Anand Classes presents the most reliable and student-friendly Probability NCERT Solutions for Exercise 13.2 Class 12 Chapter 13 (Set-2), designed to simplify complex probability concepts through clear explanations and step-by-step solutions. These Class 12 Math notes help students strengthen their understanding, improve problem-solving accuracy, and score higher in CBSE board exams. Download this high-quality PDF to access detailed solutions that follow the latest NCERT guidelines and support effective self-study. Click the print button to download study material and notes.
Access Probability NCERT Solutions for Exercise 13.2 Class 12 Chapter 13
NCERT Question.10 : Events (A) and (B) satisfy
$$P(A)=\frac12, P(B)=\frac{7}{12}, P(\text{not }A \text{ or not }B)=\frac14.$$
Determine whether (A) and (B) are independent.
Solution
We are given
$$P(A’ \cup B’)=\frac14.$$
Use the complement identity:
$$(A’ \cup B’) = (A \cap B)’.$$
Thus,
$$P((A \cap B)’)=\frac14.$$
So,
$$P(A \cap B)=1-\frac14=\frac34.$$
Now check for independence.
Events (A) and (B) are independent iff
$$P(A \cap B)=P(A)P(B).$$
Compute:
$$P(A)P(B)=\frac12 \times \frac{7}{12}=\frac{7}{24}.$$
Compare:
$$\frac{7}{24} \neq \frac34.$$
Hence,
$$P(A \cap B) \ne P(A)P(B),$$
which means that A and B are not independent.
Final Result
$$\boxed{\text{A and B are not independent}}$$
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NCERT Question 11 : Given two independent events $A$ and $B$ such that $P(A)=0.3$, $P(B)=0.6$. Find
(i) $P(A \text{ and } B)$
(ii) $P(A \text{ and not } B)$
(iii) $P(A \text{ or } B)$
(iv) $P(\text{neither A nor B})$
Solution
Given
$$P(A)=0.3$$
$$P(B)=0.6$$
Since A and B are independent,
$$P(A \cap B)=P(A)P(B)$$
(i) Probability of $A$ and $B$
$$P(A \cap B)=0.3 \times 0.6=0.18$$
(ii) Probability of $A$ and not $B$
$$P(A \cap B’)=P(A)-P(A \cap B)$$
So,
$$P(A \cap B’)=0.30-0.18=0.12$$
(iii) Probability of $A$ or $B$
$$P(A \cup B)=P(A)+P(B)-P(A \cap B)$$
Substitute:
$$P(A \cup B)=0.30+0.60-0.18=0.72$$
(iv) Probability of neither $A$ nor $B$
$$P(A’ \cap B’)=P(A \cup B)’=1 – P(A \cup B)$$
Thus,
$$P(A’ \cap B’)=1 – 0.72=0.28$$
Final Answers:
$$
\boxed{
\begin{aligned}
(i)0.18 \
(ii)0.12 \
(iii)0.72 \
(iv)0.28
\end{aligned}}
$$
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NCERT Question 12 : A die is tossed thrice. Find the probability of getting an odd number at least once.
Solution
A die is tossed thrice. Probability of getting an odd number atleast once = Probability of getting an odd number once + Probability of getting an odd number twice + Probability of getting an odd number thrice.
Probability of getting an odd number at least once
$$= 1 – \text{Probability of getting an odd number 0 times}$$
Probability of getting an odd number 0 times
$$= \text{Probability of getting an even number all 3 times}$$
Since probability of an even number on one toss is =$\dfrac{1}{2}$
$$\text{Probability of getting an even number all 3 times}=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$$
Thus,
Probability of getting an odd number at least once
$$= 1 – \text{Probability of getting an odd number 0 times}$$
Probability of getting an odd number at least once
$$=1-\text{Probability of getting an even number all 3 times}$$
$$=1 – \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{7}{8}$$
Final Answer
$$
\boxed{\frac{7}{8}}
$$
NCERT Question.13 :
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
(i) both balls are red
(ii) first ball is black and second is red
(iii) one of them is black and the other is red
Solution
Since the two balls are drawn with replacement, the two draws are independent.
Number of black balls = 10, Number red balls = 8
Total number of balls = 10 + 8 = 18
Let R = red ball, B = black ball. Drawing is with replacement, so events are independent.
