Computing Probability in a Coin problem We place three coins in a box. One coin is two-headed. Another coin is two tailed.
Only the third coin is real, but they all look alike. Then we randomly
select one coin from this box, and we flip it. If heads comes up, what is the
probability that the other side of the coin is also heads?
Let $H$ represent the event that heads comes up on tossing a coin drawn randomly. $
P(H) = \frac{1}{3} + \frac{1}{3} \frac{1}{2} = \frac{1}{2}$. Let $I$ be the event of other side being head. Do I need to compute $P(I|H)?$ 
$P(H \cap I) = P(\text{choosing the coin with 2 heads}) = \frac{1}{3}$. Then $P(I|H) = \frac{P(H \cap I)}{P(H)} = \frac{2}{3}.$ Is this correct? 
 A: Your computation is correct. I have tried to make things a bit more explicit.
Let $H$ be the event that we see heads and $C_2$ we chose the $2$-headed coin, $C_1$ we chose the normal coin, and $C_0$ we chose the $2$-tailed coin. Then
$P(H\cap C_2)=P(C_2)P(H\mid C_2)=\frac13\cdot1=\frac13$
$P(H\cap C_1)=P(C_1)P(H\mid C_1)=\frac13\cdot\frac12=\frac16$
$P(H\cap C_0)=P(C_0)P(H\mid C_0)=\frac13\cdot0=0$
Since $C_0$, $C_1$, and $C_2$ are disjoint and all-inclusive, we have, as expected, that
$P(H)=P(H\cap C_2)+P(H\cap C_1)+P(H\cap C_0)=\frac13+\frac16+0=\frac12$
Since the only way the other side can also be a head is with the $2$-headed coin, what we want is
$P(C_2\mid H)=\frac{P(H\cap C_2)}{P(H)}=\frac{1/3}{1/2}=\frac23$

Another way to look at this is to label the sides of each coin as follows
$$
\begin{array}{c}
\text{coin}&\text{A}&\text{B}\\
\hline
\text{two head}&\text{head}&\text{head}\\
\text{normal}&\text{head}&\text{tail}\\
\text{two tail}&\text{tail}&\text{tail}\\
\end{array}
$$
Each of the $6$ possibilities is equally likely. We have chosen one of the three heads, and two of those choices give the flip side as a head. Thus, the probability is $\frac23$.
