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) $\frac{7}{13}$   
 A: HINT
This is a Bayes rule problem. $$P(Biased|Heads) = \frac{P(Heads|Biased)P(Biased)}{P(Heads|Biased)P(Biased) + P(Heads|Fair)P(Fair)}.$$ $P(Heads|Biased) = 4/5$ and I'll leave the rest to you.
A: You should use Bayes Rule, but if you are new to these types of problems, here is a way to maybe intuitively see what is happening.
You need to go down both branches of what could happen on the first coin toss. When you take that first coin and flip it, there are 4 possible endpoints.  Fair coin - heads, fair coin - tails, biased coin - heads, biased coin - tails.  Their probabilities each are $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$, $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$, $\frac{1}{2}\times\frac{4}{5}=\frac{2}{5}$, and $\frac{1}{2}\times\frac{1}{5}=\frac{1}{10}$. The first $\frac{1}{2}$ in each of those probabilities is the probability of choosing that type of coin and is really $\frac{5}{10}$.
So, now that we know that, we can look at the event of getting a head on the first toss. The probability of that head being from the fair coin is $$\frac{0.25}{0.25+0.40}$$and the probability that the head was from the biased coin is $$\frac{0.40}{0.25+0.40}$$
So given that the first coin was a head, there is a $\frac{5}{13}$ chance it was the fair coin and a $\frac{8}{13}$ chance it was the biased coin.  
From there you can calculate the probability of getting a fair coin for the second coin. $$\frac{5}{13}*\frac{4}{9}+\frac{8}{13}*\frac{5}{9}$$
A: The events that the second coin is fair and that the first shows heads are conditionally independent when given the fairness of the first coin.  So by the Law of Total Probability, and letting $F_n$ represent that coin $n$ is fair and $H_n$ represent that the coin shows heads, we have:
$$\def\P{\operatorname{\sf P}} \P(F_2\mid H_1) ~{= \P(F_2\mid F_1,H_1)\P(F_1\mid H_1) + \P(F_2\mid F_1^\complement, H_1)\P(F_1^\complement\mid H_1)\\= \P(F_2\mid F_1)\P(F_1\mid H_1)+\P(F_2\mid F_1^\complement)\P(F_1^\complement\mid H_1) \\ = \dfrac{\P(F_2\mid F_1)\P(F_1)\P(H_1\mid F_1)}{\P(H_1)}+\dfrac{\P(F_2\mid F_1^\complement)\P(F_1^\complement)\P(H_1\mid F_1^\complement)}{\P(H_1)} \\ = \dfrac{\P(F_2\mid F_1)\P(F_1)\P(H_1\mid F_1)+\P(F_2\mid F_1^\complement)\P(F_1^\complement)\P(H_1\mid F_1^\complement)}{\P(F_1)\P(H_1\mid F_1)+\P(F_1^\complement)\P(H_1\mid F_1^\complement)}}$$
Evaluation of these terms is now left to the student.
A: $$P(\text{second is fair}\mid \text{first is head}) = \frac{P(\text{second is fair and first is head})}{P(\text{1st is head})}.$$
$$P(\text{first is head}) = P(\text{first is biased}) \times P(\text{head}\mid\text{biased}) + P(\text{first is unbiased}) \times P(\text{head}\mid\text{unbiased}) $$ $$= \left(\frac{4}{5} \times\frac{1}{2}\right)+\left(\frac{1}{2}\times\frac{1}{2}\right)=\frac{4}{10}+\frac{1}{4}=\frac{13}{20}.$$

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

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