Let E be the event that the doctor visits the patient late and let T1, T2, T3, T4 be the events that the doctor comes by train, bus, scooter, and other means of transport respectively.
Then P(T1) = 3/10, P(T2) = 1/5, P(T3) = 1/10 and P(T4) = 2/5
P(E|T1) = Probability that the doctor arriving late comes by train = 1/4
Similarly, P(E|T2) = 1/3, P(E|T3) = 1/12 and P(E|T4) = 0, since he is not late if he comes by other means of transport.
Therefore, by Bayes' Theorem, we have
P(T1|E) = Probability that the doctor arriving late comes by train
P(T1/E) = (P(T1)P(E/T1))/(P(T1)P(E/T1) + P(T2)P(E/T2) + P(T3)P(E/T3) + P(T4)P(R/T4))
= (3/10 x 1/4)/(3/10 x 1/4 + 1/5 x 1/3 + 1/10 x 1/12 + 2/5 x 0) = 3/40 x 120/18 = 1/2
Hence, the required probability is 1/2