Question

A person goes to office by car, scooter, bus and train, probability of which are $\frac{1}{7},\frac{3}{7},\frac{2}{7}$ and $\frac{1}{7}$, respectively. Probability that he reaches office late, if he takes car, scooter, bus or train is $\frac{2}{9},\frac{1}{9}.\frac{4}{9}$, and $\frac{1}{9}$, respectively. Given that he reached office in time, the probability that he travelled by a car, is

• A. $\frac{1}{7}$
• B. $\frac{2}{7}$
• C. $\frac{3}{7}$
• D. $\frac{4}{7}$

Answer 1

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

Let the event of person goes to office by a car, scooter, bus or train be A, B, C and D, respectively.

We, have $P\left(A\right)=\frac{1}{7},P\left(B\right)=\frac{3}{7},P\left(C\right)=\frac{2}{7}$ and $P\left(D\right)=\frac{1}{7}$

Let $E=$ He reached office in time

We have,

$P\left(\frac{\overline{E}}{A}\right)=\frac{2}{9},P\left(\frac{\overline{E}}{B}\right)=\frac{1}{9},P\left(\frac{\overline{E}}{C}\right)=\frac{4}{9}$ and $P\left(\frac{\overline{E}}{D}\right)=\frac{1}{9}$

$\therefore P\left(\frac{A}{E}\right)=\frac{P\left(A\right)\cdot P\left(\frac{E}{A}\right)}{P\left(A\right)\cdot P\left(\frac{E}{A}\right)+P\left(B\right)\cdot \left(\frac{E}{B}\right)}+P\left(C\right)\cdot P\left(\frac{E}{C}\right)+P\left(D\right)\cdot \left(\frac{E}{D}\right)$

$=\frac{\frac{1}{7}\cdot \frac{7}{9}}{\frac{1}{7}\cdot \frac{7}{9}+\frac{3}{9}\cdot \frac{8}{9}+\frac{2}{7}\cdot \frac{5}{9}+\frac{1}{7}\cdot \frac{8}{9}}$

$=\frac{7}{7+24+10+8}=\frac{7}{49}=\frac{1}{7}$