(i) Probability both balls are red
P(both balls are red) = P(first ball is red and second ball is red)
P(both balls are red) = P(first ball is red) × P(second ball is red)
$$P(\text{both red}) = P(R \cap R) = P(R) \cdot P(R)$$
$$P(\text{both red}) = \frac{8}{18} \cdot \frac{8}{18} = \frac{64}{324} = \frac{16}{81}$$
(ii) Probability first ball is black and second is red
P(first ball is black and second is red) = P(first ball is black) × P(second ball is red)
$$P(B \cap R) = P(B) \cdot P(R)$$
$$P(B \cap R) = \frac{10}{18} \cdot \frac{8}{18} = \frac{80}{324} = \frac{20}{81}$$
(iii) Probability one ball is black and the other is red
P(one of the balls is black and other is red) = P(‘first ball is black and second is red’ or ‘first ball is red and second is black’)
P(one of the balls is black and other is red)= P(first ball is black and second is red) + P(first ball is red and second is black)
P(one of the balls is black and other is red) = P(first ball is black) × P(second ball is red) + P(first ball is red) × P(second ball is black)
$$P(\text{one black, one red}) = P(B \cap R) + P(R \cap B)$$
$$P(\text{one black, one red}) = \frac{10}{18} \cdot \frac{8}{18} + \frac{8}{18} \cdot \frac{10}{18} = 2 \cdot \frac{80}{324} = \frac{160}{324} = \frac{40}{81}$$
Final Result
$$\boxed{(i)\frac{16}{81}(ii)\frac{20}{81}(iii)\frac{40}{81}}$$
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NCERT Question.14 :
Probability of solving a specific problem independently by A and B are $\dfrac{1}{2}$ and $\dfrac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that
(i) the problem is solved
(ii) exactly one of them solves the problem
Solution
Given, $P(A) = \dfrac{1}{2}$, $P(B) = \dfrac{1}{3}$
Since A and B are independent events,
(i) Probability that the problem is solved
$$P(\text{problem is solved}) = P(A \cup B) = P(A) + P(B) – P(A \cap B)$$
$$P(A \cap B) = P(A) \cdot P(B) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$$
$$P(\text{problem is solved}) = \frac{1}{2} + \frac{1}{3} – \frac{1}{6} = \frac{3}{6} + \frac{2}{6} – \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$$
(ii) Probability that exactly one of them solves the problem
P(exactly one of them solves the problem) = P(‘A solves and B does not solve’ or ‘B solves and A does not solve’)
$$P(\text{exactly one solves}) = P(A \cap B’) + P(B \cap A’)$$
$$P(\text{exactly one solves})= (P(A) – P(A \cap B)) + (P(B) – P(A \cap B))$$
$$P(\text{exactly one solves})= P(A) + P(B) – 2P(A \cap B)$$
$$P(\text{exactly one solves})= \frac{1}{2} + \frac{1}{3} – 2 \cdot \frac{1}{6} = \frac{1}{2} + \frac{1}{3} – \frac{1}{3} = \frac{1}{2}$$
Final Result
$$\boxed{(i)\frac{2}{3}(ii)\frac{1}{2}}$$
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NCERT Question.15 : One card is drawn at random from a well-shuffled deck of 52 cards. Determine in which of the following cases the events $E$ and $F$ are independent.
(i) $E$ : ‘the card drawn is a spade’, $F$ : ‘the card drawn is an ace’
(ii) $E$ : ‘the card drawn is black’, $F$ : ‘the card drawn is a king’
(iii) $E$ : ‘the card drawn is a king or queen’, $F$ : ‘the card drawn is a queen or jack’
Solution
(i) $E$: the card drawn is a spade
$$P(E) = \frac{13}{52} = \frac{1}{4}$$
$F$: the card drawn is an ace
$$P(F) = \frac{4}{52} = \frac{1}{13}$$
$(E \cap F)$: the card drawn is the ace of spades
$$P(E \cap F) = \frac{1}{52}$$
$$P(E) \cdot P(F) = \frac{1}{4} \cdot \frac{1}{13} = \frac{1}{52} = P(E \cap F)$$
Since $P(E \cap F) = P(E) \cdot P(F)$, $E$ and $F$ are independent.
