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

Answer 1

Answer: (A)

$E_{1}:$ coin is fair, $E_{2}:$ coin is biased, $A$ second toss shows tail. $P\left(E_{1} / A\right)=\frac{P\left(A / E_{1}\right) P\left(E_{1}\right)}{P\left(A / E_{1}\right) P\left(E_{1}\right)+P\left(A / E_{1}\right) P\left(E_{2}\right)}$
$=\frac{\frac{m}{N} \cdot \frac{1}{2} \cdot \frac{1}{2}}{\frac{m}{N} \cdot \frac{1}{2} \cdot \frac{1}{2}+\frac{N-m}{N} \cdot \frac{2}{3} \cdot \frac{1}{3}}$
$=\frac{9 m}{8 N+m}$