Question

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

Answer 1

Suppose that,

$T_1$ : Event that doctor come by train

$T_2$: Event that doctor come by bus

$T_3$: Event that doctor come by scooter

$T_4$: Event that doctor come by other

$T_5$: Event that doctor come by arrives late.

Now,

Probability that doctor come by train $P\left(T_1\right)=\frac{3}{10}$

Probability that doctor comes late if he comes by train

$P\left(\frac{T_5}{T_1}\right)=\frac{1}{4}$

Now,

Probability that doctor comes by Bus $P\left(T_2\right)=\frac{1}{5}$

Probability that doctor comes by scooter $=P\left(T_3\right)=\frac{1}{10}$

Probability that doctor comes late if he comes by scooter $=\frac{1}{12}$

Now, Probability that doctor comes by other means of transport $\Rightarrow P\left(T_4\right)=\frac{2}{5}$

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

$P\left(\frac{T_5}{T_4}\right)=0$

then , put the values in formula.

$P\left(\frac{T_1}{T_5}\right)=\frac{P\left(T_1\right)\times P\left(\frac{T_5}{T_1}\right)}{P\left(T_1\right)\times P\left(\frac{T_5}{T_1}\right)+P\left(T_2\right)\times P\left(\frac{T_5}{T_2}\right)+P\left(T_3\right)\times P\times \left(\frac{T_5}{T_3}\right)+P\left(T_4\right)\times P\left(\frac{T_5}{T_4}\right)}$

$=\frac{\frac{1}{10}\times \frac{1}{4}}{\frac{1}{10}\times \frac{1}{4}+\frac{1}{5}\times \frac{1}{3}+\frac{1}{10}\times \frac{1}{12}+\frac{2}{6}\times 0}$

$=\frac{\frac{1}{40}}{\frac{3+8+1}{120}}$

$=\frac{\frac{1}{40}}{\frac{12}{120}}=\frac{\frac{1}{40}}{\frac{1}{10}}=\frac{10}{40}=\frac{1}{4}$

Hence, $P\left(\frac{T_1}{T_5}\right)=\frac{1}{4}=$ Hence this is the answer.