A box contains n coins,m of which are fair an | vedclass

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(B) Let $F$ be the event that a fair coin is drawn and $B$ be the event that a biased coin is drawn. Let $E$ be the event that the first toss is a hea

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$\frac{9 m}{8 n+m}$

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(B) Let $F$ be the event that a fair coin is drawn and $B$ be the event that a biased coin is drawn. Let $E$ be the event that the first toss is a head and the second toss is a tail.Given $P(F) = \frac{m}{n}$ and $P(B) = \frac{n-m}{n}$.For a fair coin,$P(H) = \frac{1}{2}$ and $P(T) = \frac{1}{2}$. Thus,$P(E|F) = \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{4}$.For a biased coin,$P(H) = 2P(T)$. Since $P(H) + P(T) = 1$,we have $3P(T) = 1$,so $P(T) = \frac{1}{3}$ and $P(H) = \frac{2}{3}$. Thus,$P(E|B) = \frac{2}{3} 	imes \frac{1}{3} = \frac{2}{9}$.Using Bayes' theorem,the probability that the coin is fair given the event $E$ is:$P(F|E) = \frac{P(F)P(E|F)}{P(F)P(E|F) + P(B)P(E|B)}$$P(F|E) = \frac{(\frac{m}{n})(\frac{1}{4})}{(\frac{m}{n})(\frac{1}{4}) + (\frac{n-m}{n})(\frac{2}{9})}$$P(F|E) = \frac{\frac{m}{4}}{\frac{m}{4} + \frac{2(n-m)}{9}} = \frac{9m}{9m + 8(n-m)} = \frac{9m}{9m + 8n - 8m} = \frac{9m}{8n + m}$.

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