Anand Classes provides NCERT Solutions for Class 11 Maths Chapter 14 – Probability (Exercise 14.2) following the latest NCERT and CBSE syllabus (2025–2026). This exercise helps students understand addition and multiplication theorems of probability, including the concepts of mutually exclusive and independent events. Each question is solved with step-by-step explanations, examples, and definitions to make the topic easy to understand. These NCERT Class 11 Probability Solutions are essential for scoring well in CBSE Board Exams and building a strong foundation for JEE Main, JEE Advanced, NDA, and CUET. Click the print button to download study material and notes.
NCERT Question 1 : Which of the following cannot be a valid assignment of probabilities for outcomes of sample space
$$S = \lbrace \omega_1, \omega_2, \omega_3, \omega_4, \omega_5, \omega_6, \omega_7 \rbrace$$
| Assignment | ω1 | ω2 | ω3 | ω4 | ω5 | ω6 | ω7 |
|---|---|---|---|---|---|---|---|
| (a) | 0.1 | 0.01 | 0.05 | 0.03 | 0.01 | 0.2 | 0.6 |
| (b) | $\frac{1}{7}$ | $\frac{1}{7}$ | $\frac{1}{7}$ | $\frac{1}{7}$ | $\frac{1}{7}$ | $\frac{1}{7}$ | $\frac{1}{7}$ |
| (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) | $\frac{1}{14}$ | $\frac{2}{14}$ | $\frac{3}{14}$ | $\frac{4}{14}$ | $\frac{5}{14}$ | $\frac{6}{14}$ | $\frac{15}{14}$ |
Solution :
To check whether a given assignment is valid or not, we must verify two conditions:
- All given probabilities must be non-negative.
- The sum of all probabilities must be equal to 1.
(a)
$$0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1$$
Both conditions are satisfied.
✅ Therefore, the given assignment is valid.
(b)
$$\frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} = \frac{7}{7} = 1$$
Both conditions are satisfied.
✅ Therefore, the given assignment is valid.
(c)
$$0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8 > 1$$
The second condition is not satisfied.
❌ Therefore, the given assignment is not valid.
(d)
The given probabilities include negative values (e.g., $-0.1$, $-0.2$).
This violates the first condition.
❌ Therefore, the given assignment is not valid.
(e)
$$\frac{1}{14} + \frac{2}{14} + \frac{3}{14} + \frac{4}{14} + \frac{5}{14} + \frac{6}{14} + \frac{15}{14} = \frac{36}{14} > 1$$
The second condition is not satisfied.
❌ Therefore, the given assignment is not valid.
✅ Final Answer
The assignments (c), (d), and (e) cannot be valid probability distributions.
Probability distribution validity, rules of probability, NCERT Class 11 Maths Chapter 16, sample space and events, Anand Classes NCERT solutions for JEE Main, NDA, and CUET preparation.
NCERT Question 2 : A coin is tossed twice. What is the probability that at least one tail occurs?
Solution :
The possible outcomes for a coin toss are Head (H) and Tail (T).
When the coin is tossed twice, the sample space is:
$$S = \lbrace TT, HH, TH, HT \rbrace$$
Hence,
$$n(S) = 4$$
Let event A be “getting at least one tail.”
The outcomes favorable to event A are:
$$A = \lbrace TT, TH, HT \rbrace$$
So,
$$n(A) = 3$$
Now,
$$P(A) = \frac{n(A)}{n(S)}$$
Substituting the values:
$$P(A) = \frac{3}{4}$$
✅ Final Answer
$$\boxed{P(A) = \frac{3}{4}}$$
Probability of getting tails in coin toss, sample space of coin experiment, basic probability problems, NCERT Class 11 Maths Chapter 14, Anand Classes NCERT solutions for JEE Main, NDA, and CUET preparation.
NCERT 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 1 will appear.
(iv) A number more than 6 will appear.
(v) A number less than 6 will appear.
