Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Probability

Question:

Match List-I with List-I

Let A and B are two events such that $P(A) = 0.8, P(B) = 0.5, P(B|A)= 0.4$

List-I

List-II

(A) $P (A∩ B)$

(I) 0.2

(B) $P(A|B)$

(II) 0.32

(C) $P (A∪B)$

(III) 0.64

(D) $P (A')$

(IV) 0.98

Choose the correct answer from the options given below:

Options:

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

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

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

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

Correct Answer:

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

Explanation:

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

List-I

List-II

(A) $P (A∩ B)$

(II) 0.32

(B) $P(A|B)$

(III) 0.64

(C) $P (A∪B)$

(IV) 0.98

(D) $P (A')$

(I) 0.2

$P(A)=0.8,\; P(B)=0.5,\; P(B|A)=0.4$

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

$=0.4\times0.8$

$=0.32$

$(A)\rightarrow(II)$

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

$=\frac{0.32}{0.5}$

$=0.64$

$(B)\rightarrow(III)$

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

$=0.8+0.5-0.32$

$=0.98$

$(C)\rightarrow(IV)$

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

$=1-0.8$

$=0.2$

$(D)\rightarrow(I)$

Final Matching: (A)-(II), (B)-(III), (C)-(IV), (D)-(I).