Thus, r is false.
⇒ x is not an even integer.
⇒ x is an odd integer.
⇒ x = (2x +1) for some integer 4.
⇒ x² = 4x² + 4x + 1 = 4n(n + 1) + 1
⇒ x² is an odd integer, [∵ 4x (x + 1) is even]
⇒ q is false.
Thus, r is false ⇒ q is false.
Hence, p : “If q, then r” is a true statement.

Question 4.
By giving a counter example, show that the following statements are not true.
(i) p : If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle.
(ii) q : The equation x² – 1 = 0 does not have a root lying between 0 and 2.
Answer.
(i) Since the triangle is obtuse angled triangle then θ > 90°
Let θ = 100°
Also all the angles of the triangle are equal.
Sum of all angles of the triangle are 300°, which is not possible.
Thus the given compound statement is not true.

(ii) We see that x = 1 is a root of the equation x² – 1 = 0, which lies between 0 and 2.
Thus the given compound statement is not true.

Question 5.
Which of the following statements are true and which are false? In each case give a valid reason for saying so.
(i) p : Each radius of a circle is a chord of the circle.
(ii) q : The centre of a circle bisects each chord of the circle.
(iii) r : Circle is a particular case of an ellipse.
(iv) s : If x and y are integers such that x > y, then x < – y.
(v) t : √11 is a rational number.
Answer.
(i) False : The end points of radius do not lie on the circle, therefore, it is not a chord.
(ii) False : Only diameters are bisected at the centre. Other chords do not pass through the centre. Therefore centre can not bisect them.
(iii) True : Equation of ellipse is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\)
When b = a. The equation becomes \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}}\) = 1 or x² + y² = a², which is equation of the circle.
(iv) True : If x and y are integers and x > y then – x < – y. By rule of inequality.
(v) False : 11 is a prime number.
∴ √11 is irrational.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 14 Mathematical Reasoning Ex 14.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.4

Question 1.
Rewrite the following statement with “if-then” in five different ways conveying the same meaning.
If a natural number is odd, then its square is also odd.
Answer.
(i) A natural number is odd implies that its square is odd.
(ii) A natural number is odd only if its square is odd.
(iii) If the square of a natural number is not odd, then the natural number is also not odd.
(iv) For a natural number to be odd it is necessary that its square is odd.
(v) For a square of a natural number to be odd, it is sufficient that the number is odd.

Question 2.
Write the contrapositive and converse of the following statements.
(i) If x is a prime number, then x is odd.
(ii) It the two lines are parallel, then they do not intersect in the same plane.
(iii) Something is cold implies that it has low temperature.
(iv) You cannot comprehend geometry if you do not know how to reason deductively.
(v) x is an even number implies that x is divisible by 4.
Answer.
(i) Contrapositive statement: If a number x is not odd, then xis not a prime number.
Converse statement: If x is odd, then it is a prime number.

(ii) Contrapositive statement: If two straight lines intersect in a plane than the lines are not parallel.
Converse statement: If two lines do not intersect in the sample plane, then the two lines are parallel.

(iii) Contrapositive statement : If the temperature of something is not low, then it is not cold.
Converse statement : If something has low temperature, then it is cold.

(iv) Contrapositive statement: If you know how to reason deductively, then you comprehend geometry.

(v) Converse statement: If you do not know how to reason deductively, then you can not comprehend geometry.
The given statement can be written as :
“If x is an even number, then x is divisible by 4”.
Contrapositive statement: If x is not divisible by 4, then x is not an even number.
Converse statement: If x is divisible by 4 then x is an even number.

Question 3.
Write each of the following statement in the form “if-then”.
(i) You get a job implies that your credentials are good.
(ii) The Banana trees will bloom if it stays warm for a month.
(iii) A quadrilateral is a parallelogram if its diagonals bisect each other.
(iv) To get A+ in the class, it is necessary that you do the exercises of the book.
Answer.
(i) If you get a job, then your credential are good.
(ii) If the banana trees,stays warm for a month, then it will bloom.
(iii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
(iv) If you get A+ in the class, then you do all exercises in the book.

Question 4.
Given statements in (a) and (b). Identify the statements given below as contrapositive or converse of each other.
(a) If you live in Delhi, then you have winter clothes.
(i) If you do not have winter clothes, then you do not live in Delhi.
(ii) If you have winter clothes, then you live in Delhi.

(b) If a quadrilateral is a parallelogram, then its diagonals bisect each other.
(i) If the diagonals of a quadrilateral do not bisect each other, then the quadrilateral is not a parallelogram.
(ii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
Answer.
(a) (i) Contrapositive statement
(ii) Converse statement.

(b) (i) Contrapositive statement
(ii) Converse statement.