Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Probability

Question:

Let A and B are two independent events such that $P(A) =\frac{3}{5}$ and $P(B)=\frac{4}{9}$.

Match List-I with List-II

List-I

List-II

(A) $P(A∩B)$

(I) $\frac{2}{5}$

(B) $P(A|B)$

(II) $\frac{4}{15}$

(C) $P(A'|B)$

(III) $\frac{3}{5}$

(D) $P(A'∩B')$

(IV) $\frac{2}{9}$

Choose the correct answer from the options given below:

Options:

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

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

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

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

Correct Answer:

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

Explanation:

The correct answer is Option (2) → (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

(A) $P(A∩B)=P(A)×P(B)$  [A and B are independent]

$=\frac{3}{5}×\frac{4}{9}=\frac{12}{45}=\frac{4}{15}$

(B) $P(A|B)=\frac{P(A∩B)}{P(B)}=\frac{\frac{4}{15}}{\frac{4}{9}}=\frac{3}{5}$

(C) $P(A'|B)=\frac{P(A'∩B)}{P(B)}=\frac{\frac{2}{5}×\frac{4}{9}}{\frac{4}{9}}=\frac{2}{5}$

(D) $P(A'∩B')=P(A')P(B')$

$=\frac{2}{5}×\frac{5}{9}=\frac{2}{9}$  [A' and B' are also independent events]