A bag contains four balls colours of which are unknown. Two balls are drawn and are found to be white. The probability that all the balls in the bag are white is

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$}\right.\right)=\frac{3}{5}$