Question

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 $\frac{1}{2}$, 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. The first time it shows head and the second time it shows tail. The probability that the coin drawn is fair is

• A. $\frac{8m}{8N+m}$
• B. $\frac{m}{8N+m}$
• C. $\frac{9m}{8N+m}$
• D. $\frac{9N}{8N+m}$

Answer 1

The correct option is C $\frac{9m}{8N+m}$

Let $E$ be the event that coin tossed twice , show head at first
time and tail at second time and $F$ be the event that coin dawn is fair.
$P\left(\frac{F}{E}\right)=\frac{P\left(\frac{E}{F}\right).P\left(F\right)}{P\left(\frac{E}{F}\right).P\left(F\right)+P\left(\frac{E}{F^{\prime }}\right).P\left(F^{\prime }\right)}$
$=\frac{\frac{1}{2}.\frac{1}{2}.\frac{m}{N}}{\frac{1}{2}.\frac{1}{2}.\frac{m}{N}+\frac{2}{3}.\frac{1}{3}.\frac{N-m}{N}}$
$=\frac{\frac{m}{4}}{\frac{m}{4}+\frac{2(N-m)}{9}}=\frac{9m}{8N+m}$