NCERT Solutions for Class 11 Maths Chapter 14 – Probability Exercise 14.2

NCERT Solutions for Class 11 Maths Chapter 14 – Probability Exercise 14.2

Class 11 Maths Chapter 14 – Probability, Exercise 14.2 moves from event description to actual probability calculation. This exercise covers the classical definition of probability, the addition theorem for two events, and the complement rule. Students learn to compute probabilities for experiments like tossing coins, throwing dice, drawing cards, and real-life scenarios involving groups of people. The problems also involve checking whether given probability assignments are valid and determining whether events are consistently defined.

NCERT Solutions for Class 11 Maths Chapter 14 – Probability Exercise 14.2

Probability

Easy

Q.

$\begin{array}{l}\mathbf{\text{Which of the following cannot be valid assignment of probabilities}}\\ \mathbf{\text{for outcomes of sample Space S = {}}{\mathbf{\omega }}_{\mathbf{1}}\mathbf{\text{, }}{\mathbf{\omega }}_{\mathbf{2}}\mathbf{\text{, }}{\mathbf{\omega }}_{\mathbf{3}}\mathbf{\text{, }}{\mathbf{\omega }}_{\mathbf{4}}\mathbf{\text{, }}{\mathbf{\omega }}_{\mathbf{5}}\mathbf{\text{, }}{\mathbf{\omega }}_{\mathbf{6}}\mathbf{\text{, }}{\mathbf{\omega }}_{\mathbf{7}}\mathbf{\text{}}}\\ \begin{array}{cccccccc}\mathbf{Assignment}& {\mathbf{\omega }}_{\mathbf{1}}& {\mathbf{\omega }}_{\mathbf{2}}& {\mathbf{\omega }}_{\mathbf{3}}& {\mathbf{\omega }}_{\mathbf{4}}& {\mathbf{\omega }}_{\mathbf{5}}& {\mathbf{\omega }}_{\mathbf{6}}& {\mathbf{\omega }}_{\mathbf{7}}\\ \mathbf{\left(}\mathbf{a}\mathbf{\right)}& \mathbf{0}\mathbf{.1}& \mathbf{0}\mathbf{.01}& \mathbf{0}\mathbf{.05}& \mathbf{0}\mathbf{.03}& \mathbf{0}\mathbf{.01}& \mathbf{0}\mathbf{.2}& \mathbf{0}\mathbf{.6}\\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}& \frac{\mathbf{1}}{\mathbf{7}}& \frac{\mathbf{1}}{\mathbf{7}}& \frac{\mathbf{1}}{\mathbf{7}}& \frac{\mathbf{1}}{\mathbf{7}}& \frac{\mathbf{1}}{\mathbf{7}}& \frac{\mathbf{1}}{\mathbf{7}}& \frac{\mathbf{1}}{\mathbf{7}}\\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}& \mathbf{0}\mathbf{.1}& \mathbf{0}\mathbf{.2}& \mathbf{0}\mathbf{.3}& \mathbf{0}\mathbf{.4}& \mathbf{0}\mathbf{.5}& \mathbf{0}\mathbf{.6}& \mathbf{0}\mathbf{.7}\\ \mathbf{\left(}\mathbf{d}\mathbf{\right)}& \mathbf{-}\mathbf{0}\mathbf{.1}& \mathbf{0}\mathbf{.2}& \mathbf{0}\mathbf{.3}& \mathbf{0}\mathbf{.4}& \mathbf{-}\mathbf{0}\mathbf{.2}& \mathbf{0}\mathbf{.1}& \mathbf{0}\mathbf{.3}\\ \mathbf{\left(}\mathbf{e}\mathbf{\right)}& \frac{\mathbf{1}}{\mathbf{14}}& \frac{\mathbf{2}}{\mathbf{14}}& \frac{\mathbf{3}}{\mathbf{14}}& \frac{\mathbf{4}}{\mathbf{14}}& \frac{\mathbf{5}}{\mathbf{14}}& \frac{\mathbf{6}}{\mathbf{14}}& \frac{\mathbf{15}}{\mathbf{14}}\end{array}\end{array}$

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Probability

Easy

Q.

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)

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Probability

Medium

Q.

A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35.
Find (i) P (A∪B)               (ii) P(A´∩B´)             (iii) P(A ∩ B´)           (iv) P(B ∩ A´)

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Probability

Medium

Q.

Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.

