A man has 5 coins in his pocket... A man has 5 coins in his pocket.  Two are double-headed, one is double-tailed, and two are normal.  The coins cannot be distinguished unless one looks at them.  
a) The man shuts his eyes, chooses a coin at random and tosses it.  What is the probability that the lower face of the coin is heads?
b) He opens his eyes and sees that the upper face of the coin is a head.  What is the probability that the lower face is a head?
 A: This is a variant of the Bertrand's box "paradox".
a) The probability of getting heads down is the probability of choosing a double headed coin ($\frac{2}{5} \times 1$) plus the probability of getting a normal coin and heads landing down ($\frac{2}{5} \times \frac{1}{2}$).  So the total probability is $\frac{3}{5}$.
b) The man has 6 "heads" in his pocket, four of which have heads on the other side, and two of which have tails on the other side.  So if the upper face is a head, there is a $\frac{4}{6} = \frac{2}{3}$ probability the lower face is a head also.  You can also work this out by conditional probabilities if you are thus inclined.
A: 
A man has 5 coins in his pocket. Two are double-headed, one is double-tailed, and two are normal. The coins cannot be distinguished unless one looks at them. 

Let $D_T$ be the event of drawing a double tailed coin, $D_H$ the event of drawing a double head, and $D_0$ be the event of drawing an unbiased coin.
$$\mathsf P(D_H)=\frac 2 5, \mathsf P(D_T)=\frac 1 5, \mathsf P(D_0)=\frac 2 5$$
Let $H$ be the event of flipping a head, and $T$ be that of flipping a tail.
$$\mathsf P(H\mid D_T) = 0, \mathsf P(H\mid D_H)=1, \mathsf P(H\mid D_0)=\frac 12$$

a) The man shuts his eyes, chooses a coin at random and tosses it. What is the probability that the lower face of the coin is heads?

This is a straight application of the Law of Total Probability
$$\begin{align}
\mathsf P(H) & = \mathsf P(D_H)\;\mathsf P(H\mid D_H) + \mathsf P(D_T)\;\mathsf P(H\mid D_T) + \mathsf P(D_0)\;\mathsf P(H\mid D_0)
\\[1ex] & = \frac 2 5\times 1 + \frac 1 5\times 0 + \frac 2 5\times\frac 1 2
\\[1ex] & = \frac 3 5
\end{align}$$

b) He opens his eyes and sees that the upper face of the coin is a head. What is the probability that the lower face is a head?

Here we apply Baye's Theorem of conditional probability.
$$\begin{align}
\mathsf P(D_H\mid H) & = \frac{\mathsf P(D_H)\;\mathsf P(H\mid D_H)}{\mathsf P(H)}
\\[1ex] & = \frac{\frac 2 5\times 1}{\frac 3 5}
\\[1ex] & = \frac 2 3
\end{align}$$
A: There are $10$ coin faces in the man's pocket: four heads attached to heads, two heads attached to tails, and four tails (whose attachments we ignore here since they will not matter).
The procedure results in $10$ possible events, each equally likely (by symmetry). We don't need to analyze the coin selection and coin tossing as separate events.
Of the $10$ events, $6$ have a head face down (since each event selects a different face in this position), hence the probability of a face-down head is $3/5.$
In the second part of the problem, we are restricted to the $6$ events that have a head
facing up.  The face on the bottom is whatever was attached to the top face.  Each of the heads is equally likely to be the top face; four of those are attached to heads, two are not, so the probability that the bottom face is heads is $2/3.$
