A box contains $n$ coins, $m$ of which are fair and the rest are biased. When a biased coin is tossed, the…
A box contains $n$ coins, $m$ of which are fair and the rest are biased. When a biased coin is tossed, the probability of getting a head is twice as likely as tail. A coin is drawn from the box at random and is tossed twice. It is found that first time it shows head and the second time it shows tail. Then, the probability that the coin drawn is fair is
$\frac{7 m}{8 n+m}$
$\frac{9 m}{8 n+m}$
$\frac{7 m}{8 m+n}$
$\frac{9 m}{8 m+n}$
Solution
Let the event $A$ and $B$
$A=$ fair coin tossed
$B=$ biased coin tossed
$E=$ first time head second time tail
$P(A)=\frac{m}{n}, P(B)=\frac{n-m}{n}$
$P(E / A)=\frac{1}{4}, P(E / B)=\frac{2}{9}$
Required probability
$P(A / E)=\frac{P(A) \times P(E / A)}{P(A) \times P(E / A)+P(B) \times P(E / B)}$
$=\frac{\frac{m}{n} \times \frac{1}{4}}{\frac{m}{n} \times \frac{1}{4}+\frac{n-m}{n}\left(\frac{2}{9}\right)}$
$=\frac{9 m}{9 m+8 n-8 m}=\frac{9 m}{8 n+m}$