Solution: 
$\Large\frac{1}{2} \Large $

Step 1: Identify what needs to be found and write it out symbolically. You have to find the probability that the doctor takes the train given that he is late.
i.e. $P(A_{1 } / B)$
Step 2: Apply Bayes Theorem for the given problem.
$P(A_{1 } / B)= \Large \frac{P(A_{1})P( B/ A_{1})}{P(A_{1})P( B/ A_{1})+P(A_{2})P( B/ A_{2 })+P(A_{3})P( B/ A_{3 })+P(A_{4})P( B/ A_{4 })}\Large$
= $\Large \frac{P(A_{1}\cap B)}{P(A_{1}\cap B)+P(A_{2}\cap B)+P(A_{3}\cap B)+P(A_{4}\cap B)} \Large$
Note that $P(A_{1} / B)= \Large \frac{P(A_{1}\cap B)}{P(B)} \Large $.
Step 3:
$P(A_{1}\cap B)$ = the probability that the doctor takes the train and he is late.
= $ \Large \frac{3}{10} \Large. \Large \frac{1}{4} \Large = \Large \frac{3}{40}\Large$
$P(A_{2}\cap B)= $the probability that the doctor takes the bus and he is late.
= $ \Large \frac{2}{10} \Large. \Large \frac{1}{3} \Large = \Large \frac{1}{15} \Large $
$P(A_{3}\cap B)= $the probability that the doctor takes the scooter and he is late.
= $Large \frac{1}{10} \Large . \Large \frac{1}{12} \Large= \Large \frac{1}{120} \Large $
$P(A_{4}\cap B)=$ the probability that the doctor takes other means of transport and he is late
= 0, because in this case the doctor is not late
Step 4: Substitute the values in the formula to arrive at $P(A_{1 } / B)$
Numerator = $\Large \frac{3}{40} \Large $
Denominator = $ \Large \frac{3}{40}\Large + \Large\frac{1}{15}\Large +\Large\frac{1}{120} \Large +0 = \Large \frac{9+8+1}{120} \Large $
Therefore, $P(A_{1 } / B)$ = $ \Large \frac{3}{40} \Large . \Large \frac{120}{18}\Large $= $\Large \frac{9}{18 \Large }$ = $\Large \frac{1}{2} \Large$