A man has three routes from his home and he is equally likely in choosing these three ways.the chance that he will miss the train on choosing these three ways are respectively .25,.3,.35.one day he missed train find the probability that he had chosen the first route

Let $A$ denote the event ''he had chosen the first route''.

Let $B$ denote the event ''he had chosen the second route''.

Let $C$ denote the event ''he had chosen the third route''.

Let $M$ denote the event ''he missed train''.

Given $P(A)=P(B)=P(C)=1/3,$

$P(M|A)=0.25, P(M|B)=0.30, P(M|C)=0.35.$

Then by the Bayes' Theorem

$P(A|M)$

$=\dfrac{P(A)P(M|A)}{P(A)P(M|A)+P(A)P(M|A)+P(A)P(M|A)}$

$=\dfrac{\dfrac{1}{3}(0.25)}{\dfrac{1}{3}(0.25)+\dfrac{1}{3}(0.30)+\dfrac{1}{3}(0.35)}=\dfrac{5}{18}$