Let A and B are two independent events such that \(P(A)=\displaystyle\frac{3}{5}\) and \(P(B)=\displaystyle\frac{4}{9}\).
Match List-I with List-II
| List-I | List-II | ||
| (A) | \(P(A\cap B)\) | (I) | \(\displaystyle\frac{2}{5}\) |
| (B) | \(P(A|B)\) | (II) | \(\displaystyle\frac{4}{15}\) |
| (C) | \(P(A'|B)\) | (III) | \(\displaystyle\frac{3}{5}\) |
| (D) | \(P(A'\cap B')\) | (IV) | \(\displaystyle\frac{2}{9}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
For independent events, \(P(A\cap B) = P(A)P(B)\), and the complements A′, B′ are also independent. Here
\( \displaystyle P(A') = 1 - \frac{3}{5} = \frac{2}{5},\quad P(B') = 1 - \frac{4}{9} = \frac{5}{9} \)
(A) \(P(A\cap B)\)
This matches (II) \(\frac{4}{15}\).
\( \displaystyle P(A\cap B) = \frac{3}{5}\times\frac{4}{9} = \frac{12}{45} = \frac{4}{15} \)
(B) \(P(A|B)\)
This matches (III) \(\frac{3}{5}\).
\( \displaystyle P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{4/15}{4/9} = \frac{3}{5} \)
(For independent events this is just P(A).)
(C) \(P(A'|B)\)
This matches (I) \(\frac{2}{5}\).
\( \displaystyle P(A'|B) = 1 - P(A|B) = 1 - \frac{3}{5} = \frac{2}{5} \)
(D) \(P(A'\cap B')\)
This matches (IV) \(\frac{2}{9}\).
\( \displaystyle P(A'\cap B') = P(A')P(B') = \frac{2}{5}\times\frac{5}{9} = \frac{2}{9} \)
So the matching is (A)-(II), (B)-(III), (C)-(I), (D)-(IV).