NCERT Solutions Class 11 - Chapter 14 Probability - Exercise 14.2

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

Assignment:	ω1	ω2	ω3	ω4	ω5	ω6	ω7
(a)	0.1	0.01	0.05	0.03	0.01	0.2	0.6
(b)	1/7	1/7	1/7	1/7	1/7	1/7	1/7
(c)	0.1	0.2	0.3	0.4	0.5	0.6	0.7
(d)	-0.1	0.2	0.3	0.4	-0.2	0.1	0.3
(e)	1/14	2/14	3/14	4/14	5/14	6/14	15/14

Solution:

(a) To check whether a given assignment is valid or not we should check 2 conditions:

i) Given numbers should be positive.

ii) sum of probabilities is 1.

0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1

Therefore, the given assignment is valid.

(b) To check whether a given assignment is valid or not we should check 2 conditions:

i) Given numbers should be positive.

ii) sum of probabilities is 1.

= (1/7) + (1/7) + (1/7) + (1/7) + (1/7) + (1/7) + (1/7)

= 7/7

= 1

Therefore, the given assignment is valid.

(c) To check whether a given assignment is valid or not we should check 2 conditions:

i) Given numbers should be positive.

ii) sum of probabilities is 1.

= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7

= 2.8 > 1

Therefore, the 2nd condition is not satisfied

So, the given assignment is not valid.

(d) Since, the given numbers are negative, which doesn't satisfy the 1st condition.

So, the assignment is not valid.

(e) To check whether a given assignment is valid or not we should check 2 conditions:

i) Given numbers should be positive.

ii) sum of probabilities is 1.

= (1/14) + (2/14) + (3/14) + (4/14) + (5/14) + (6/14) + (7/14)

= (28/14) ≥ 1

The second condition doesn’t hold true so the assignment is not valid.

Question 2. A coin is tossed twice, what is the probability that at least one tail occurs?

Solution:

The possible outcomes are Head(H) and Tail(T).

Here coin is tossed twice, then sample space is S = (TT, HH, TH, HT),  n(S) = 4.

Let A be the event of getting at least one tail

n (A) = 3

P(Event) = Number of outcomes favorable to event/ Total number of possible outcomes

P(A) = n(A)/n(S)

= 3/4.

Question 3. A die is thrown, find the probability of the 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.

Solution:

Here, S = {1, 2, 3, 4, 5, 6}

n(S) = 6

(i) A = {2, 3, 5}; n(A) = 3

P(A) = n(A)/n(S)

= 3/6

(ii) A = {3, 4, 5, 6}; n(A) = 4

P(A) = n(A)/n(S)

= 4/6

(iii) A = {1}; n (A) = 1

P(A) = n(A)/n(S)

= 1/6

(iv) A = {0}; n (A) = 0

P(A) = n(A)/n(S)

= 0/6 = 0

(v) A= {1, 2, 3, 4, 5}; n (A) = 5

P(A) = n(A)/n(S)

= 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

Solution:

(a) Number of points in the sample space = 52 

n(S) = 52

(b) Let us assume ‘A’ be the event of drawing an ace of spades.

A = 1

Then, n (A) = 1

P(A) = n(A)/n(S)

= 1/52

(c) Let us assume ‘A’ be the event of drawing an ace. There are four aces.

Then, n (A) = 4

P(A) = n(A)/n(S)

= 4/52

= 1/13

(d) Let us assume ‘A’ be the event of drawing a black card. There are 26 black cards.

Then, n (A) = 26

P(A) = n(A)/n(S)

= 26/52

= 1/2

Question 5. A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is (i) 3 (ii) 12

Solution:

1,2,3,4,5,6 are the possible outcomes

sample space 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)}

n(S) = 12

(i) A = {(1, 2)}; n (A) = 1

P(A) = n(A)/n(S)

= 1/12

(ii) A = {(6, 6)}; n (A) = 1

P(A) = n(A)/n(S)

= 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?

Solution:

Total members in the council = 4 + 6 = 10; n (S) = 10

Number of women are 6

n (A) = 6

P(A) = n(A)/n(S)

= 6/10

Question 7. A fair coin is tossed four times, and a person wins Rs 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 having each of these amounts.