(ii) $E$: the card drawn is black
$$P(E) = \frac{26}{52} = \frac{1}{2}$$
$F$: the card drawn is a king
$$P(F) = \frac{4}{52} = \frac{1}{13}$$
$E \cap F$: the card drawn is a black king
$$P(E \cap F) = \frac{2}{52} = \frac{1}{26}$$
$$P(E) \cdot P(F) = \frac{1}{2} \cdot \frac{1}{13} = \frac{1}{26} = P(E \cap F)$$
Hence, $E$ and $F$ are independent.
(iii) $E$: the card drawn is a king or queen
$$P(E) = \frac{8}{52} = \frac{2}{13}$$
$F$: the card drawn is a queen or jack
$$P(F) = \frac{8}{52} = \frac{2}{13}$$
$E \cap F$: the card drawn is a queen
$$P(E \cap F) = \frac{4}{52} = \frac{1}{13}$$
$$P(E) \cdot P(F) = \frac{2}{13} \cdot \frac{2}{13} = \frac{4}{169} \neq \frac{1}{13}$$
Since $$P(E \cap F) \neq P(E) \cdot P(F)$$,
$E$ and $F$ are not independent.
Final Result
$$\boxed{\text{(i) Independent, (ii) Independent, (iii) Not Independent}}$$
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NCERT Question.16 : In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper, and 20% read both Hindi and English newspapers. A student is selected at random. Find:
(a) the probability that the student reads neither Hindi nor English newspapers
(b) the probability that the student reads English newspaper, given that she reads Hindi newspaper
(c) the probability that the student reads Hindi newspaper, given that she reads English newspaper
Solution
Let $H$ = event that a student reads Hindi newspaper, $E$ = event that a student reads English newspaper
Given:
$$P(H) = 0.6, P(E) = 0.4, P(H \cap E) = 0.2$$
(a) Probability that the student reads neither Hindi nor English newspapers:
$$P(H’ \cap E’) = 1 – P(H \cup E)$$
$$P(H \cup E) = P(H) + P(E) – P(H \cap E) = 0.6 + 0.4 – 0.2 = 0.8$$
$$P(H’ \cap E’) = 1 – 0.8 = 0.2$$
(b) Probability that the student reads English newspaper, given that she reads Hindi newspaper:
$$P(E|H) = \frac{P(E \cap H)}{P(H)} = \frac{0.2}{0.6} = \frac{1}{3}$$
(c) Probability that the student reads Hindi newspaper, given that she reads English newspaper:
$$P(H|E) = \frac{P(H \cap E)}{P(E)} = \frac{0.2}{0.4} = \frac{1}{2}$$
Final Result
$$\boxed{(a) 0.2, (b) \frac{1}{3}, (c) \frac{1}{2}}$$
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NCERT Question.17 : The Probability of obtaining an even prime number on each die, when a pair of dice is rolled is:
(A) $0$ (B) $\frac{1}{3}$ (C) $\frac{1}{12}$ (D) $\frac{1}{36}$
Solution
A pair of dice is rolled. The only even prime number is $2$.
Getting an even prime number on each die means the outcome is $(2, 2)$.
The total number of outcomes when two dice are rolled is $6 \times 6 = 36$.
So, the probability:
$$P(\text{both dice show 2}) = \frac{1}{36}$$
Final Result : Hence, the correct option is (D).
$$\boxed{\frac{1}{36}}$$
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NCERT Question.18 : Two events A and B will be independent, if:
(A) A and B are mutually exclusive
(B) $P(A′ \cap B′) = [1 – P(A)] [1 – P(B)]$
(C) $P(A) = P(B)$
(D) $P(A) + P(B) = 1$
Solution
Two events A and B are independent if:
$$P(A \cap B) = P(A) \cdot P(B)$$
Replacing A with $A′$ and B with $B′$, we get:
$$P(A′ \cap B′) = P(A′) \cdot P(B′) = [1 – P(A)] \cdot [1 – P(B)]$$
Hence, the correct option is (B).
Final Result
$$\boxed{P(A′ \cap B′) = [1 – P(A)] [1 – P(B)]}$$
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Summary
Exercise 13.2 focuses on conditional probability and its applications. It covers the concept of dependent and independent events, the multiplication rule of probability for dependent events, and how to calculate conditional probabilities. Students learn to solve problems involving real-life scenarios where the occurrence of one event affects the probability of another event.
