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Match List-I with List-II

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

List-IList-II
(A) P(A∩B)P(A \cap B)(I) 0.2
(B) P(A∣B)P(A \mid B)(II) 0.32
(C) P(A∪B)P(A \cup B)(III) 0.64
(D) P(A′)P(A')(IV) 0.98

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 2

Given P(A)=0.8P(A) = 0.8, P(B)=0.5P(B) = 0.5, P(B∣A)=0.4P(B|A) = 0.4


(A) P(A∩B)P(A \cap B)

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

P(A∩B)=P(B∣A)×P(A)P(A \cap B) = P(B|A) \times P(A)

=0.4×0.8= 0.4 \times 0.8

=0.32⟶(II)= 0.32 \longrightarrow \text{(II)}


(B) P(A∣B)P(A|B)

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

=0.320.5= \dfrac{0.32}{0.5}

=0.64⟶(III)= 0.64 \longrightarrow \text{(III)}


(C) P(A∪B)P(A \cup B)

P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

=0.8+0.5−0.32= 0.8 + 0.5 - 0.32

=0.98⟶(IV)= 0.98 \longrightarrow \text{(IV)}


(D) P(A′)P(A')

P(A′)=1−P(A)P(A') = 1 - P(A)

=1−0.8= 1 - 0.8

=0.2⟶(I)= 0.2 \longrightarrow \text{(I)}

Therefore, the correct answer is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).

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