Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Probability

Question:

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) $P(A ∩ B)$

(I) 0.5

(B) $P(A|B)$

(II) 0.8

(C) $P(A ∪ B)$

(III) 0.25

(D) $P(A')$

(IV) 0.1

Choose the correct answer from the options given below:

Options:

(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 Answer:

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

Explanation:

The correct answer is Option (1) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

List-I

List-II

(A) $P(A ∩ B)$

(IV) 0.1

(B) $P(A|B)$

(III) 0.25

(C) $P(A ∪ B)$

(I) 0.5

(D) $P(A')$

(II) 0.8

Given:

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

(A) $P(A \cap B) = P(A) \cdot P(B|A) = 0.2 \cdot 0.5 = 0.1$ ⇒ Match with (IV)

(B) $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.1}{0.4} = 0.25$ ⇒ Match with (III)

(C) $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.2 + 0.4 - 0.1 = 0.5$ ⇒ Match with (I)

(D) $P(A') = 1 - P(A) = 1 - 0.2 = 0.8$ ⇒ Match with (II)