How to update a probability that a defendant is guilty after testimonies of multiple unreliable witnesses? 
a) Your initial belief is that a defendant in a court case is guilty with probability 0.5. A
  witness comes forward claiming he saw the defendant committed the crime. You
  know the witness is not totally reliable and tells the truth with probability p. Calculate
  the posterior probability that the defendant is guilty, based on the witness’s
  evidence.
b) A second witness, equally unreliable, comes forward and claims she saw the
  defendant committed the crime. Assuming the witnesses are not colluding, what is
  your posterior probability of guilt?
c) In total, $n$ equally unreliable witnesses claim that they saw the defendant committed
  the crime. If there is no collusion among them, what is your posterior probability of
  guilt?
d) Compare the answers to a), b) and c). How do you explain this result?

So far I know I have to make a probability tree have and I two marginal probabilities which are $P(\mathrm{Guilty})=0.5$ and $P(\mathrm{NotGuilty})=0.5$. How to extend the tree branches from these two marginal probabilities?
 A: Instead of a tree, use a contingency table (which is the same thing, but lays out the calculations in a more convenient form).  Instead of probabilities, perform the calculations in terms of odds.

Because the problem eventually concerns multiple witnesses, let's address the case where after $n-1$ witnesses have come forward our belief in guilt has probability $q$, say (and therefore our belief in lack of guilt is $1-q$).  Because the next witness will split each row into two parts in the proportion $p:(1-p)$, the table must divide up like this:
               Witness:     Guilty Not guilty  Total
                        ---------- ---------- |-----
Prior belief is guilty:         pq     (1-p)q |    q
        ... not guilty: (1-p)(1-q)     p(1-q) |  1-q

The updated belief in guilt when the witness testifies to guilt will be found in the proportions of the "Guilty" column appearing in the first row.  (The second column becomes irrelevant.)  Since the total in the "Guilty" column is $pq + (1-p)(1-q)$, this proportion equals
$$\frac{pq}{pq + (1-p)(1-q)}.$$
A neater way to express this is in terms of the odds, assuming neither $p$ nor $q$ equals $1$. (When either does equal $1$, the odds become infinite but the calculations in terms of probabilities are easy.)  The odds are found by dividing the belief in guilt (first row) given the witness is testifying to guilt (left column) by the belief in not guilty (second row) within the same column.  The final odds are therefore 
$$\frac{pq}{(1-p)(1-q)} = \frac{p}{1-p} \frac{q}{1-q}.$$
We see that the initial odds of $q/(1-q)$ have been updated to odds of $p/(1-p)\times q/(1-q)$.  In other words,

The posterior odds of guilt are the prior odds of guilt $q/(1-q)$ multiplied by the witness's odds of guilt $p(1-p)$.

This updating will happen each time a witness comes forward.  (That explains how to extend the probability tree, if one really wanted to: it consists of a sequence of identical steps, each leading to the same calculations.)  Therefore, after $n$ witnesses appear, the initial odds $q_0 = (1/2)/(1 - 1/2) = 1$ have been multiplied by $\left(p/(1-p)\right)^n$.  That is, with $n$ witnesses the odds of our belief in guilt should be updated to
$$\text{Odds(Guilty)} = \left(\frac{p}{1-p}\right)^n \frac{q_0}{1-q_0} = \left(\frac{p}{1-p}\right)^n.$$
This forms a geometric sequence with initial value $q_0/(1-q_0)=1$ and common ratio $\lambda=p/(1-p)$.  When $\lambda\gt 1$ (equivalently, $p\gt 1/2$), which occurs when the witnesses are somewhat reliable, the sequence increases without bound, showing that the probability of guilt is growing large.  When $\lambda\lt 1$ (i.e., $p\lt 1/2$), which occurs when the witnesses are more likely to lie than not, the sequence converges to zero, showing that the probability of guilt is becoming small: a pack of lying bastards should convince you the defendant is innocent!  When $\lambda=1$ (which corresponds to witnesses who are doing no better than randomly guessing), the odds remain the same as always.
A: P(guilty) = 0.5
P(witness says guilty | guilty) = p    (Witness is telling the truth)
P(witness says guilty | not guilty) = 1- p  (Witness is telling a lie)
P(guilty and witness says guilty) = 0.5p 
P(not guilty and witness says guilty) = 0.5(1-p) 
P(witness says guilty) = 0.5[p + (1-p)] = 0.5 
by bayes' rule 
P(guilty | witness says guilty)= P(guilty and witness says guilty)/P(witness says guilty) = 0.5p/0.5 = p
Source: http://slaystats.com/adms2320assignment.html
