Given P(A)=0.2, P(B)=0.4, P(B∣A)=0.5
Using the conditional probability formula:
P(B∣A)=P(A)P(A∩B)
P(A∩B)=P(B∣A)×P(A)
=0.5×0.2
=0.1
⇒ Matches (IV)
P(A∣B)=P(B)P(A∩B)
=0.40.1
=0.25
⇒ Matches (III)
P(A∪B)=P(A)+P(B)−P(A∩B)
=0.2+0.4−0.1
=0.5
⇒ Matches (I)
P(A′)=1−P(A)
=1−0.2
=0.8
⇒ Matches (II)
| List-I |
Value |
List-II |
| (A) P(A∩B) |
0.1 |
(IV) |
| (B) P(A∣B) |
0.25 |
(III) |
| (C) P(A∪B) |
0.5 |
(I) |
| (D) P(A′) |
0.8 |
(II) |
Correct Answer: (A)-(IV), (B)-(III), (C)-(I), (D)-(II)