Question

There are $n$ white and $n+1$ black balls in urn $A$, there are $n+1$ white balls and $n$ black balls in urn $B$. One ball is drawn from urn $A$ and put into urn $B$. Then two balls are drawn from urn $B$ and put into urn $A$. When the operation is completed, the probability that urn $A$ contains same number of white and black balls is $\frac{13}{25}$. Then the number of balls in urn $A$ at the start of the operation was

Answer 1

Case 1:
When one white ball is transferred from urn $A$, then the probability of same number of white and black balls in urn $A$ will be,
$P_1=\frac{{}^n{C}_1}{{}^{2n+1}{C}_1}\cdot \frac{{}^{n+2}{C}_2}{{}^{2n+2}{C}_2}$

Case 2:
When one black ball is transferred from urn $A$, then the probability of same number of white and black balls in urn $A$ will be,
$P_2=\frac{{}^{n+1}{C}_1}{{}^{2n+1}{C}_1}\cdot \frac{{}^{n+1}{C}_1{}^{n+1}{C}_1}{{}^{2n+2}{C}_2}$

So, $P_1+P_2=\frac{13}{25}$
$\Rightarrow 29n^2-46n-24=0\Rightarrow (n-2)(29n+12)=0$
As $n$ is an integer, so $n=2$

Hence, the number of balls in the urn $A$ before the operation begins was $5$.