Question

A bag contains $4$ balls. Two balls are drawn at random and are found to be white. what is the probability that all the balls in the bag are white?

Answer 1

$E_1:2$ balls are white, $2-$ are non white
$E_2:3$ balls are white, $1$ is not white
$E_3:4$ balls are white
$A:$ Two balls are white
$P\left(E_1\right)=\frac{1}{3},P\left(E_2\right)=\frac{1}{3},P\left(E_3\right)=\frac{1}{3}$
$P\left(A|E_1\right)=\frac{2C_2}{4C_2}=\frac{1}{6}$
$P\left(A|E_2\right)=\frac{3C_2}{4C_2}=\frac{3}{6}=\frac{1}{2},P\left(A|E_3\right)=\frac{4C_2}{4C_2}=1$
$\therefore P\left(E_3|A\right)=\frac{P\left(A|E_3\right)P\left(E_3\right)}{P\left(A|E_1\right)P\left(E_1\right)+P\left(A|E_2\right)P\left(E_2\right)+P\left(A|E_3\right)P\left(E_3\right)}$
$=\frac{1\times \frac{1}{3}}{\frac{1}{6}\times \frac{1}{3}+\frac{1}{2}\times \frac{1}{3}+\frac{1}{3}\times 1}=\frac{1}{\frac{1}{6}+\frac{1}{2}+1}=\frac{6}{10}=\frac{3}{5}$