Let $A_1,A_2,A_3$ denote the events that bag contains 2, 3, 4 white balls respectively and ‘E’ be the event that two white balls are drawn from the bag.

$P\left(A_1\right)=\frac{1}{3},P\left(A_2\right)=\frac{1}{3},P\left(A_3\right)=\frac{1}{3}$

$\begin{array}{l}P\left(\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{$A_1$}\right.\right)=\frac{{}^{\text{2}}{\text{C}}_2}{{}^4{\text{C}}_2}=\frac{1}{6}\\ P\left(\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{$A_2$}\right.\right)=\frac{{}^3{\text{C}}_2}{{}^4{\text{C}}_2}=\frac{1}{2}\\ P\left(\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{$A_3$}\right.\right)=\frac{{}^4{\text{C}}_2}{{}^4{\text{C}}_2}=1\end{array}$

From Baye’s theorem

$P\left(\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{$A_3$}\right.\right)=\frac{3}{5}$