There are three boxes. First box has $2$ white, $3$ black and $4$ red balls. Second box has $3$ white, $2$ black and $2$ red balls. Third box has $4$ white, $1$ black and $3$ red balls. A box is chosen at random and $2$ balls are drawn out of which $1$ is white, and $1$ is red. What is the probability that the balls came from first box?

Let the boxes be $B_1, B_2, B_3$ and event $E$ be drawing $1$ white and $1$ red ball.

Since one box is chosen randomly,

$P(B_1)=P(B_2)=P(B_3)=\frac{1}{3}$

For first box,

$P(E|B_1)=\frac{\binom{2}{1}\binom{4}{1}}{\binom{9}{2}}=\frac{8}{36}=\frac{2}{9}$

For second box,

$P(E|B_2)=\frac{\binom{3}{1}\binom{2}{1}}{\binom{7}{2}}=\frac{6}{21}=\frac{2}{7}$

For third box,

$P(E|B_3)=\frac{\binom{4}{1}\binom{3}{1}}{\binom{8}{2}}=\frac{12}{28}=\frac{3}{7}$

Using Bayes theorem,

$P(B_1|E)=\frac{P(B_1)P(E|B_1)}{P(B_1)P(E|B_1)+P(B_2)P(E|B_2)+P(B_3)P(E|B_3)}$

$P(B_1|E)=\frac{\frac{1}{3}\times\frac{2}{9}}{\frac{1}{3}\times\frac{2}{9}+\frac{1}{3}\times\frac{2}{7}+\frac{1}{3}\times\frac{3}{7}}$

$P(B_1|E)=\frac{\frac{2}{9}}{\frac{2}{9}+\frac{2}{7}+\frac{3}{7}}$

$P(B_1|E)=\frac{\frac{2}{9}}{\frac{59}{63}}=\frac{14}{59}=0.237$