What is the probability that the coin drawn is fair A box contains $n$ number of coins, $m$ of which are fair and the rest are biased. The probability of getting a head when a fair coin is tossed is $1/2$, while it is $2/3$ when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows a head and the second time it shows tail. What is the probability that the coin drawn is fair.
 A: Let 


*

*$A$ = Even that coin selected is fair

*$B$ = Coin selected is biased

*$C$= Even that first results in a head, second results in a tail.
Now use Bayes Theorem to get the answer. Answer is  $\frac{9m}{8n+m}$
You have:


*

*$P(A)=\frac{m}{n}$ and $P(B)=\frac{n-m}{n}$

*$P(C/A) = \frac{1}{4}$ and $P(C/B) = \frac{2}{3} \times \frac{1}{3} = \frac{1}{9}$.

*By Bayes theorem $$P(A/C) = \frac{P(A) \times P(C/A)}{P(A) \times P(C/A) + P(B) \times P(C/B)}=\frac{9m}{8n+m}$$
A: What have you done so far?  Hint:  Without the flips, what is the chance you draw a fair coin?  If you draw a fair coin, what is the chance you get the flips specified?  If you draw a biased coin, what is the chance?  
A: I know this isn't really answering your question, but a coin toss, is rarely ever fair.  And I'm not talking a few thousandths of a percent, but more like 52% vs 48%.  Below makes for fascinating reading:
http://www.codingthewheel.com/archives/the-coin-flip-a-fundamentally-unfair-proposition
And remember, always be the tosser!
