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)
For independent events, \(P(A \cap B) = P(A)P(B)\), and A and B′ (as well as A′ and B′) are also independent. Here
\(P(A') = 0.7,\quad P(B') = 0.4\)
(A) P(A and B)
This matches (II) 0.18:
\(P(A \cap B) = 0.3 \times 0.6 = 0.18\)
(B) P(A and not B)
This matches (III) 0.12:
\(P(A \cap B') = P(A)P(B') = 0.3 \times 0.4 = 0.12\)
(C) P(A or B)
This matches (IV) 0.72, by the addition rule:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(= 0.3 + 0.6 - 0.18 = 0.72\)
(D) P(neither A nor B)
This matches (I) 0.28:
\(P(A' \cap B') = 1 - P(A \cup B) = 1 - 0.72 = 0.28\)
As a check, \(P(A')P(B') = 0.7 \times 0.4 = 0.28\).
So the matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).
NCERT (2026-27 edition):
Independent Events
