Question ID: #473
Bag $B_{1}$ contains 6 white and 4 blue balls, Bag $B_{2}$ contains 4 white and 6 blue balls, and Bag $B_{3}$ contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability, that the ball is drawn from Bag $B_{2}$, is:
- (1) $\frac{6}{15}$
- (2) $\frac{4}{15}$
- (3) $\frac{5}{15}$
- (4) $\frac{2}{5}$
Solution:
Let $E_{1}, E_{2}, E_{3}$ be the events of selecting Bag $B_{1}, B_{2}, B_{3}$ respectively.
Since one bag is selected at random, $P(E_{1}) = P(E_{2}) = P(E_{3}) = \frac{1}{3}$.
Let $A$ be the event that the drawn ball is white.
The conditional probabilities of drawing a white ball from each bag are:
$P(A|E_{1}) = \frac{6}{10}$ (from 6W, 4B)
$P(A|E_{2}) = \frac{4}{10}$ (from 4W, 6B)
$P(A|E_{3}) = \frac{5}{10}$ (from 5W, 5B)
Using Bayes’ Theorem to find the probability that the ball came from Bag $B_{2}$ given it is white:
$$P(E_{2}|A) = \frac{P(E_{2})P(A|E_{2})}{P(E_{1})P(A|E_{1}) + P(E_{2})P(A|E_{2}) + P(E_{3})P(A|E_{3})}$$
$$P(E_{2}|A) = \frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{6}{10} + \frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}}$$
$$P(E_{2}|A) = \frac{\frac{4}{30}}{\frac{6}{30} + \frac{4}{30} + \frac{5}{30}}$$
$$P(E_{2}|A) = \frac{4}{6 + 4 + 5}$$
$$P(E_{2}|A) = \frac{4}{15}$$
Ans. (2)
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