Suppose that a bag A contains n red and 2 bla | vedclass

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(B) Let $E_1$ be the event that bag $A$ is selected and $E_2$ be the event that bag $B$ is selected. Let $R$ be the event that the two balls drawn are

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$4$

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(B) Let $E_1$ be the event that bag $A$ is selected and $E_2$ be the event that bag $B$ is selected. Let $R$ be the event that the two balls drawn are red.Given $P(E_1) = P(E_2) = \frac{1}{2}$.The probability of drawing $2$ red balls from bag $A$ is $P(R|E_1) = \frac{^nC_2}{^{n+2}C_2} = \frac{n(n-1)}{(n+2)(n+1)}$.The probability of drawing $2$ red balls from bag $B$ is $P(R|E_2) = \frac{^2C_2}{^{n+2}C_2} = \frac{1 	imes 2}{(n+2)(n+1)} = \frac{2}{(n+2)(n+1)}$.By Bayes' theorem,$P(E_1|R) = \frac{P(E_1)P(R|E_1)}{P(E_1)P(R|E_1) + P(E_2)P(R|E_2)} = \frac{6}{7}$.Substituting the values:$\frac{\frac{1}{2} \cdot \frac{n(n-1)}{(n+2)(n+1)}}{\frac{1}{2} \cdot \frac{n(n-1)}{(n+2)(n+1)} + \frac{1}{2} \cdot \frac{2}{(n+2)(n+1)}} = \frac{6}{7}$.$\frac{n(n-1)}{n(n-1) + 2} = \frac{6}{7}$.$7(n^2 - n) = 6(n^2 - n + 2)$.$7n^2 - 7n = 6n^2 - 6n + 12$.$n^2 - n - 12 = 0$.$(n-4)(n+3) = 0$.Since $n > 0$,we have $n = 4$.

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