Get detailed solutions for NCERT Class 12 Maths Exercise 13.2 (Probability). Learn to identify and solve problems based on Independent Events using the formula $P(A \cap B) = P(A) \cdot P(B)$. Covers finding probabilities for card draws (with and without replacement), system reliability, and checking for independence using the Multiplication Rule. Essential for Chapter 13.

This exercise focuses on the concept of Independent Events, where the occurrence of one event does not affect the probability of the other. The key formula for independent events $A$ and $B$ is:

Ex. 13.1

Ex. 13.2

Ex. 13.3

Ch 13 Misc.

Rbse Solutions Class 12 Maths Exercise 13.2 | Independent Events and Multiplication Rule

$$P(A \cap B) = P(A) \cdot P(B)$$

1. If $P(A) = 3/5$ and $P(B) = 1/5$, find $P(A \cap B)$ if $A$ and $B$ are independent events.

Since $A$ and $B$ are independent:

$$P(A \cap B) = P(A) \cdot P(B) = \frac{3}{5} \times \frac{1}{5} = \mathbf{3/25}$$

2. Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Let $A$ be the event that the first card is black, and $B$ be the event that the second card is black.

Total cards = 52. Black cards = 26.

$P(A)$: Probability the first card is black.$$P(A) = 26/52 = 1/2$$
$P(B|A)$: Probability the second card is black given the first was black (without replacement).After the first draw, there are 51 cards left, with 25 black cards.$$P(B|A) = 25/51$$
$P(A \cap B)$: Probability both are black.$$P(A \cap B) = P(A) \cdot P(B|A) = \frac{26}{52} \times \frac{25}{51} = \frac{1}{2} \times \frac{25}{51} = \mathbf{25/102}$$

3. A box contains 15 oranges (12 good, 3 bad). Three oranges are drawn without replacement. The box is approved if all three are good. Find the probability of approval.

Let $G_1, G_2, G_3$ be the events that the first, second, and third oranges drawn are good, respectively. We need $P(G_1 \cap G_2 \cap G_3)$.

$P(G_1)$: Probability the first is good.$$P(G_1) = 12/15$$
$P(G_2|G_1)$: Probability the second is good given the first was good.(11 good remaining out of 14 total).$$P(G_2|G_1) = 11/14$$
$P(G_3|G_1 \cap G_2)$: Probability the third is good given the first two were good.(10 good remaining out of 13 total).$$P(G_3|G_1 \cap G_2) = 10/13$$

$$P(\text{Approval}) = P(G_1) \cdot P(G_2|G_1) \cdot P(G_3|G_1 \cap G_2)$$

$$P(\text{Approval}) = \frac{12}{15} \times \frac{11}{14} \times \frac{10}{13} = \frac{4}{5} \times \frac{11}{14} \times \frac{10}{13}$$

$$P(\text{Approval}) = \frac{4 \times 11 \times 2}{14 \times 13} = \frac{88}{182} = \mathbf{44/91}$$

4. A fair coin ($S_C = \{H, T\}$) and an unbiased die ($S_D = \{1, 2, 3, 4, 5, 6\}$) are tossed. $S = S_C \times S_D$, $|S|=12$.

$A$: ‘head appears on the coin’ ($A = \{H1, H2, \dots, H6\}$)

$B$: ‘3 on the die’ ($B = \{H3, T3\}$)

Check whether $A$ and $B$ are independent.

$P(A) = 6/12 = 1/2$
$P(B) = 2/12 = 1/6$
$A \cap B$: ‘Head appears AND 3 on the die’ $\implies A \cap B = \{H3\}$
$P(A \cap B) = 1/12$

Check the condition $P(A \cap B) = P(A) \cdot P(B)$:

$$P(A) \cdot P(B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$$

Since $P(A \cap B) = 1/12$, $A$ and $B$ are independent events.

5. A die marked 1, 2, 3 in red ($R$) and 4, 5, 6 in green ($G$) is tossed.

$S = \{1, 2, 3, 4, 5, 6\}$.

$A$: ‘the number is even’ ($A = \{2, 4, 6\}$)

$B$: ‘the number is red’ ($B = \{1, 2, 3\}$)

Are $A$ and $B$ independent?

$P(A) = 3/6 = 1/2$
$P(B) = 3/6 = 1/2$
$A \cap B$: ‘Even AND Red’ $\implies A \cap B = \{2\}$
$P(A \cap B) = 1/6$

Check the condition $P(A \cap B) = P(A) \cdot P(B)$:

$$P(A) \cdot P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$

Since $P(A \cap B) = 1/6 \ne 1/4$, $A$ and $B$ are not independent.

