A die is thrown. If it shows 1 or 2, a ball is drawn at random from Box I, which contains 3 white and 2 black balls; otherwise a ball is drawn at random from Box II, which contains 2 white and 3 black balls. Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | Probability that Box I is used | (I) | \(\displaystyle\frac{3}{7}\) |
| (B) | Probability that the ball drawn is white | (II) | \(\displaystyle\frac{7}{15}\) |
| (C) | Probability that Box I was used, given that the ball is white | (III) | \(\displaystyle\frac{3}{4}\) |
| (D) | Probability that Box II was used, given that the ball is black | (IV) | \(\displaystyle\frac{1}{3}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
Let \(E_1\) and \(E_2\) be the events that Box I and Box II are used, and W, B the events that the ball is white, black.
(A) Probability that Box I is used
This matches (IV) \(\frac{1}{3}\).
Box I is used when the die shows 1 or 2, that is, 2 of the 6 faces:
\( \displaystyle P(E_1) = \frac{2}{6} = \frac{1}{3},\quad P(E_2) = \frac{2}{3} \)
(B) Probability that the ball drawn is white
This matches (II) \(\frac{7}{15}\).
By the theorem of total probability:
\( \displaystyle P(W) = \frac{1}{3}\times\frac{3}{5} + \frac{2}{3}\times\frac{2}{5} = \frac{3}{15}+\frac{4}{15} = \frac{7}{15} \)
(C) Probability that Box I was used, given that the ball is white
This matches (I) \(\frac{3}{7}\).
By Bayes’ theorem:
\( \displaystyle P(E_1|W) = \frac{\frac{3}{15}}{\frac{7}{15}} = \frac{3}{7} \)
(D) Probability that Box II was used, given that the ball is black
This matches (III) \(\frac{3}{4}\).
\( \displaystyle P(B) = 1-\frac{7}{15} = \frac{8}{15},\quad P(E_2)P(B|E_2) = \frac{2}{3}\times\frac{3}{5} = \frac{6}{15} \)
\( \displaystyle P(E_2|B) = \frac{\frac{6}{15}}{\frac{8}{15}} = \frac{3}{4} \)