probablity - recursion and random balls In each of $k$ urns numbered $1,\ldots,k$ are
$b$ white balls and $c$ black balls. Balls are randomly selected (always one ball from one urn) from the urn and throw to the next urn. From 1. to 2., 2. to 3....
What is probablity that from last urn ($k$) we select white ball ?
Look at my attempt and correct me when I am wrong.
$B_k$ - from $k^{th}$ urn we selected white ball.
$C_k$ - from $k^{th}$ urn we selected black ball.  
$$P(B_k) = P(B_k|B_{k-1}) P(B_{k-1}) + 
P(B_k|C_{k-1})P(C_{k-1}) = 
\frac{b+1}{b+c+1}P(B_{k-1}) + \frac{b}{b+c+1} (1-P(B_{k-1})=
\frac{1}{b+c+1}P(B_{k-1}) + \frac{b}{b+c+1} $$
$$Pr(B_1) = \frac{b}{b+c}$$
So we have:
$$Pr(B_k) = \frac{\frac{b}{b+c}}{(b+c+1)^{k-1}} + \frac{(k-1)b}{b+c+1}$$
Is it ok ? 

All OK until you solve the recurrence relation. I think it should be:
\begin{eqnarray*}
P(B_k) &=& \dfrac{1}{b+c+1}P(B_{k-1}) + \dfrac{b}{b+c+1} \\
&& \\
&=& \dfrac{1}{b+c+1}\left(\dfrac{1}{b+c+1}P(B_{k-2}) + \dfrac{b}{b+c+1}\right) + \dfrac{b}{b+c+1} \\
&& \\
&=& \left(\dfrac{1}{b+c+1}\right)^2P(B_{k-2}) + \dfrac{b}{b+c+1}\left(1+\dfrac{1}{b+c+1}\right) \\
&& \\
&& \cdots \\
&=& \left(\dfrac{1}{b+c+1}\right)^{k-1}P(B_1) + \dfrac{b}{b+c+1}\left(1+\dfrac{1}{b+c+1}+\cdots +\left(\dfrac{1}{b+c+1}\right)^{k-1}\right) \\
&& \\
&=& \left(\dfrac{1}{b+c+1}\right)^{k-1}\dfrac{b}{b+c} + \dfrac{b}{b+c+1}\dfrac{1-\left(\dfrac{1}{b+c+1}\right)^k}{1-\dfrac{1}{b+c+1}} \\
&& \\
&=& \left(\dfrac{1}{b+c+1}\right)^{k-1}\dfrac{b}{b+c} + \dfrac{b-b\left(\dfrac{1}{b+c+1}\right)^k}{b+c}.
\end{eqnarray*}