Solution :
The sample space for a die throw is:
$$S = \lbrace 1, 2, 3, 4, 5, 6 \rbrace$$
Hence,
$$n(S) = 6$$
(i) A prime number will appear
Prime numbers between 1 and 6 are:
$$A = \lbrace 2, 3, 5 \rbrace$$
Thus,
$$n(A) = 3$$
$$P(A) = \frac{n(A)}{n(S)} = \frac{3}{6} = \frac{1}{2}$$
✅ Therefore, the probability of getting a prime number is $\frac{1}{2}$.
(ii) A number greater than or equal to 3 will appear
$$A = \lbrace 3, 4, 5, 6 \rbrace$$
$$n(A) = 4$$
$$P(A) = \frac{n(A)}{n(S)} = \frac{4}{6} = \frac{2}{3}$$
✅ Therefore, the probability is $\frac{2}{3}$.
(iii) A number less than or equal to 1 will appear
$$A = \lbrace 1 \rbrace$$
$$n(A) = 1$$
$$P(A) = \frac{n(A)}{n(S)} = \frac{1}{6}$$
✅ Therefore, the probability is $\frac{1}{6}$.
(iv) A number more than 6 will appear
Since no number on a die is greater than 6,
$$A = \varnothing$$
$$n(A) = 0$$
$$P(A) = \frac{n(A)}{n(S)} = \frac{0}{6} = 0$$
❌ Therefore, the probability is 0.
(v) A number less than 6 will appear
$$A = \lbrace 1, 2, 3, 4, 5 \rbrace$$
$$n(A) = 5$$
$$P(A) = \frac{n(A)}{n(S)} = \frac{5}{6}$$
✅ Therefore, the probability is $\frac{5}{6}$.
✅ Final Answers
| Event | Probability |
|---|---|
| (i) A prime number appears | $\frac{1}{2}$ |
| (ii) Number ≥ 3 | $\frac{2}{3}$ |
| (iii) Number ≤ 1 | $\frac{1}{6}$ |
| (iv) Number > 6 | $0$ |
| (v) Number < 6 | $\frac{5}{6}$ |
Probability of events in dice throw, prime number probability, basic probability formulas, NCERT Class 11 Maths Chapter 14, Anand Classes NCERT solutions for JEE Main, NDA, and CUET preparation.
NCERT 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) a black card
Solution :
(a)
The total number of cards in a pack is 52.
Hence,
$$n(S) = 52$$
✅ Therefore, the number of points in the sample space is 52.
(b)
Let event A be “drawing an ace of spades.”
There is only one ace of spades in the pack.
$$n(A) = 1$$
Hence,
$$P(A) = \frac{n(A)}{n(S)} = \frac{1}{52}$$
✅ Therefore, the probability of drawing an ace of spades is $\frac{1}{52}$.
(c)(i)
Let event A be “drawing any ace.”
There are 4 aces in a deck of 52 cards.
$$n(A) = 4$$
Hence,
$$P(A) = \frac{n(A)}{n(S)} = \frac{4}{52} = \frac{1}{13}$$
✅ Therefore, the probability of drawing an ace is $\frac{1}{13}$.
(c)(ii)
Let event A be “drawing a black card.”
There are 26 black cards (13 spades + 13 clubs).
$$n(A) = 26$$
Hence,
$$P(A) = \frac{n(A)}{n(S)} = \frac{26}{52} = \frac{1}{2}$$
✅ Therefore, the probability of drawing a black card is $\frac{1}{2}$.
✅ Final Answers
| Event | Probability |
|---|---|
| (a) Number of points in sample space | 52 |
| (b) Ace of spades | $\frac{1}{52}$ |
| (c)(i) Any ace | $\frac{1}{13}$ |
| (c)(ii) Black card | $\frac{1}{2}$ |
Probability of drawing cards from a deck, ace of spades probability, playing cards sample space, NCERT Class 11 Maths Chapter 14, Anand Classes NCERT solutions for JEE Main, NDA, and CUET preparation.
