Exercise 13.2 Class 12 Maths Probability

NCERT Solutions Class 12 Maths Chapter 13 Ex 13.2 | LearnCBSEHub

💡 Multiplication Theorem & Independence

Independent Events: Two events A and B are independent if and only if:

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

Note: “Independent” is different from “Mutually Exclusive”. Mutually Exclusive means $P(A \cap B) = 0$.

Question 01

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

For independent events:
$P(A \cap B) = P(A) \cdot P(B) = \frac{3}{5} \times \frac{1}{5} = \frac{3}{25}$.

Answer: 3/25

Question 02

Two cards drawn without replacement. Find prob. both are black.

Total cards = 52. Black cards = 26.
$P(\text{Both Black}) = P(\text{1st Black}) \times P(\text{2nd Black | 1st Black})$
$= \frac{26}{52} \times \frac{25}{51} = \frac{1}{2} \times \frac{25}{51} = \frac{25}{102}$.

Answer: 25/102

Question 03

Box of 15 oranges (12 Good, 3 Bad). 3 drawn without replacement. Find prob. box is approved (all 3 Good).

$P(\text{All 3 Good}) = P(G_1) \times P(G_2|G_1) \times P(G_3|G_1 G_2)$
$= \frac{12}{15} \times \frac{11}{14} \times \frac{10}{13}$
$= \frac{4}{5} \times \frac{11}{14} \times \frac{10}{13} = \frac{44}{91}$.

Answer: 44/91

Question 04

Coin and Die tossed. A: Head, B: 3 on die. Check independence.

Sample Space size = $2 \times 6 = 12$.
$A = \{H1, H2, H3, H4, H5, H6\} \implies P(A) = 6/12 = 1/2$.
$B = \{H3, T3\} \implies P(B) = 2/12 = 1/6$.
$A \cap B = \{H3\} \implies P(A \cap B) = 1/12$.
Check: $P(A) \cdot P(B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$.
Since $P(A \cap B) = P(A) \cdot P(B)$, events are Independent.

Question 05

Die: 1,2,3 (Red); 4,5,6 (Green). A: Even, B: Red. Independent?

Sample Space $S = \{1, 2, 3, 4, 5, 6\}$.
$A (\text{Even}) = \{2, 4, 6\} \implies P(A) = 3/6 = 1/2$.
$B (\text{Red}) = \{1, 2, 3\} \implies P(B) = 3/6 = 1/2$.
$A \cap B (\text{Even AND Red}) = \{2\} \implies P(A \cap B) = 1/6$.
Check: $P(A) \cdot P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
Since $1/6 \neq 1/4$, NOT Independent.

Question 06

Given $P(E)=3/5, P(F)=3/10, P(E \cap F)=1/5$. Independent?

$P(E) \cdot P(F) = \frac{3}{5} \times \frac{3}{10} = \frac{9}{50}$.
$P(E \cap F) = \frac{1}{5} = \frac{10}{50}$.
Since $9/50 \neq 10/50$, NOT Independent.

Question 07

Find $p$ if $P(A)=1/2, P(A \cup B)=3/5, P(B)=p$.

(i) Mutually Exclusive

$P(A \cap B) = 0$.
$P(A \cup B) = P(A) + P(B) \implies \frac{3}{5} = \frac{1}{2} + p$.
$p = \frac{3}{5} – \frac{1}{2} = \frac{6-5}{10} = \frac{1}{10}$.

(ii) Independent

$P(A \cap B) = P(A)P(B) = \frac{1}{2}p$.
$P(A \cup B) = P(A) + P(B) – P(A \cap B)$
$\frac{3}{5} = \frac{1}{2} + p – \frac{p}{2} \implies \frac{3}{5} – \frac{1}{2} = \frac{p}{2} \implies \frac{1}{10} = \frac{p}{2}$.
$p = \frac{2}{10} = \frac{1}{5}$.

Question 08

Independent events A, B. $P(A)=0.3, P(B)=0.4$.

(i) $P(A \cap B) = 0.3 \times 0.4 = 0.12$
(ii) $P(A \cup B) = 0.3 + 0.4 – 0.12 = 0.58$
(iii) $P(A|B) = P(A) = 0.3$ (Since independent)
(iv) $P(B|A) = P(B) = 0.4$

Question 09

Find $P(\text{not A and not B})$.