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Probability

Medium

Q.

In a certain lottery 10,000 tickets are sold and ten equal prizes are awarded.
What is the probability of not getting a prize if you buy    (a) one ticket     (b) two tickets     (c) 10 tickets.

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Probability

Medium

Q.

A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with number ‘3’. If die is rolled once, determine

(i) P(2)
(ii) P(1 or 3)
(iii) P(not 3)

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Probability

Medium

Q.

A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that

(i) all will be blue?
(ii) at least one will be green?

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Probability

Difficult

Q.

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.

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Probability

Medium

Q.

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?

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Probability

Medium

Q.

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?

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Probability

Easy

Q.

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.

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Probability

Easy

Q.

$\begin{array}{l}\mathbf{\text{If E and F are events such that P(E) = }}\frac{\mathbf{\text{1}}}{\mathbf{\text{4}}}\mathbf{\text{,\hspace{0.17em}P(F) = }}\frac{\mathbf{\text{1}}}{\mathbf{\text{2}}}\mathbf{\text{ and }}\\ \mathbf{\text{P(E and F) = }}\frac{\mathbf{\text{1}}}{\mathbf{\text{8}}}\mathbf{\text{, find }}\left(\text{i}\right)\mathbf{\text{P}}\left(\text{E or F}\right)\mathbf{\text{, }}\left(\text{ii}\right)\mathbf{\text{P}}\left(\text{not E and not F}\right)\mathbf{\text{.}}\end{array}$

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Probability

Easy

Q.

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

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Probability

Easy

Q.

$\begin{array}{l}\mathbf{\text{Given P(A) = }}\frac{\mathbf{\text{3}}}{\mathbf{\text{5}}}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}and P(B) = }}\frac{\mathbf{\text{1}}}{\mathbf{\text{5}}}\mathbf{\text{. Find P(A or B), if A and B are }}\\ \mathbf{\text{mutually exclusive events.}}\end{array}$

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Probability

Easy

Q.

$\begin{array}{l}\mathbf{\text{Fill in the blanks in following table:}}\\ \begin{array}{ccccc}& \mathbf{\text{P(A)}}& \mathbf{\text{P(B)}}& \mathbf{\text{P(A}}\mathbf{\cap }\mathbf{\text{B)}}& \mathbf{\text{P(A}}\mathbf{\cup }\mathbf{\text{B)}}\\ \mathbf{\text{(i)}}& \frac{\mathbf{\text{1}}}{\mathbf{\text{3}}}& \frac{\mathbf{\text{1}}}{\mathbf{\text{5}}}& \frac{\mathbf{\text{1}}}{\mathbf{\text{15}}}& \mathbf{.}\mathbf{......}\\ \mathbf{\text{(ii)}}& \mathbf{\text{0.35}}& \mathbf{.}\mathbf{......}& \mathbf{\text{0.25}}& \mathbf{\text{0.6}}\\ \mathbf{\text{(iii)}}& \mathbf{\text{0.5}}& \mathbf{\text{0.35}}& \mathbf{.}\mathbf{......}& \mathbf{\text{0.7}}\end{array}\end{array}$

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Probability

Medium

Q.

In a lottery, a person choses 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?

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Probability

Easy

Q.

A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is
(i) a vowel    (ii) a consonant

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Probability

Easy

Q.

$\mathbf{\text{If }}\frac{\mathbf{\text{2}}}{\mathbf{\text{11}}}\mathbf{\text{ is the probability of an event, what is the probability of the event 'not A'.}}$

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Probability

Medium

Q.

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

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Probability

Easy

Q.

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?

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Probability

Medium

Q.

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

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Probability

Medium

Q.

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.

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Probability

Medium

Q.

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.

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Probability

Medium

Q.

The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e., from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?

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Exercise 14.2 is one of the most important exercises in Class 11 Maths from a board exam perspective. The addition theorem P(A ∪ B) = P(A) + P(B) – P(A ∩ B) is a frequently tested formula, and questions on coin toss, card draw, and word-based probability problems appear regularly in CBSE exams. JEE aspirants must also master this exercise as it forms the foundation for conditional probability and Bayes' theorem in Class 12.

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

Answer:

(a) P(E) = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1 All values are positive and sum = 1. Valid assignment.

(b) P(E) = 1/7 × 7 = 1 All values are positive and sum = 1. Valid assignment.