NCERT 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 :
The possible outcomes for the coin are 1 and 6.
The possible outcomes for the die are 1, 2, 3, 4, 5, 6.
Hence, the sample space is:
$$S = \lbrace (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), $$ $$(6,1), (6,2), (6,3), (6,4), (6,5), (6,6) \rbrace$$
Therefore,
$$n(S) = 12$$
(i) The sum of numbers is 3
We need $(\text{coin}, \text{die})$ such that the sum equals 3.
$$1 + 2 = 3$$
Thus,
$$A = \lbrace (1, 2) \rbrace$$
$$n(A) = 1$$
Hence,
$$P(A) = \frac{n(A)}{n(S)} = \frac{1}{12}$$
✅ Therefore, the probability that the sum is 3 is $\frac{1}{12}$.
(ii) The sum of numbers is 12
We need $(\text{coin}, \text{die})$ such that the sum equals 12.
$$6 + 6 = 12$$
Thus,
$$A = \lbrace (6, 6) \rbrace$$
$$n(A) = 1$$
Hence,
$$P(A) = \frac{n(A)}{n(S)} = \frac{1}{12}$$
✅ Therefore, the probability that the sum is 12 is $\frac{1}{12}$.
✅ Final Answers
| Event | Probability |
|---|---|
| (i) Sum = 3 | $\frac{1}{12}$ |
| (ii) Sum = 12 | $\frac{1}{12}$ |
Sum of numbers from coin and die, probability of combined events, sample space of two experiments, NCERT Class 11 Maths Chapter 14, Anand Classes NCERT solutions for JEE Main, NDA, and CUET preparation.
NCERT 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:
$$4 + 6 = 10$$
Hence,
$$n(S) = 10$$
Number of women:
$$n(A) = 6$$
Now,
$$P(A) = \frac{n(A)}{n(S)} = \frac{6}{10} = \frac{3}{5}$$
✅ Therefore, the probability that the selected member is a woman is $\frac{3}{5}$.
Probability of selecting a woman, selection from council problems, basic probability in real-life contexts, NCERT Class 11 Maths Chapter 14, Anand Classes NCERT solutions for JEE Main, NDA, and CUET preparation.
NCERT Question 7 : A fair coin is tossed four times, and a person wins ₹1 for each head and loses ₹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 :
The possible outcomes for each toss are Head (H) and Tail (T).
Hence, for four tosses, the sample space is:
$$S = \lbrace HHHH, HHHT, HHTH, HTHH, THHH, $$ $$HHTT, HTHT, THHT, HTTH, THTH, TTHH, $$ $$TTTH, TTHT, THTT, HTTT, TTTT \rbrace$$
So,
$$n(S) = 16$$
(i) For 4 Heads (HHHH):
Earnings = $1 + 1 + 1 + 1 = ₹4$
✅ He wins ₹4.
Number of outcomes = 1
(ii) For 3 Heads and 1 Tail:
Earnings = $1 + 1 + 1 – 1.50 = 3 – 1.50 = ₹1.50$
✅ He wins ₹1.50.
Number of outcomes = ${}^4C_3 = 4$
(iii) For 2 Heads and 2 Tails:
Earnings = $1 + 1 – 1.50 – 1.50 = 2 – 3 = -₹1$
❌ He loses ₹1.
Number of outcomes = ${}^4C_2 = 6$
(iv) For 1 Head and 3 Tails:
Earnings = $1 – 1.50 – 1.50 – 1.50 = 1 – 4.50 = -₹3.50$
❌ He loses ₹3.50.
Number of outcomes = ${}^4C_1 = 4$
(v) For 4 Tails (TTTT):
Earnings = $-1.50 – 1.50 – 1.50 – 1.50 = -₹6$
❌ He loses ₹6.
