Suppose that a bag A contains n red and 2 black balls and another bag B contains 2 red and n black balls. One of the two bags is selected at random and two balls are drawn from it at a time. When it is known that the two balls drawn are red, if the probability that those two balls drawn are from bag A is 6/7, then n =

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answer is B.

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Detailed Solution

	A	B
Red	n	2
Black	2	n

$E_{1} =$ Bag A Selected $P (E_{1}) = \frac{1}{2}$

$E_{2} =$ Bag A Selected $P (E_{2}) = \frac{1}{2}$

A=two balls are drawn it is red

$\begin{array}{rcl} P (\frac{A}{E_{1}}) & = & \frac{{}^{n}{C_{2}}}{{}^{(n + 2)}{C_{2}}} \end{array}$

$\begin{array}{rcl} P (\frac{A}{E_{2}}) & = & \frac{{}^{n}{C_{2}}}{{}^{(n + 2)}{C_{2}}} \end{array}$

$\begin{array}{rcl} P (\frac{E_{1}}{A}) & = & \frac{\frac{1}{2} \times \frac{{}^{n}{C_{2}}}{{}^{(n + 2)}{C_{2}}}}{\frac{1}{2} \times \frac{{}^{n}{C_{2}}}{{}^{(n + 2)}{C_{2}}} + \frac{1}{2} \frac{{}^{2}{C_{2}}}{{}^{(n + 2)}{C_{2}}}} = \frac{6}{7} \end{array}$