Let \(E_1\) and \(E_2\) be two mutually exclusive and exhaustive events with \(P(E_1) = P(E_2) = \displaystyle\frac{1}{2}\), and let A be an event with \(P(A|E_1) = \displaystyle\frac{1}{5}\) and \(P(A|E_2) = \displaystyle\frac{3}{5}\). Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(P(E_1|A)\) | (I) | \(\displaystyle\frac{3}{4}\) |
| (B) | \(P(E_2|A)\) | (II) | \(\displaystyle\frac{2}{5}\) |
| (C) | P(A) | (III) | \(\displaystyle\frac{1}{4}\) |
| (D) | \(P(E_1|A')\) | (IV) | \(\displaystyle\frac{2}{3}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
Since \(E_1\) and \(E_2\) are mutually exclusive and exhaustive, the theorem of total probability gives
\( \displaystyle P(A) = \frac{1}{2}\times\frac{1}{5}+\frac{1}{2}\times\frac{3}{5} \)
\( \displaystyle P(A) = \frac{1}{10}+\frac{3}{10} = \frac{2}{5} \)
(A) \(P(E_1|A)\)
This matches (III) \(\frac{1}{4}\).
By Bayes’ theorem:
\( \displaystyle P(E_1|A) = \frac{\frac{1}{2}\times\frac{1}{5}}{\frac{2}{5}} = \frac{\frac{1}{10}}{\frac{4}{10}} = \frac{1}{4} \)
(B) \(P(E_2|A)\)
This matches (I) \(\frac{3}{4}\).
\( \displaystyle P(E_2|A) = \frac{\frac{1}{2}\times\frac{3}{5}}{\frac{2}{5}} = \frac{\frac{3}{10}}{\frac{4}{10}} = \frac{3}{4} \)
(C) P(A)
This matches (II) \(\frac{2}{5}\), the value of the total probability.
(D) \(P(E_1|A')\)
This matches (IV) \(\frac{2}{3}\).
Given \(E_1\), A fails with probability \(1-\frac{1}{5} = \frac{4}{5}\), and \(P(A') = 1-\frac{2}{5} = \frac{3}{5}\):
\( \displaystyle P(E_1|A') = \frac{\frac{1}{2}\times\frac{4}{5}}{\frac{3}{5}} = \frac{\frac{2}{5}}{\frac{3}{5}} = \frac{2}{3} \)