Let $A_i(i=1,2,3,4)$ be the event that the urn contains 2, 3, 4 or 5 white ball and the event that two white ball are drawn.
We have to find $P\left(A_4 / B\right)$
Since the four events $A_1, A_2, A_3, A_4$ are equally likely, we have $P\left(A_1\right)=P\left(A_2\right)=P\left(A_3\right)=P\left(A_4\right)=\frac{1}{4}$
$P\left(B / A_1\right)$ is the probability of event that the urn contains 2 white balls and both have been drawn.
Hence,
$P\left(B / A_1\right)=\frac{2 C_2}{5 C_2}=\frac{1}{10}$
Similarly, $P\left(B / A_2\right)=\frac{3 C_2}{5 C_2}=\frac{3}{10}$
$P\left(B / A_3\right)=\frac{4 C_2}{5 C_2}=\frac{6}{10}=\frac{3}{5} \Rightarrow P\left(B / A_4\right)=\frac{5 C_2}{5 C_2}=1$
$\therefore$ From the Baye's theorem
$P\left(A_4 / B\right)=\frac{P\left(A_4\right) P\left(B / A_4\right)}{\sum_{i=1}^4 P\left(A_1\right) P\left(B / A_1\right)}=\frac{(1 / 4) \cdot 1}{\frac{1}{4}\left(\frac{1}{10}+\frac{3}{10}+\frac{3}{5}+1\right)}$
$P\left(A_4 / B\right)=\frac{1}{2}$