Question

A person goes to office either by car, scooter, bus or train, the probability of which being $\frac{1}{7},\frac{3}{7},\frac{2}{7}and\frac{1}{7}$ respectively. The probabilities that he reaches the office late, if he takes a car, scooter, bus or train are $\frac{2}{9},\frac{1}{9},\frac{4}{9}and\frac{1}{9}$, respectively. Given that he reaches the office on time, then what is the probability that he travelled by car?

• A. $\frac{1}{7}$
• B. $\frac{2}{7}$
• C. $\frac{2}{9}$
• D. $\frac{4}{21}$

Answer 1

The correct option is A $\frac{1}{7}$

As the statement shows problem is to be related to Bayes' theorem.
Let C, S, B, T be the events when person is going by car, scooter, bus or train, respectively.
$\therefore P\left(C\right)=\frac{1}{7},P\left(S\right)=\frac{3}{7},P\left(B\right)=\frac{2}{7},P\left(T\right)=\frac{1}{7}$
Again, L be the event of the person reaching office late.
$\therefore \overline{L}$ be the event of the person reaching office in time.
Then, $P\left(\frac{\overline{L}}{C}\right)=\frac{7}{9},P\left(\frac{\overline{L}}{S}\right)=\frac{8}{9},P\left(\frac{\overline{L}}{B}\right)=\frac{5}{9}$
and $P\left(\frac{\overline{L}}{T}\right)=\frac{8}{9}$
$\therefore P\left(\frac{C}{L}\right)=\frac{P\left(\frac{\overline{L}}{C}\right).P\left(C\right)}{P\left(\frac{\overline{L}}{C}\right).P\left(C\right)+P\left(\frac{\overline{L}}{S}\right).P\left(S\right)+P\left(\frac{\overline{L}}{B}\right).P\left(B\right)+P\left(\frac{\overline{L}}{T}\right).P\left(T\right)}$
$=\frac{\frac{7}{9}\times \frac{1}{7}}{\frac{7}{9}\times \frac{1}{7}+\frac{8}{9}\times \frac{3}{7}+\frac{5}{9}\times \frac{2}{7}+\frac{8}{9}\times \frac{1}{7}}=\frac{1}{7}$