Anand Classes provides well-structured and exam-oriented NCERT Solutions for Probability Exercise 13.2 of Class 12 Chapter 13, helping students strengthen concepts and score higher in board exams. These detailed solutions break down every question using simple logic, step-by-step explanations, and CBSE-focused methods to ensure complete clarity. Students can easily revise important formulas, solved examples, and practice questions with these high-quality Class 12 Probability notes. Click the print button to download study material and notes.
Access NCERT Solutions for Probability Exercise 13.2 of Class 12 Chapter 13
NCERT Question.1 : If $P(A)=\dfrac{3}{5}$ and $P(B)=\dfrac{1}{5}$, find $P(A \cap B)$ if A and B are independent events.
Solution
Since A and B are independent events,
$$P(A \cap B)=P(A)P(B)$$
Given
$$P(A)=\frac{3}{5}$$
$$P(B)=\frac{1}{5}$$
Substituting,
$$P(A \cap B)=\frac{3}{5}\times\frac{1}{5}=\frac{3}{25}$$
Final Result
$$\boxed{\frac{3}{25}}$$
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NCERT Question.2 : Two cards are drawn at random and without replacement from a pack of $52$ playing cards. Find the probability that the cards are black.
Solution
We know that there are 26 black cards in a pack of 52 cards. Since the first card drawn is not replaced (given: without replacement), there are 25 (= 26 – 1) black cards in a pack of 51 (= 52 – 1) cards.
Probability that both cards drawn are black:
Probability both cards drawn are black = Probability first card drawn is black x Probability second black if first is black
$$P(\text{both black}) = P(\text{1st black}) P(\text{2nd black }| \text{1st black})$$
Thus,
$$P(\text{both black}) = \frac{26}{52} \times \frac{25}{51}$$
Simplifying,
$$P(\text{both black}) = \frac{25}{102}$$
Final Result
$$\boxed{\frac{25}{102}}$$
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NCERT Question.3 : A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale otherwise it is rejected. Find the probability that a box containing $15$ oranges out of which $12$ are good and $3$ are bad will be approved for sale.
Solution
Total oranges in the box: $15$
Good oranges: $12$
Bad oranges: $3$
The box will be approved only if all three selected oranges are good.
Probability that the first orange selected is good:
$$\frac{12}{15}$$
After removing one good orange, good oranges left $= 11$ and total oranges left $= 14$.
Probability that the second selected orange is good:
$$\frac{11}{14}$$
After removing another good orange, good oranges left $= 10$ and total oranges left $= 13$.
Probability that the third selected orange is good:
$$\frac{10}{13}$$
Therefore, probability that all three selected oranges are good:
$$P(\text{approved}) = \frac{12}{15} \times \frac{11}{14} \times \frac{10}{13}$$
Simplifying:
$$P(\text{approved}) = \frac{44}{91}$$
Final Result
$$\boxed{\frac{44}{91}}$$
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NCERT Question.4 : A fair coin and an unbiased die are tossed. Let $A$ be the event “head appears on the coin” and $B$ be the event “3 on the die”. Check whether $A$ and $B$ are independent events or not.
Solution
A fair coin and an unbiased die are tossed.
Sample space:
$$S = \{(H1); (H2); (H3); (H4); (H5); (H6); (T1); (T2); (T3); (T4); (T5); (T6)\}$$
Event $A$: head appears on the coin
$$A = \{(H1); (H2); (H3); (H4); (H5); (H6)\}$$
$$P(A) = \frac{6}{12} = \frac{1}{2}$$
Event $B$: outcome on die is $3$
$$B = \{(H3); (T3)\}$$
$$P(B) = \frac{2}{12} = \frac{1}{6}$$
Intersection of the events:
$$A \cap B = \{(H3)\}$$
$$P(A \cap B) = \frac{1}{12}$$
Now check independence:
$$P(A) \times P(B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$$
Since
$$P(A \cap B) = P(A) \times P(B)$$
the events $A$ and $B$ are independent.
Final Result
$$\boxed{\text{A and B are independent events}}$$
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NCERT Question.5 : A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let $A$ be the event “number is even” and $B$ be the event “number is red”. Are $A$ and $B$ independent?
Solution
Sample space:
$$S = \{1; 2; 3; 4; 5; 6\}$$
Event $A$: number is even
$$A = \{2; 4; 6\}$$
$$P(A) = \frac{3}{6} = \frac{1}{2}$$
Event $B$: number is red
$$B = \{1; 2; 3\}$$
$$P(B) = \frac{3}{6} = \frac{1}{2}$$
Intersection:
$$(A \cap B) = {1}$$
$$P(A \cap B) = \frac{1}{6}$$
Check independence:
$$P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$
Since
$$P(A \cap B) \neq P(A) \times P(B)$$
the events $A$ and $B$ are not independent.
Final Result
$$\boxed{\text{A and B are not independent}}$$
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NCERT Question.6 : Let $E$ and $F$ be events with $P(E)=\frac{3}{5}$, $P(F)=\frac{3}{10}$ and $P(E \cap F)=\frac{1}{5}$. Are $E$ and $F$ independent?
