A box contain $N$ coins, m of which are fair and rest are biased. The probability of getting a head when a fair coin is tossed is $1 / 2,$ while it is $2 / 3$ when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. The probability that the coin drawn is fair is

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$\frac{8 m}{8 N + m}$

$\frac{m}{8 N + m}$

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

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

answer is C.

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Detailed Solution

Let $E_{1}, E_{2}$ and A denote the following events:
$E_{1} :$ coin selected is fair
$E_{2} :$ coin selected is biased
$A :$ the first toss results in a head and the second toss results in a tail.

$\begin{array}{l} P (E_{1}) = \frac{m}{N}, P (E_{2}) = \frac{N - m}{N}, \\ P (A I E_{1}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}, P (A I E_{2}) = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9} \end{array}$

$\begin{array}{l} P (E_{1} ∣ A) = \frac{P (E_{1}) P (A I E_{1})}{P (E_{1}) P (A I E_{1}) + P (E_{2}) P (A I E_{2})} \\ P (E_{1} ∣ A) = \frac{\frac{m}{N} (\frac{1}{4})}{\frac{m}{N} (\frac{1}{4}) + \frac{N - m}{N} (\frac{2}{9})} = \frac{9 m}{9 m + 8 N - 8 m} \\ = \frac{9 m}{8 N + m} . \end{array}$