$P(A’ \cap B’) = P((A \cup B)’) = 1 – P(A \cup B)$.
$P(A \cup B) = P(A) + P(B) – P(A \cap B) = \frac{1}{4} + \frac{1}{2} – \frac{1}{8} = \frac{2+4-1}{8} = \frac{5}{8}$.
Result $= 1 – \frac{5}{8} = \frac{3}{8}$.

Question 10

Check Independence given $P(\text{not A or not B}) = 1/4$.

$P(A’ \cup B’) = 1/4 \implies P((A \cap B)’) = 1/4 \implies 1 – P(A \cap B) = 1/4$.
$P(A \cap B) = 3/4$.
Check: $P(A) \cdot P(B) = \frac{1}{2} \times \frac{7}{12} = \frac{7}{24}$.
$3/4 \neq 7/24$. NOT Independent.

Question 11

Independent A, B. $P(A)=0.3, P(B)=0.6$.

(i) $P(A \cap B) = 0.3 \times 0.6 = 0.18$
(ii) $P(A \cap B’) = P(A) \cdot P(B’) = 0.3 \times (1-0.6) = 0.3 \times 0.4 = 0.12$
(iii) $P(A \cup B) = 0.3 + 0.6 – 0.18 = 0.72$
(iv) $P(A’ \cap B’) = P(A’) \cdot P(B’) = 0.7 \times 0.4 = 0.28$

Question 12

Die tossed thrice. Prob of odd number at least once.

$P(\text{Odd at least once}) = 1 – P(\text{None Odd}) = 1 – P(\text{All Even})$.
P(Even in one toss) $= 3/6 = 1/2$.
$P(\text{All 3 Even}) = (1/2)^3 = 1/8$.
Result $= 1 – 1/8 = 7/8$.

Question 13

Balls drawn with replacement (10 Black, 8 Red).

$P(R) = 8/18 = 4/9$. $P(B) = 10/18 = 5/9$.
(i) Both Red: $P(R) \cdot P(R) = \frac{4}{9} \times \frac{4}{9} = \frac{16}{81}$.
(ii) First Black, Second Red: $P(B) \cdot P(R) = \frac{5}{9} \times \frac{4}{9} = \frac{20}{81}$.
(iii) One Black, Other Red: (BR or RB) $= \frac{20}{81} + \frac{20}{81} = \frac{40}{81}$.

Question 14

Problem solved by A (1/2) and B (1/3) independently.

(i) Problem Solved

$P(\text{Solved}) = 1 – P(\text{None solved}) = 1 – P(A’)P(B’)$.
$P(A’) = 1/2, P(B’) = 2/3$.
$= 1 – (\frac{1}{2} \times \frac{2}{3}) = 1 – \frac{1}{3} = \frac{2}{3}$.

(ii) Exactly One Solves

$P(A)P(B’) + P(A’)P(B)$
$= (\frac{1}{2} \times \frac{2}{3}) + (\frac{1}{2} \times \frac{1}{3}) = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$.

Question 15

Check Independence for card events.

(i) E: Spade, F: Ace

$P(E) = 13/52 = 1/4$. $P(F) = 4/52 = 1/13$.
$E \cap F$ (Ace of Spades) $= 1/52$.
$1/4 \times 1/13 = 1/52$. Independent.

(ii) E: Black, F: King

$P(E) = 26/52 = 1/2$. $P(F) = 4/52 = 1/13$.
$E \cap F$ (Black King) $= 2/52 = 1/26$.
$1/2 \times 1/13 = 1/26$. Independent.

Question 16

Newspaper: 60% Hindi (H), 40% English (E), 20% Both.

(a) Neither Hindi nor English

$P(H \cup E) = P(H) + P(E) – P(H \cap E) = 0.6 + 0.4 – 0.2 = 0.8$.
$P(\text{Neither}) = 1 – 0.8 = 0.2$ (20%).

(b) Reads E given she reads H

$P(E|H) = \frac{P(E \cap H)}{P(H)} = \frac{0.2}{0.6} = \frac{1}{3}$.

$P(H|E) = \frac{P(H \cap E)}{P(E)} = \frac{0.2}{0.4} = \frac{1}{2}$.

Question 17

Prob of even prime on each die (Pair of dice).

Even prime number is 2.
Outcome: $\{(2, 2)\}$. Only 1 outcome out of 36.

Correct Option: (D) 1/36

Question 18

Two events A and B are independent if…

If A and B are independent, then A’ and B’ are also independent.
$P(A’ \cap B’) = P(A’) P(B’) = [1-P(A)][1-P(B)]$.

Correct Option: (B)