Solution
Given:
$$P(E)=\frac{3}{5}$$
$$P(F)=\frac{3}{10}$$
$$P(E \cap F)=\frac{1}{5}$$
Check independence condition:
$$P(E) \times P(F)=\frac{3}{5} \times \frac{3}{10}=\frac{9}{50}$$
Since
$$P(E \cap F)=\frac{1}{5}=\frac{10}{50}$$
and
$$\frac{10}{50} \neq \frac{9}{50}$$
we conclude that the events $E$ and $F$ are not independent.
Final Result
$$\boxed{\text{E and F are not independent}}$$
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NCERT Question.7 : Given that the events $A$ and $B$ are such that
$$P(A)=\frac{1}{2}, \quad P(A \cup B)=\frac{3}{5}, \quad P(B)=p$$
Find $p$ if they are
(i) mutually exclusive,
(ii) independent.
Solution
(i) When A and B are mutually exclusive
Mutual exclusiveness means:
$$P(A \cap B)=0$$
Use the formula:
$$P(A \cup B)=P(A)+P(B)-P(A \cap B)$$
Substituting values:
$$\frac{3}{5}=\frac{1}{2}+p-0$$
So,
$$p=\frac{3}{5}-\frac{1}{2}$$
Compute:
$$p=\frac{6-5}{10}=\frac{1}{10}$$
Thus,
$$\boxed{p=\frac{1}{10}}$$
(ii) When A and B are independent
Independence means:
$$P(A \cap B)=P(A)P(B)=\frac{1}{2}p$$
Again use:
$$P(A \cup B)=P(A)+P(B)-P(A \cap B)$$
Substitute values:
$$\frac{3}{5}=\frac{1}{2}+p-\frac{1}{2}p$$
Combine like terms:
$$\frac{3}{5}=\frac{1}{2}+p\left(1-\frac{1}{2}\right)$$
$$\frac{3}{5}=\frac{1}{2}+\frac{p}{2}$$
Now solve for $p$:
$$\frac{3}{5}-\frac{1}{2}=\frac{p}{2}$$
$$\frac{6-5}{10}=\frac{p}{2}$$
$$\frac{1}{10}=\frac{p}{2}$$
$$p=\frac{2}{10}=\frac{1}{5}$$
Thus,
$$\boxed{p=\frac{1}{5}}$$
Final Result
$$
\boxed{
\begin{aligned}
\text{(i) If mutually exclusive: } & p=\frac{1}{10} \
\text{(ii) If independent: } & p=\frac{1}{5}
\end{aligned}}
$$
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NCERT Question.8 : Let $A$ and $B$ be independent events with
$$P(A)=0.3 \quad \text{and} \quad P(B)=0.4$$
Find:
(i) $P(A \cap B)$
(ii) $P(A \cup B)$
(iii) $P(A \mid B)$
(iv) $P(B \mid A)$
Solution
Since $A$ and $B$ are independent:
(i) Find $P(A \cap B)$
For independent events:
$$P(A \cap B)=P(A)P(B)$$
Thus,
$$P(A \cap B)=0.3 \times 0.4=0.12$$
(ii) Find $P(A \cup B)$
Use:
$$P(A \cup B)=P(A)+P(B)-P(A \cap B)$$
Substitute values:
$$P(A \cup B)=0.3+0.4-0.12$$
$$P(A \cup B)=0.58$$
(iii) Find $P(A \mid B)$
For independent events:
$$P(A \mid B)=P(A)$$
Thus,
$$P(A \mid B)=0.3$$
(iv) Find $P(B \mid A)$
For independent events:
$$P(B \mid A)=P(B)$$
Thus,
$$P(B \mid A)=0.4$$
Final Result
$$
\boxed{
\text{(i) } P(A \cap B)=0.12 \
\text{(ii) } P(A \cup B)=0.58 \\[1em]
\text{(iii) } P(A \mid B)=0.3 \
\text{(iv) } P(B \mid A)=0.4
}
$$
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NCERT Question.9 : If
$$P(A)=\frac{1}{4}, \quad P(B)=\frac{1}{2}, \quad P(A \cap B)=\frac{1}{8}$$
find
$$P(\text{not }A \text{ and not }B).$$
Solution
We need to find:
$$P(\text{not }A \text{ and not }B)=P(A’ \cap B’).$$
Using the complement rule:
$$P(A’ \cap B’) = P((A \cup B)’) = 1 – P(A \cup B).$$
Now compute $P(A \cup B)$:
$$
P(A \cup B)=P(A)+P(B)-P(A \cap B)
$$
Substitute values:
$$
P(A \cup B)=\frac{1}{4}+\frac{1}{2}-\frac{1}{8}
$$
Thus,
$$
P(A \cup B)=\frac{2}{8}+\frac{4}{8}-\frac{1}{8}=\frac{5}{8}
$$
Therefore,
$$
P(A’ \cap B’)=1-\frac{5}{8}=\frac{3}{8}
$$
Final Result
$$
\boxed{\frac{3}{8}}
$$
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