If A and B are independent events and \(P(A) = \displaystyle\frac{1}{2}, P(B) = \displaystyle\frac{1}{3}\) then
Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(P(A\cap B)\) | (I) | \(\displaystyle\frac{1}{2}\) |
| (B) | \(P(\bar A)P(B) + P(A)P(\bar B)\) | (II) | \(\displaystyle\frac{1}{3}\) |
| (C) | \(P(A \mid B) + P(B \mid A)\) | (III) | \(\displaystyle\frac{1}{6}\) |
| (D) | \(P(A\cap\bar B)\) | (IV) | \(\displaystyle\frac{5}{6}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
For independent events, the probability of both is the product of the probabilities. Also
\(\displaystyle P(\bar A) = 1 - \frac{1}{2} = \frac{1}{2},\quad P(\bar B) = 1 - \frac{1}{3} = \frac{2}{3}\)
(A) \(P(A\cap B)\)
This matches (III) \(\frac{1}{6}\).
\(\displaystyle P(A\cap B) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\)
(B) \(P(\bar A)P(B) + P(A)P(\bar B)\)
This matches (I) \(\frac{1}{2}\).
\(\displaystyle \frac{1}{2} \times \frac{1}{3} + \frac{1}{2} \times \frac{2}{3} = \frac{1}{6} + \frac{2}{6} = \frac{1}{2}\)
(C) \(P(A \mid B) + P(B \mid A)\)
This matches (IV) \(\frac{5}{6}\).
For independent events \(P(A \mid B) = P(A)\) and \(P(B \mid A) = P(B)\):
\(\displaystyle P(A \mid B) + P(B \mid A) = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}\)
(D) \(P(A\cap\bar B)\)
This matches (II) \(\frac{1}{3}\).
A and \(\bar B\) are also independent:
\(\displaystyle P(A\cap\bar B) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}\)
So the matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II).