Question

A visiting lecturer visits the college for taking his class. From the past experience, it is known that the probabilities 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 if he comes by other means of transport, then he will not be late. When he arrives the college, he is late. Then the probability that he comes by train is

• A. $\frac{2}{5}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is D $\frac{1}{2}$

Let $E$ be the event that visiting lecturer visit the college late and let $E_1,E_2,E_3,E_4$ be the events that the lecturer comes by train, bus, scooter and other means of transport respectively.
$\therefore P\left(E_1\right)=\frac{3}{10},P\left(E_2\right)=\frac{1}{5},P\left(E_3\right)=\frac{1}{10},P\left(E_4\right)=\frac{2}{5}$
$P(\left(\frac{E}{E_1}\right)=$ probability that the lecturer arriving late by coming train $=\frac{1}{4}$
$P\left(\frac{E}{E_2}\right)=\frac{1}{3},P\left(\frac{E}{E_3}\right)=\frac{1}{12}$ and $P\left(\frac{E}{E_4}\right)=0$, same he is not late if he comes by other means of transport.
$\therefore$ By Bayes’ theorem
$\therefore P\left(\frac{E_1}{E}\right)=\frac{P\left(E_1\right).P\left(\frac{E}{E_1}\right)}{P\left(E_1\right)\left(\frac{E}{E_1}\right)+P\left(E_2\right).P\left(\frac{E}{E_2}\right)+P\left(E_3\right).P\left(\frac{E}{E_3}\right)+P\left(E_4\right).P\left(\frac{E}{E_4}\right)}$
$=\frac{\frac{3}{10}\times \frac{1}{4}}{\frac{3}{10}\times \frac{1}{4}+\frac{1}{5}\times \frac{1}{3}+\frac{1}{10}\times \frac{1}{12}+\frac{2}{5}\times 0}$
$=\frac{\frac{3}{40}}{\frac{3}{40}+\frac{1}{15}+\frac{1}{120}}=\frac{\frac{3}{40}}{\frac{9+8+1}{120}}$
$=\frac{3}{40}\times \frac{120}{18}=\frac{1}{2}$