Solution:

Here, Head(H) and Tail(T) are the possible outcomes.

S = (HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THHT, 

        HTTH, THTH, TTHH, TTTH, TTHT, THTT, HTTT, TTTT)

(i) For 4 heads = 1 + 1 + 1 + 1 = Rs 4

So, he wins Rs 4

(ii) For 3 heads and 1 tail = 1 + 1 + 1 – 1.50

= 3 – 1.50

= Rs 1.50

So, he will be winning Rs 1.50

(iii) For 2 heads and 2 tails = 1 + 1 – 1.50 – 1.50

= 2 – 3

= – Rs 1

So, he will be losing Rs 1

(iv) For 1 head and 3 tails = 1 – 1.50 – 1.50 – 1.50

= 1 – 4.50

= – Rs 3.50

So, he will be losing Rs. 3.50

(v) For 4 tails = – 1.50 – 1.50 – 1.50 – 1.50

= – Rs 6

So, he will be losing Rs. 6

Now the sample space is

S = {4, 1.50, 1.50, 1.50, 1.50, – 1, – 1, – 1,

         – 1, – 1, – 1, – 3.50, – 3.50, – 3.50, – 3.50, – 6}

Then, n (S) = 16

P (winning Rs 4) = 1/16

P (winning Rs 1.50) = 4/16 

= 1/4

P (winning Rs 1) = 6/16

= 3/8

P (winning Rs 3.50) = 4/16 

= 1/4

P (winning Rs 6) = 1/16

= 3/8

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

Solution:

Here, Head(H) and Tail(T) are the possible outcomes.

S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}; n(S) = 8

(i) Possibility of getting 3 heads is 1; n(A) = 1

P(A) = n(A)/n(S)

= 1/8

(ii) Possibility of getting 2 heads is 3; n(A) = 3

P(A) = n(A)/n(S)

= 3/8

(iii) Possibility of getting at least 2 heads is 4; n(A) = 4

P(A) = n(A)/n(S)

= 4/8

(iv) Possibility of getting at most 2 heads is 7; n(A) = 7

P(A) = n(A)/n(S)

= 7/8

(v) Possibility of getting no head is 1; n(A) = 1

P(A) = n(A)/n(S)

= 1/8

(vi) Possibility of getting 3 tails is 1; n(A) = 1

P(A) = n(A)/n(S)

= 1/8

(vii) Possibility of getting 2 tails is 3; n(A) = 3

P(A) = n(A)/n(S)

= 3/8

(viii) Possibility of getting no tail is 1; n(A) = 1

P(A) = n(A)/n(S)

= 1/8

(ix) Possibility of getting at most 2 tails is 7; n(A) = 7

P(A) = n(A)/n(S)

= 7/8

Question 9. If 2/11 is the probability of an event, what is the probability of the event ‘not A’.

Solution:

2/11 is the probability of an event A

P (A) = 2/11

P (not A) = 1 – P (A)

= 1 – (2/11)

= (11 – 2)/11

= 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

Solution:

Total letters in the given word = 13

Number of vowels in the given word = 6

Number of consonants in the given word = 7

Then, the sample space n(S) = 13

(i) a vowel

sample space n(S) = 6

P(A) = n(A)/n(S)

= 6/13

(ii) a consonant

n(A) = 7

P(A) = n(A)/n(S)

= 7/13

Question 11. In a lottery, a person chooses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already 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.]

Solution:

Total numbers of numbers in the draw = 20

Numbers to be selected = 6

So, n(S) = ²⁰C₆

Now, let us assume the X be the event that chosen six numbers to 

match with the six numbers already fixed by the lottery committee

n(A) =  ⁶C₆ = 1

So, the probability of winning the prize is

P(A) = n(A)/n(S) = ⁶C₆/²⁰C₆

= 6×5×4×3×2×1×14! / 20×19×18×17×16×15×14!

= 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

Solution:

(i) P(A ∩ B) > P(A)

Here, the given probabilities are not consistently defined.

(ii) 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

Therefore, P(A ∩ B) < P(A) and P(A ∩ B) < P(B)

So, the given probabilities are consistently defined.

