Three urns A, B and C contain 7 red, 5 black, 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urns is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn A is

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Solution

Q.
Three urns A, B and C contain 7 red, 5 black, 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urns is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn A is __________ . [2024]

1 $\frac{5}{16}$

2 $\frac{4}{17}$

3 $\frac{5}{18}$

4 $\frac{7}{18}$

Ans.
(3)

Let A, B, C and E be the events defined as follows

A: First urn is chosen

B: Second urn is chosen

C: Third urn is chosen

E: Ball drawn is black

$∴$ $P (A) = P (B) = P (C) = \frac{1}{3}$

$∴$ $P (E | A) = \frac{5}{12}, P (E | B) = \frac{7}{12}, P (E | C) = \frac{6}{12}$

By Bayes' Theorem,

$P (A | E) = \frac{P (A) \cdot P (E | A)}{P (A) \cdot P (E | A) + P (B) \cdot P (E | B) + P (C) \cdot P (E | C)}$

$= \frac{\frac{1}{3} \times \frac{5}{12}}{\frac{1}{3} \times \frac{5}{12} + \frac{1}{3} \times \frac{7}{12} + \frac{1}{3} \times \frac{6}{12}} = \frac{5}{5 + 7 + 6} = \frac{5}{18}$

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