Question

A doctor is to visit a patient. From the past experience, it is known that the probabilites that he will come by train, bus, scooter or by train, bus, scooter or by means of transport are respectively $\frac{3}{10},\frac{1}{5},\frac{1}{10}\text{and}\frac{2}{5}$. The probabilites that he will be late are $\frac{1}{4},\frac{1}{3}\text{and}\frac{1}{12}$ if he comes by train, bus, and scooter respectively,but if he comes by other means of transport, then he will not be late. When he arrives, he is late. What is the probability that he comes by train?

Answer 1

$A_1:$ Event that doctor comes by Train.
$A_2:$ Event that doctor comes by Bus.
$A_3:$ Event that doctor comes by Scooter.
$A_4:$ Event that doctor comes by other means of the transport.
$B:$ Event that doctor arrives late.

CASE-1

Probability that doctor comes by train

$P\left(A_1\right)=\frac{3}{10}$ ...$\left(1\right)$

Probability that doctor arrives late if he comes by train

$P\left(B|A_1\right)=\frac{1}{4}$ ...$\left(2\right)$

CASE-2

Probability that doctor comes by bus

$P\left(A_2\right)=\frac{1}{5}$ ...$\left(3\right)$

Probability that doctor arrives late if he comes by bus

$P\left(B|A_2\right)=\frac{1}{3}$ ...$\left(4\right)$


CASE-3

Probability that doctor comes by scooter

$P\left(A_3\right)=\frac{1}{10}$ ...$\left(5\right)$

Probability that doctor arrives late if he comes by scooter

$P\left(B|A_3\right)=\frac{1}{12}$ ...$\left(6\right)$

CASE-4

Probability that doctor comes by other means of transport

$P\left(A_4\right)=\frac{2}{5}$ ...$\left(7\right)$

Probability that doctor arrives late if he comes by other means of transport

$P\left(B|A_4\right)=0$ ...$\left(8\right)$

By Bayes' theorem,

$P\left(A_1|B\right)=\frac{P\left(A_1\right)\times P\left(B|A_1\right)}{P\left(A_1\right)\times P\left(B|A_1\right)+p\left(A_2\right)\times P\left(B|A_2\right)+P\left(A_3\right)\times P\left(B|A_3\right)+P\left(A_4\right)\times P\left(B|A_4\right)}$ ...$\left(9\right)$

Subsituting the value of $\left(1\right),\left(2\right),\left(3\right),\left(4\right),\left(5\right),\left(6\right),\left(7\right),\left(8\right)\text{in}\left(9\right)$,

$P\left(A_1|B\right)=\frac{\frac{3}{10}\times \frac{1}{4}}{\frac{3}{10}\times \frac{1}{4}+\frac{1}{5}\times \frac{1}{3}+\frac{1}{10}\times \frac{1}{12}+\frac{2}{5}\times 0}$


$P\left(A_1|B\right)=\frac{\frac{3}{40}}{\frac{3}{40}+\frac{1}{15}+\frac{1}{120}+0}$

$P\left(A_1|B\right)=\frac{\frac{3}{40}}{\frac{18}{120}}$

$P\left(A_1|B\right)=\frac{3}{40}\times \frac{120}{18}$

$P\left(A_1|B\right)=\frac{360}{720}$

$\therefore P\left(A_1|B\right)=\frac{1}{2}$

Therefore, probability of doctor coming by train if he arrives late=$\frac{1}{2}$