(c) P(E) = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8 > 1 Sum exceeds 1. Not a valid assignment.

(d) Two values (–0.1 and –0.2) are negative. Probability can never be negative. Not a valid assignment.

(e) P(E) = 1/14 + 2/14 + 3/14 + 4/14 + 5/14 + 6/14 + 15/14 = 36/14 > 1 Also, 15/14 > 1 individually. Not a valid assignment.

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

Answer:

Sample space: S = {HH, HT, TH, TT} n(S) = 4

Let A = event of getting at least one tail A = {HT, TH, TT} n(A) = 3

P(A) = n(A) / n(S) = 3/4

Question 3. A die is thrown. Find the probability of the following events:

Answer:

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

(i) A prime number will appear A = {2, 3, 5}, n(A) = 3 P(A) = 3/6 = 1/2

(ii) A number greater than or equal to 3 will appear B = {3, 4, 5, 6}, n(B) = 4 P(B) = 4/6 = 2/3

(iii) A number less than or equal to 1 will appear A = {1}, n(A) = 1 P(A) = 1/6 = 1/6

(iv) A number more than 6 will appear A = φ, n(A) = 0 P(A) = 0/6 = 0

(v) A number less than 6 will appear A = {1, 2, 3, 4, 5}, n(A) = 5 P(A) = 5/6

Question 4. A card is selected from a pack of 52 cards.

Answer:

(a) How many points are there in the sample space? n(S) = 52

(b) Probability that the card is an ace of spades n(A) = 1 P(A) = 1/52 = 1/52

(c)(i) Probability that the card is an ace There are 4 aces. P(A) = 4/52 = 1/13

(c)(ii) Probability that the card is a black card There are 26 black cards. P(A) = 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:

Answer:

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) Sum = 3 A = {(1,2)}, n(A) = 1 P(A) = 1/12

(ii) Sum = 12 A = {(6,6)}, n(A) = 1 P(A) = 1/12

Question 6. There are four men and six women on the city council. If one council member is selected at random, how likely is it that it is a woman?

Answer:

Total members = 4 + 6 = 10, so n(S) = 10 Number of women = 6

P(selecting a woman) = 6/10 = 3/5

Question 7. A fair coin is tossed four times. A person wins Rs 1 for each head and loses Rs 1.50 for each tail. Find the probability of having each possible amount.

Answer:

n(S) = 16

Question 8. Three coins are tossed once. Find the probability of getting:

Answer:

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

(i) 3 heads: n(A) = 1, P(A) = 1/8

(ii) 2 heads: n(A) = 3, P(A) = 3/8

(iii) At least 2 heads: n(A) = 4, P(A) = 4/8 = 1/2

(iv) At most 2 heads: n(A) = 7, P(A) = 7/8

(v) No head: n(A) = 1, P(A) = 1/8

(vi) 3 tails: n(A) = 1, P(A) = 1/8

(vii) Exactly two tails: n(A) = 3, P(A) = 3/8

(viii) No tail: n(A) = 1, P(A) = 1/8

(ix) At most two tails: n(A) = 7, P(A) = 7/8

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

Answer:

P(A) = 2/11

P(not A) = 1 – P(A) = 1 – 2/11 = 9/11

Question 10. A letter is chosen at random from the word 'ASSASSINATION'. Find the probability that the letter is:

Answer:

Total letters = 13, so n(S) = 13 (A appears 3 times, S appears 4 times, I appears 2 times, N appears 2 times, T and O appear once each)

(i) A vowel Vowels: A, A, A, I, I, O → n(A) = 6 P(vowel) = 6/13

(ii) A consonant Consonants: S, S, S, S, N, N, T → n(A) = 7 P(consonant) = 7/13

Question 11. In a lottery, a person chooses six different natural numbers at random from 1 to 20. What is the probability of winning the prize?

Answer:

Total ways to choose 6 numbers from 20: n(S) = ²⁰C₆ = 20! / (6! × 14!) = 38760

Only 1 combination matches the winning numbers: n(A) = 1

P(winning) = 1/38760 = 1/38760

Question 12. Check whether the following probabilities P(A) and P(B) are consistently defined:

Answer:

(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6 P(A ∩ B) = 0.6 > P(A) = 0.5, which is impossible since P(A ∩ B) ≤ P(A). Not consistently defined.

(ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8 P(A ∩ B) = P(A) + P(B) – P(A ∪ B) = 0.5 + 0.4 – 0.8 = 0.1 P(A ∩ B) = 0.1 ≤ P(A) and P(A ∩ B) ≤ P(B). Consistently defined.

Question 13. Fill in the blanks in the following table:

Answer:

(i) P(A ∪ B) = 1/3 + 1/5 – 1/15 = 5/15 + 3/15 – 1/15 = 7/15

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

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

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

Answer:

Since A and B are mutually exclusive, P(A ∩ B) = 0

P(A ∪ B) = P(A) + P(B) = 3/5 + 1/5 = 4/5

Question 15. If E and F are events such that P(E) = 1/4, P(F) = 1/2 and P(E and F) = 1/8, find:

Answer:

(i) P(E or F) P(E ∪ F) = P(E) + P(F) – P(E ∩ F) = 1/4 + 1/2 – 1/8 = 2/8 + 4/8 – 1/8 = 5/8

(ii) P(not E and not F) P(not E and not F) = P(E' ∩ F') = P(E ∪ F)' = 1 – P(E ∪ F) = 1 – 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.

Answer:

P(not E or not F) = P(E' ∪ F') = P(E ∩ F)' = 0.25

1 – P(E ∩ F) = 0.25 P(E ∩ F) = 0.75 ≠ 0

Since P(E ∩ F) ≠ 0, E and F are not mutually exclusive.

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:

Answer:

(i) P(not A) P(not A) = 1 – 0.42 = 0.58

(ii) P(not B) P(not B) = 1 – 0.48 = 0.52

(iii) P(A or B) P(A ∪ B) = 0.42 + 0.48 – 0.16 = 0.74

Question 18. In a class, 40% study Mathematics, 30% study Biology, 10% study both. Find the probability that a randomly selected student studies Mathematics or Biology.

Answer:

P(A) = 40/100 = 2/5 P(B) = 30/100 = 3/10 P(A ∩ B) = 10/100 = 1/10

P(A ∪ B) = 2/5 + 3/10 – 1/10 = 4/10 + 3/10 – 1/10 = 6/10 = 3/5

Question 19. The probability of passing the first examination is 0.8, second is 0.7, and at least one is 0.95. What is the probability of passing both?

Answer:

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. Probability of passing both English and Hindi = 0.5, passing neither = 0.1, passing English = 0.75. Find probability of passing Hindi.

Answer:

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

P(A ∪ B) = 1 – P(A' ∩ B') = 1 – 0.1 = 0.9

P(B) = P(A ∪ B) – P(A) + P(A ∩ B) = 0.9 – 0.75 + 0.5 = 0.65

Question 21. In a class of 60 students, 30 opted for NCC, 32 for NSS, 24 for both. Find:

Answer:

n(S) = 60, n(A) = 30, n(B) = 32, n(A ∩ B) = 24

P(A) = 30/60 = 1/2 P(B) = 32/60 = 8/15 P(A ∩ B) = 24/60 = 2/5

(i) P(NCC or NSS) P(A ∪ B) = 1/2 + 8/15 – 2/5 = 15/30 + 16/30 – 12/30 = 19/30

(ii) P(neither NCC nor NSS) P(A' ∩ B') = 1 – P(A ∪ B) = 1 – 19/30 = 11/30

(iii) P(NSS but not NCC) n(B – A) = n(B) – n(A ∩ B) = 32 – 24 = 8 P(B – A) = 8/60 = 2/15

FAQs – Chapter 14 Probability Exercise 14.2

Q1. What are the conditions for a valid probability assignment?

Each probability value must be non-negative (≥ 0), each value must be at most 1, and the sum of all probabilities over the sample space must equal exactly 1.

Q2. What is the addition theorem of probability?

For any two events A and B: P(A ∪ B) = P(A) + P(B) – P(A ∩ B). If A and B are mutually exclusive, P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B).

Q3. How do you find P(not A)?

P(not A) = P(A') = 1 – P(A). This is the complement rule, one of the most frequently used formulas in this exercise.

Q4. What does "at least one" mean in probability problems?

"At least one" means one or more occurrences. It is easier to calculate using the complement: P(at least one) = 1 – P(none).

Q5. How is P(A ∩ B) calculated when P(A ∪ B) is given?

Rearrange the addition theorem: P(A ∩ B) = P(A) + P(B) – P(A ∪ B). This is used in Questions 12, 13, 19, and 20 of this exercise.