The probabilities of having a defective toy in three cartons A ,   B ,   C are 1 3 ,   1 4 ,…

Consider $E_1, E_2$ and $E_3$ be the events of selecting defective
toys in cartons$A, B$ and $C$.

From Bays rule, the probability that a selected carton is drawn
from carton $B$,

$P\mathit{=}\frac{P\left(E_{\mathit{2}}\right)\mathit{·}P\left({\displaystyle \frac{A}{E_{\mathit{2}}}}\right)}{P\left(E_{\mathit{1}}\right)\mathit{·}P\left({\displaystyle \frac{A}{E_1}}\right)\mathit{+}P\left(E_{\mathit{2}}\right)P\left({\displaystyle \frac{A}{E_{\mathit{2}}}}\right)\mathit{+}P\left(E_{\mathit{3}}\right)P\left({\displaystyle \frac{A}{E_3}}\right)}$

$=\frac{{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{1}{4}}}{{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{2}{5}}}$

$=\frac{{\displaystyle \frac{1}{12}}}{{\displaystyle \frac{1}{9}}+{\displaystyle \frac{1}{12}}+{\displaystyle \frac{2}{15}}}=\frac{15}{59}$