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, when first toss head, second toss tail is:

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

Answer 1

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

Box contains $m$ fair coins, $N-m$ biased coins.
For a fair coin $P\left(H\right)=P\left(T\right)=\frac{1}{2}$
For a biased coin $P\left(H\right)=\frac{2}{3},P\left(T\right)=\frac{1}{3}$
Let $E_1=$ event of drawing a fair coin
and $E_2=$ event of drawing biased coin
Let $A=$ event of getting head on $1st$ toss, tail on $2nd$ toss
$P\left(E_1\right)=\frac{m}{N},P\left(E_2\right)=\frac{N-m}{N}$
$P\left(\frac{A}{E_1}\right)=\frac{1}{2}\times \frac{1}{2}\text{and}P\left(\frac{A}{E_2}\right)=\frac{2}{3}\times \frac{1}{3}$
Using Baye's theorem, we have:
$P\left(E_1/A\right)=\frac{P\left(E_1\right).P\left(A/E_1\right)}{P\left(E_1\right).P\left(A/E_1\right)+P\left(E_2\right).P\left(A/E_2\right)}$
$=\left[\frac{\frac{m}{N}\times (\frac{1}{2}.\frac{1}{2})}{\frac{m}{N}\times (\frac{1}{2}.\frac{1}{2})+\frac{N-m}{N}\times (\frac{2}{3}.\frac{1}{3})}\right]=\frac{9m}{m+8N}$