Conditional Probability. Drawing the ball from the urn.

Question

Reading the paper I have encountered the following example:

The urn that have been filled with black and white balls as follows. First, a ball is placed in the urn according to the outcome of a fair coin toss. If the coin toss for an urn produces heads, the “coin ball” placed in the urn is black; it is white otherwise. In addition to a coin ball, the urn contains n “noncoin balls,” of which exactly n/2 are black and n/2 are white.

The agent is invited to bet on the colour of the two coin balls. Any bet (on a ball of some colour drawn from some urn) pays one dollar if the ball has the desired colour and zero otherwise.

Suppose now that one draw from the urn is observed and that a ball drawn is black. Calculate the conditional probability of a black coin ball.

The correct answer is (n/2+1)/(n+1). But I can't understand how this result was obtained. I kindly ask for the help from the community.

Answer 1

To have an easy exposition, i will will denote by $k$ the natural number $k:=n/2$. We consider the given two-steps experiment with the probabilities given by the tree model.

                                       Black ===> Relevant, event E 
                                      /
                                     / (k+1)/(2k+1)
                                    /
      Black coin ball, (k+1)B and kW
     /                              \
    /                                \ etc.
   / 1/2                              \
  /                                    White
 /
*
 \                                   
  \                                    Black ===> Relevant, event F
   \ 1/2                              /
    \                                / k/(2k+1)
     \                              /
      White coin ball, kB and (k+1)W
                                    \
                                     \ etc
                                      \
                                       White

Later edit: In the above diagram, we have the following compositions of the urn after the first coin ball:

• if we add a black coin ball, then there are $(k+1)$ B(lack) balls and $k$ W(hite) balls in the urn,
• if we add a white coin ball, then there are $k$ B(lack) balls and $(k+1)$ W(hite) balls in the urn.

So we are on the relevant side, and compute the probability to land "up", in the event $E$, considering that we already "sit" in the event $E\cup F$.

The corresponding probability is: $$ \frac{P(E)}{P(E\cup F)} = \frac{P(E)}{P(E)+ P(F)} = \frac {\displaystyle \frac 12\cdot\frac{k+1}{2k+1}} {\displaystyle \frac 12\cdot\frac{k+1}{2k+1} + \frac 12\cdot\frac{k}{2k+1}} = \frac{(k+1)}{(k+1)+k}\ . $$