Number of outcomes = ${}^4C_0 = 1$
Different Amounts of Money and Corresponding Probabilities
| Amount (₹) | No. of Outcomes | Probability |
|---|---|---|
| +4 | 1 | $\frac{1}{16}$ |
| +1.50 | 4 | $\frac{4}{16} = \frac{1}{4}$ |
| -1 | 6 | $\frac{6}{16} = \frac{3}{8}$ |
| -3.50 | 4 | $\frac{4}{16} = \frac{1}{4}$ |
| -6 | 1 | $\frac{1}{16}$ |
✅ Final Results
The person can have five different possible net amounts after four tosses:
$$\lbrace 4, 1.5, -1, -3.5, -6 \rbrace$$
and their corresponding probabilities are:
$$
P(₹4) = \frac{1}{16}, \
P(₹1.5) = \frac{1}{4}, \
P(-₹1) = \frac{3}{8}, \\
P(-₹3.5) = \frac{1}{4}, \
P(-₹6) = \frac{1}{16}.
$$
Expected value in probability, coin toss outcomes, probability distribution of gains and losses, NCERT Class 11 Maths Chapter 14, Anand Classes NCERT solutions for JEE Main, NDA, and CUET preparation.
NCERT 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.
The sample space is
$$S = \lbrace HHH, HHT, HTH, THH, TTH, HTT, TTT, THT \rbrace$$
and
$$n(S) = 8$$
(i) 3 heads
Possibility of getting 3 heads is $1$; $n(A)=1$.
$$P(A)=\frac{n(A)}{n(S)}=\frac{1}{8}$$
(ii) 2 heads
Possibility of getting 2 heads is $3$; $n(A)=3$.
$$P(A)=\frac{n(A)}{n(S)}=\frac{3}{8}$$
(iii) at least 2 heads
Possibility of getting at least 2 heads is $4$; $n(A)=4$.
$$P(A)=\frac{n(A)}{n(S)}=\frac{4}{8}=\frac{1}{2}$$
(iv) at most 2 heads
Possibility of getting at most 2 heads is $7$; $n(A)=7$.
$$P(A)=\frac{n(A)}{n(S)}=\frac{7}{8}$$
(v) no head
Possibility of getting no head is $1$; $n(A)=1$.
$$P(A)=\frac{n(A)}{n(S)}=\frac{1}{8}$$
(vi) 3 tails
Possibility of getting 3 tails is $1$; $n(A)=1$.
$$P(A)=\frac{n(A)}{n(S)}=\frac{1}{8}$$
(vii) Exactly two tails
Possibility of getting 2 tails is $3$; $n(A)=3$.
$$P(A)=\frac{n(A)}{n(S)}=\frac{3}{8}$$
(viii) no tail
Possibility of getting no tail is $1$; $n(A)=1$.
$$P(A)=\frac{n(A)}{n(S)}=\frac{1}{8}$$
(ix) at most two tails
Possibility of getting at most 2 tails is $7$; $n(A)=7$.
$$P(A)=\frac{n(A)}{n(S)}=\frac{7}{8}$$
Download detailed NCERT solutions by Anand Classes for Class 11 Maths, JEE Main, NDA, and CUET preparation.
Additional practice: three-coin probability examples, sample-space enumeration for beginners, probability distributions for small experiments.
NCERT Question 9 : If $\dfrac{2}{11}$ is the probability of an event, what is the probability of the event ‘not A’?
Solution :
Given:
$$P(A) = \frac{2}{11}$$
The probability of the complementary event ‘not A’ is given by
$$P(\text{not }A) = 1 – P(A)$$
Substitute the given value:
$$P(\text{not }A) = 1 – \frac{2}{11}$$
$$P(\text{not }A) = \frac{11 – 2}{11} = \frac{9}{11}$$
✅ Final Answer
$$\boxed{P(\text{not }A) = \frac{9}{11}}$$
Complementary probability formula, basic probability rules, NCERT Class 11 Maths Chapter 16, Anand Classes NCERT solutions for JEE Main, NDA, and CUET preparation.