Question 13. Fill in the blanks in following table: 

P(A)	P(B)	P(A∩B)	P(A∪B)
(i)	1/3	1/5	1/15	.......
(ii)	0.35	.......	0.25	0.6
(iii)	0.5	0.35	.......	0.7

Solution:

(i) P(A) = 1/3, P(B) = 1/5, P(A ∩ B) = 1/15, P(A ∪ B) = ?

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= (1/3) + (1/5) – (1/15)

= ((5 + 3)/15) – (1/15)

= 7/15

(ii) P(A) = 0.35, P(B) = ?, P(A ∩ B) = 0.25, P(A ∪ B) = 0.6

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

0.6 = 0.35 + P(B) – 0.25

P(B) = 0.6 + 0.25 – 0.35

= 0.5

(iii) P(A) = 0.5, P(B) = 0.35, P(A ∪ B) = 0.7, P(A ∩ B) = ?

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

0.7 = 0.5 + 0.35 – P(A ∩ B)

P(A ∩ B) = 0.85 – 0.7

= 0.15

Question 14. Given P(A) = 5/3 and P(B) = 1/5. Find P(A or B), if A and B are mutually exclusive events.

Solution:

P(A) = 5/3 and P(B) = 1/5

P(A ∪ B) or P(A or B) = P(A) + P(B)

= (3/5) + (1/5)

= 4/5

Question 15. If E and F are events such that P(E) = ¼, P(F) = ½ and P(E and F) = 1/8, find
(i) P(E or F), (ii) P(not E and not F)

Solution:

(i) P(E∪F) = P(E) + P(F) – P(E ∩ F)

= 1/4 + 1/2 - (1/8)

= 5/8

(ii) P(E' ∩ F') = P((E U F)') = 1 – P(E U F )

= 1 – (5/8)

= (8 – 5)/8

= 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.

Solution:

P(E' U F') = 0.25

P((E ∩ F)') = 0.25

1  – P(E ∩ F) = 0.25

P(E ∩ F) = 0.75

P(E ∩ F) is not equal to 0

So, E and F are not mutually exclusive events.

Question 17. A and B are 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) and (iii) P(A or B)

Solution:

(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) P(A not B) = P(A ∪ B) = P(A) + P(B) – P(A ∩ 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.

Solution:

P(A) = 40/100 = 2/5

P(B) = 30/100 = 3/10

P(A ∩ B) = 10/100 = 1/10

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 2/5 + 3/10 – 1/10

P(A ∪ B) = 3/5

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?

Solution:

P(A ∪ B) = 0.95, P(A) = 0.8, P(B) = 0.7

P(A ∪ B) = P(A) + P(B) – P(A∩B)

0.95 = 0.8 + 0.7 – P(A∩B)

P(A ∩ B) = 1.5 – 0.95

= 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?

Solution:

Given that, P(A) = 0.75, P(A ∩ B) =0.5, P(A' ∩ B') = 0.1

P(A' ∩ B') = 1 – P(A ∪ B)

Then, P(A ∪ B) = 1 – P(A' ∩ B')

= 1 – 0.1

= 0.9

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

0.9 = 0.75 + P(B) – 0.5

P(B) = 0.9 + 0.5 – 0.75

= 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.

Solution:

The total number of students in class = 60

n(S) = 60

Assume NCC be ‘A’ and NSS be ‘B’

n(A) = 30, n(B) = 32 , n(A∩B) = 24

P(A) = n(A)/n(S)

= 30/60

= 1/2

P(B) = n(B)/n(S)

= 32/60

= 8/15

P(A ∩ B) = n(A ∩ B)/n(S)

= 24/60

= 2/5

(i) The student opted for NCC or NSS.

P (A or B) = P(A) + P(B) – P(A and B)

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= (1/2)+ (8/15) – (2/5)

= 19/30 

(ii) P(student opted neither NCC nor NSS)

P(not A and not B) = P(A' ∩ B')

We know that, P(A' ∩ B') = 1 – P(A ∪ B)

= 1 – (19/30)

= 11/30

(iii) P(student opted NSS but not NCC)

n(B – A) = n(B) – n (A ∩ B)

32 – 24 = 8

= (8/60) = 2/15