6. Let $E$ and $F$ be events with $P(E) = 3/5$, $P(F) = 3/10$ and $P(E \cap F) = 1/5$. Are $E$ and $F$ independent?

Check the condition $P(E \cap F) = P(E) \cdot P(F)$:

$$P(E) \cdot P(F) = \frac{3}{5} \times \frac{3}{10} = \frac{9}{50}$$

Given $P(E \cap F) = 1/5 = 10/50$.

Since $P(E \cap F) = 10/50 \ne 9/50$, $E$ and $F$ are not independent.

7. Given $P(A) = 1/2$, $P(A \cup B) = 3/5$ and $P(B) = p$. Find $p$ if they are (i) mutually exclusive (ii) independent.

General Formula: $P(A \cup B) = P(A) + P(B) – P(A \cap B)$

$$\frac{3}{5} = \frac{1}{2} + p – P(A \cap B)$$

(i) Mutually Exclusive: $A \cap B = \emptyset \implies P(A \cap B) = 0$.

$$\frac{3}{5} = \frac{1}{2} + p – 0$$

$$p = \frac{3}{5} – \frac{1}{2} = \frac{6 – 5}{10} = \mathbf{1/10}$$

(ii) Independent: $P(A \cap B) = P(A) \cdot P(B) = \frac{1}{2} p$.

$$\frac{3}{5} = \frac{1}{2} + p – \frac{1}{2} p$$

$$\frac{3}{5} = \frac{1}{2} + \frac{1}{2} p$$

$$\frac{1}{2} p = \frac{3}{5} – \frac{1}{2} = \frac{6 – 5}{10} = \frac{1}{10}$$

$$p = 2 \times \frac{1}{10} = \mathbf{1/5}$$

8. Let $A$ and $B$ be independent events with $P(A) = 0.3$ and $P(B) = 0.4$. Find:

(i) $P(A \cap B)$

Since $A$ and $B$ are independent:

$$P(A \cap B) = P(A) \cdot P(B) = 0.3 \times 0.4 = \mathbf{0.12}$$

(ii) $P(A \cup B)$

$$P(A \cup B) = P(A) + P(B) – P(A \cap B) = 0.3 + 0.4 – 0.12 = 0.7 – 0.12 = \mathbf{0.58}$$

(iii) $P(A|B)$

Since $A$ and $B$ are independent, $P(A|B) = P(A)$.

$$P(A|B) = \mathbf{0.3}$$

(Alternatively: $P(A|B) = P(A \cap B) / P(B) = 0.12 / 0.4 = 0.3$)

(iv) $P(B|A)$

Since $A$ and $B$ are independent, $P(B|A) = P(B)$.

$$P(B|A) = \mathbf{0.4}$$

9. If $P(A) = 1/4$, $P(B) = 1/2$ and $P(A \cap B) = 1/8$, find $P(\text{not } A \text{ and not } B)$, which is $P(A’ \cap B’)$.

Using De Morgan’s Law: $P(A’ \cap B’) = P((A \cup B)’)$

$$P(A’ \cap B’) = 1 – P(A \cup B)$$

Find $P(A \cup B)$:$$P(A \cup B) = P(A) + P(B) – P(A \cap B) = \frac{1}{4} + \frac{1}{2} – \frac{1}{8} = \frac{2}{8} + \frac{4}{8} – \frac{1}{8} = \frac{5}{8}$$
Find $P(A’ \cap B’)$:$$P(A’ \cap B’) = 1 – \frac{5}{8} = \mathbf{3/8}$$

(Note: Since $P(A) \cdot P(B) = (1/4)(1/2) = 1/8 = P(A \cap B)$, $A$ and $B$ are independent. If $A$ and $B$ are independent, then $A’$ and $B’$ are also independent. $P(A’ \cap B’) = P(A’) P(B’) = (1 – 1/4)(1 – 1/2) = (3/4)(1/2) = 3/8$.)

10. $P(A) = 1/2$, $P(B) = 7/12$ and $P(\text{not } A \text{ or not } B) = 1/4$. State whether $A$ and $B$ are independent?

$P(\text{not } A \text{ or not } B) = P(A’ \cup B’) = 1/4$.

Using De Morgan’s Law: $P(A’ \cup B’) = P((A \cap B)’)$

$$P((A \cap B)’) = 1/4$$

$$1 – P(A \cap B) = 1/4$$

$$P(A \cap B) = 1 – 1/4 = 3/4$$

Check the condition for independence $P(A \cap B) = P(A) \cdot P(B)$:

$$P(A) \cdot P(B) = \frac{1}{2} \times \frac{7}{12} = \frac{7}{24}$$

Since $P(A \cap B) = 3/4 = 18/24 \ne 7/24$, $A$ and $B$ are not independent.

