Three boxes $B_{1}, B_{2}, B_{3}$ contain balls with different colours as follows:
White Black Red
$B_{1}$ 2 1 2
$B_{2}$ 3 2 4
$B_{3}$ 4 3 2
A dice is thrown. If $1$ or $2$ turns up on the dice, box $B_{1}$ is selected; if $3$ or $4$ turns up, $B_{2}$ is selected if $5$ or $6$ turns up, then $B_{3}$ is selected. If a box is selected like this, a ball is drawn from that box. If the ball is red, then find the probability that it was drawn from $B_{2}$ .

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Solution

Probability of $B_{2}$ if it is red $= \frac{Probability of red from B_{2}}{Probability of red from B_{1} + Probability of red from B_{2} + Probability of red from B_{3}}$
$= \frac{\frac{2}{6} \times \frac{4}{9}}{\frac{2}{6} \times \frac{2}{5} + \frac{2}{6} \times \frac{4}{9} + \frac{2}{6} \times \frac{2}{9}}$
$= \frac{\frac{4}{27}}{\frac{2}{15} + \frac{4}{27} + \frac{2}{27}} = \frac{\frac{4}{27}}{\frac{2}{15} + \frac{2}{9}}$
[Probability of $B_{n}^{'} s$ selection $\times$ Probability than red is selected from $B_{n}$ .]
$= \frac{\frac{4}{27}}{\frac{6 + 10}{45}} = \frac{45}{27 \times 16}$
$= \frac{5}{48}$

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