A box A contains 2 white, 3 red and 2 black balls. Another box B contains 4 white, 2 red and 3 black balls.…

Probability that box $A$ is selected $P\left(A\right)=\frac{1}{2}$

Probability that box $B$ is selected $P\left(B\right)=\frac{1}{2 }$

Let $E$ be the event that one ball is white while the other is red.

$\therefore P\left(E\right)=P\left(A\right)·P\left(E/A\right)+P\left(B\right)·P\left(E/B\right)$

$=\frac{1}{2}\left[\frac{2·3}{{}^7{C}_2}+\frac{4·2}{{}^9{C}_2}\right]=\frac{1}{2}\left[\frac{6}{21}+\frac{8}{36}\right]=\frac{1}{2}\left[\frac{2}{7}+\frac{2}{9}\right]=\frac{16}{63}$

Thus, $P\left(B/E\right)=\frac{P\left(B\right) P\left(E/B\right)}{P\left(E\right)}=\frac{{\displaystyle \frac{1}{9}}}{{\displaystyle \frac{16}{63}}}=\frac{7}{16}$