11. Given $A$ and $B$ are independent events, $P(A) = 0.3$, $P(B) = 0.6$. Find:

(i) $P(A \text{ and } B) = P(A \cap B)$

$$P(A \cap B) = P(A) \cdot P(B) = 0.3 \times 0.6 = \mathbf{0.18}$$

(ii) $P(A \text{ and not } B) = P(A \cap B’)$

Since $A$ and $B$ are independent, $A$ and $B’$ are independent.

$$P(A \cap B’) = P(A) \cdot P(B’) = P(A) \cdot (1 – P(B))$$

$$P(A \cap B’) = 0.3 \times (1 – 0.6) = 0.3 \times 0.4 = \mathbf{0.12}$$

(iii) $P(A \text{ or } B) = P(A \cup B)$

$$P(A \cup B) = P(A) + P(B) – P(A \cap B) = 0.3 + 0.6 – 0.18 = 0.9 – 0.18 = \mathbf{0.72}$$

(Alternatively: $P(A \cup B) = 1 – P(A’ \cap B’) = 1 – P(A’)P(B’) = 1 – (0.7)(0.4) = 1 – 0.28 = 0.72$)

(iv) $P(\text{neither } A \text{ nor } B) = P(A’ \cap B’)$

Since $A$ and $B$ are independent, $A’$ and $B’$ are independent.

$$P(A’ \cap B’) = P(A’) \cdot P(B’) = (1 – P(A)) \cdot (1 – P(B))$$

$$P(A’ \cap B’) = (1 – 0.3) \cdot (1 – 0.6) = 0.7 \times 0.4 = \mathbf{0.28}$$

12. A die is tossed thrice. Find the probability of getting an odd number at least once.

Let $E$ be the event of getting an odd number (1, 3, 5). $P(E) = 3/6 = 1/2$.

Let $E’$ be the event of getting an even number (2, 4, 6). $P(E’) = 3/6 = 1/2$.

Let $A$ be the event of getting an odd number at least once in three tosses.

It is easier to find $P(A’)$: the probability of NO odd numbers in three tosses, meaning all three tosses are even.

Since the tosses are independent:

$$P(A’) = P(\text{Even on 1st}) \cdot P(\text{Even on 2nd}) \cdot P(\text{Even on 3rd})$$

$$P(A’) = P(E’) \cdot P(E’) \cdot P(E’) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$$

$$P(A) = 1 – P(A’) = 1 – 1/8 = \mathbf{7/8}$$

13. Box contains 10 black (B) and 8 red (R) balls. Two balls are drawn with replacement.

Total balls = 18. $P(B) = 10/18 = 5/9$, $P(R) = 8/18 = 4/9$.

Since the drawing is with replacement, the events are independent.

(i) Probability that both balls are red.

$$P(R_1 \cap R_2) = P(R_1) \cdot P(R_2) = \frac{4}{9} \times \frac{4}{9} = \mathbf{16/81}$$

(ii) Probability that first ball is black and second is red.

$$P(B_1 \cap R_2) = P(B_1) \cdot P(R_2) = \frac{5}{9} \times \frac{4}{9} = \mathbf{20/81}$$

(iii) Probability that one of them is black and the other is red.

This means (Black, Red) OR (Red, Black).

$$P(B_1 R_2 \cup R_1 B_2) = P(B_1 R_2) + P(R_1 B_2)$$

$$P(B_1 R_2) = 20/81 \quad (\text{from part ii})$$

$$P(R_1 B_2) = P(R_1) P(B_2) = \frac{4}{9} \times \frac{5}{9} = 20/81$$

$$P(\text{one B, one R}) = 20/81 + 20/81 = \mathbf{40/81}$$

14. $P(A \text{ solves}) = P(A) = 1/2$, $P(B \text{ solves}) = P(B) = 1/3$. $A$ and $B$ are independent.

(i) Probability that the problem is solved.

The problem is solved if $A$ solves OR $B$ solves OR both solve: $P(A \cup B)$.

$$P(A \cup B) = P(A) + P(B) – P(A \cap B)$$

Since independent, $P(A \cap B) = P(A) P(B) = (1/2)(1/3) = 1/6$.

$$P(\text{Solved}) = \frac{1}{2} + \frac{1}{3} – \frac{1}{6} = \frac{3}{6} + \frac{2}{6} – \frac{1}{6} = \frac{4}{6} = \mathbf{2/3}$$

(Alternatively: $P(\text{Solved}) = 1 – P(\text{neither solve}) = 1 – P(A’ \cap B’) = 1 – P(A’) P(B’) = 1 – (1/2)(2/3) = 1 – 1/3 = 2/3$.)

(ii) Probability that exactly one of them solves the problem.

