Consider two independent events A and B such that P(A) = 0.3, P(B) = 0.6.
Match List-I with List-II
|
List-I |
List-II |
|
(A) P(A and B) |
(I) 0.28 |
|
(B) P(A and not B) |
(II) 0.18 |
|
(C) P(A or B) |
(III) 0.12 |
|
(D) P(neither A nor B) |
(IV) 0.72 |
Choose the correct answer from the options given below.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I) **
|
List-I |
List-II |
|
(A) P(A and B) |
(II) 0.18 |
|
(B) P(A and not B) |
(III) 0.12 |
|
(C) P(A or B) |
(IV) 0.72 |
|
(D) P(neither A nor B) |
(I) 0.28 |
Given: $P(A)=0.3,\ P(B)=0.6$, and A, B are independent.
(A) $P(A\text{ and }B)=P(A)P(B)=0.3\times0.6=0.18$ → (II)
(B) $P(A\text{ and not }B)=P(A)P(B')=0.3\times(1-0.6)=0.3\times0.4=0.12$ → (III)
(C) $P(A\text{ or }B)=P(A)+P(B)-P(A)P(B)=0.3+0.6-0.18=0.72$ → (IV)
(D) $P(\text{neither A nor B})=1-P(A\text{ or }B)=1-0.72=0.28$ → (I)
Matching:
(A) → (II), (B) → (III), (C) → (IV), (D) → (I)