Algebra

Probability

Easy

core

Match List-I with List-II

Let $A$ and $B$ be two events such that $P(A) = 0.2$ , $P(B) = 0.4$ , $P(B|A) = 0.5$

List-I List-II
(A) $P(A \cap B)$ (I) $0.5$
(B) $P(A\vert B)$ (II) $0.8$
(C) $P(A \cup B)$ (III) $0.25$
(D) $P(A')$ (IV) $0.1$

Choose the correct answer from the options given below:

List-I	List-II
(A) $P(A \cap B)$	(I) $0.5$
(B) $P(A\vert B)$	(II) $0.8$
(C) $P(A \cup B)$	(III) $0.25$
(D) $P(A')$	(IV) $0.1$

(A) - (IV), (B) - (III), (C) - (I), (D) - (II)

(A) - (III), (B) - (IV), (C) - (I), (D) - (II)

(A) - (I), (B) - (II), (C) - (IV), (D) - (III)

(A) - (II), (B) - (IV), (C) - (III), (D) - (I)

✅ Correct Option: 1

Given $P(A) = 0.2$ , $P(B) = 0.4$ , $P(B|A) = 0.5$

Using the conditional probability formula:

$P(B|A) = \dfrac{P(A \cap B)}{P(A)}$

$P(A \cap B) = P(B|A) \times P(A)$

$= 0.5 \times 0.2$

$= 0.1$

$\Rightarrow$ Matches (IV)

$P(A|B) = \dfrac{P(A \cap B)}{P(B)}$

$= \dfrac{0.1}{0.4}$

$= 0.25$

$\Rightarrow$ Matches (III)

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

$= 0.2 + 0.4 - 0.1$

$= 0.5$

$\Rightarrow$ Matches (I)

$P(A') = 1 - P(A)$

$= 1 - 0.2$

$= 0.8$

$\Rightarrow$ Matches (II)

List-I	Value	List-II
(A) $P(A \cap B)$	$0.1$	(IV)
(B) $P(A	B)$	$0.25$
(C) $P(A \cup B)$	$0.5$	(I)
(D) $P(A')$	$0.8$	(II)

Correct Answer: (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

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