<s>[INST] A Conditional Probability problem with Biased Coins I need explanation for solving the following problem - exactly how one goes about solving it ? Any pointers or hints will certainly be appreciated:
A box contains 5 fair coins and 5 biased coins. Each biased coin has probability of a head as $\frac{4}{5}$. A coin is drawn at random from the box and tossed. Then a second coin is drawn at random from the box (without replacing the first one). Given that the first coin has shown head, the conditional probability that the second coin is fair, is:  
A) $\frac{20}{39}$  B) $\frac{20}{37}$  C) $\frac{1}{2}$  D) $\frac{7}{13}$   
 [/INST] $$P(\text{second is fair}\mid \text{first is head}) = \frac{P(\text{second is fair and first is head})}{P(\text{1st is head})}.$$
$$P(\text{first is head}) = P(\text{first is biased}) \times P(\text{head}\mid\text{biased}) + P(\text{first is unbiased}) \times P(\text{head}\mid\text{unbiased}) $$ $$= \left(\frac{4}{5} \times\frac{1}{2}\right)+\left(\frac{1}{2}\times\frac{1}{2}\right)=\frac{4}{10}+\frac{1}{4}=\frac{13}{20}.$$

$$P(\text{second is fair and first is head})$$ $$= P(\text{second is fair and first is fair and is head}) + P(\text{second is fair and first is unfair and is head}) $$ $$=
\left(\frac{1}{2}\times\frac{1}{2}\times\frac{4}{9}\right)+\left(\frac{4}{5}\times\frac{1}{2}\times\frac{5}{9}\right) = \frac{4}{36}+\frac{20}{90}=\frac{60}{180}=\frac{1}{3}.$$

Hence, the given probability equals $$\frac{1}{3} \div \frac{13}{20}= \boxed{\frac{20}{39}}.$$
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