Three boxes $B_{1}, B_{2}, B_{3}$ contains balls with different colour as follows.
White Black Red
B₁ 2 1 2
B₂ 3 2 4
B₃ 4 3 2
A die is thrown box B₁ is choosen if either 1 (or) 2 turns up box B₂ is choosen if 3 (or) 4 turns up and box B₃ is choosen if 5 (or) 6 turns up. In this way a box is choosen and a ball is drawn from it. If the ball drawn is found to be red then the probability that it is drawn from box B₂

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$\frac{3}{26}$

$\frac{7}{12}$

$\frac{5}{12}$

$\frac{1}{12}$

answer is B.

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Detailed Solution

$\begin{array}{l} P (B_{1}) = P (B_{2}) = P (B_{3}) = \frac{1}{3} \\ P (\frac{E}{B_{1}}) = \frac{2}{5} P (\frac{E}{B_{2}}) = \frac{4}{9} P (\frac{E}{B_{3}}) = \frac{2}{9} \end{array}$
$P (\frac{B_{2}}{E}) = \frac{P (B_{2}) . P (\frac{E}{B_{2}})}{P (B_{1}) P (\frac{E}{B_{1}}) + P (B_{2}) P (\frac{E}{B_{2}}) + P (B_{3}) P (\frac{E}{B_{3}})} = \frac{5}{12}$

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	White	Black	Red
B₁	2	1	2
B₂	3	2	4
B₃	4	3	2