Question
A box contains $$N$$ coins, $$m$$ of which are fair and the rest are biased. The probability of getting a head when a fair coin is tossed is $$\frac{1}{2},$$ while it is $$\frac{2}{3}$$ when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. Then the probability that the coin drawn is fair, is :
A.
$$\frac{{9m}}{{8N + m}}$$
B.
$$\frac{{9m}}{{8N - m}}$$
C.
$$\frac{{9m}}{{8m - N}}$$
D.
$$\frac{{9m}}{{8m + N}}$$
Answer :
$$\frac{{9m}}{{8N + m}}$$
Solution :
$$\eqalign{
& {E_1}\,:{\text{coin is fair, }}{E_2}\,:{\text{coin is biased,}} \cr
& A{\text{ second toss shows tail}}{\text{.}} \cr
& P\left( {\frac{{{E_1}}}{A}} \right) = \frac{{P\left( {\frac{A}{{{E_1}}}} \right)P\left( {{E_1}} \right)}}{{P\left( {\frac{A}{{{E_1}}}} \right)P\left( {{E_1}} \right) + P\left( {\frac{A}{{{E_2}}}} \right)P\left( {{E_2}} \right)}} \cr
& = \frac{{\frac{m}{N}.\frac{1}{2}.\frac{1}{2}}}{{\frac{m}{N}.\frac{1}{2}.\frac{1}{2} + \frac{{N - m}}{N}.\frac{2}{3}.\frac{1}{3}}} \cr
& = \frac{{9m}}{{8N + m}} \cr} $$