This means ($A$ solves AND $B$ doesn’t) OR ($B$ solves AND $A$ doesn’t).

$$P(\text{Exactly one}) = P(A \cap B’) + P(A’ \cap B)$$

Since independent:

$$P(\text{Exactly one}) = P(A) P(B’) + P(A’) P(B)$$

$$P(\text{Exactly one}) = \left(\frac{1}{2}\right)\left(1 – \frac{1}{3}\right) + \left(1 – \frac{1}{2}\right)\left(\frac{1}{3}\right)$$

$$P(\text{Exactly one}) = \left(\frac{1}{2}\right)\left(\frac{2}{3}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{3}\right) = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \mathbf{1/2}$$

15. One card is drawn. Check independence of $E$ and $F$. $P(A \cap B) = P(A)P(B)$.

Total cards = 52.

(i) $E$: ‘the card drawn is a spade’ ($13$ spades), $F$: ‘the card drawn is an ace’ ($4$ aces)

$P(E) = 13/52 = 1/4$
$P(F) = 4/52 = 1/13$
$E \cap F$: The card is the ‘Ace of Spades’ (1 card).
$P(E \cap F) = 1/52$
Check: $P(E) P(F) = (1/4)(1/13) = 1/52$.Since $P(E \cap F) = P(E) P(F)$, $E$ and $F$ are independent.

(ii) $E$: ‘the card drawn is black’ ($26$ cards), $F$: ‘the card drawn is a king’ ($4$ kings)

$P(E) = 26/52 = 1/2$
$P(F) = 4/52 = 1/13$
$E \cap F$: The card is a ‘Black King’ (King of Spades, King of Clubs: 2 cards).
$P(E \cap F) = 2/52 = 1/26$
Check: $P(E) P(F) = (1/2)(1/13) = 1/26$.Since $P(E \cap F) = P(E) P(F)$, $E$ and $F$ are independent.

(iii) $E$: ‘the card drawn is a king or queen’ ($4+4=8$ cards), $F$: ‘the card drawn is a queen or jack’ ($4+4=8$ cards)

$P(E) = 8/52 = 2/13$
$P(F) = 8/52 = 2/13$
$E \cap F$: The card is a ‘Queen’ (4 cards).
$P(E \cap F) = 4/52 = 1/13$
Check: $P(E) P(F) = (2/13)(2/13) = 4/169$Since $P(E \cap F) = 1/13 \ne 4/169$, $E$ and $F$ are not independent.

16. Hostel student newspaper data.

Let $H$ be Hindi readers, $E$ be English readers.

$P(H) = 0.60$, $P(E) = 0.40$, $P(H \cap E) = 0.20$.

(a) Find the probability that she reads neither Hindi nor English newspapers, $P(H’ \cap E’)$

$P(H’ \cap E’) = P((H \cup E)’)$

$$P(H \cup E) = P(H) + P(E) – P(H \cap E) = 0.60 + 0.40 – 0.20 = 0.80$$

$$P(H’ \cap E’) = 1 – P(H \cup E) = 1 – 0.80 = \mathbf{0.20}$$

(b) If she reads Hindi newspaper, find the probability that she reads English newspaper, $P(E|H)$

$$P(E|H) = \frac{P(E \cap H)}{P(H)} = \frac{0.20}{0.60} = \mathbf{1/3}$$

$$P(H|E) = \frac{P(H \cap E)}{P(E)} = \frac{0.20}{0.40} = \mathbf{1/2}$$

Multiple Choice Questions

17. The probability of obtaining an even prime number on each die, when a pair of dice is rolled is

The prime numbers are $\{2, 3, 5\}$.
The even prime number is $\{2\}$.
Let $E_1$ be the event that the first die shows 2, and $E_2$ be the event that the second die shows 2.
$P(E_1) = 1/6$, $P(E_2) = 1/6$. Since the rolls are independent:$$P(E_1 \cap E_2) = P(E_1) P(E_2) = \frac{1}{6} \times \frac{1}{6} = \mathbf{1/36}$$

Answer: (D) $1/36$

18. Two events $A$ and $B$ will be independent, if

(A) Mutually exclusive means $P(A \cap B) = 0$. This implies independence only if $P(A)=0$ or $P(B)=0$. False in general.

(B) $P(A’ \cap B’) = P((A \cup B)’)$ by De Morgan’s Law. If $A$ and $B$ are independent, then $A’$ and $B’$ are independent, so $P(A’ \cap B’) = P(A’) P(B’) = (1 – P(A))(1 – P(B))$. This is the definition of independence for complements.

(D) $P(A)+P(B)=1$ does not imply $P(A \cap B) = P(A)P(B)$.

Answer: (B) $P(A’ \cap B’) = [1 – P(A)][1 – P(B)]$