Question

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

• A. 1/7
• B. 2/7
• C. 3/7
• D. 4/7

Answer 1

Answer: (A)

Let the event of person goes to office by a car, scooter, bus or train be $A, B, C$ and $D$, respectively.
We have, $P(A)=\frac{1}{7}, P(B)=\frac{3}{7}, P(C)=\frac{2}{7}$
and $P(D)=\frac{1}{7}$
Let $E=$ He reached office in time We have,
$P\left(\frac{\bar{E}}{A}\right)=\frac{2}{9}, P\left(\frac{\bar{E}}{B}\right)$
$=\frac{1}{9}, P\left(\frac{\bar{E}}{C}\right)=\frac{4}{9}$
and $P\left(\frac{\bar{E}}{D}\right)=\frac{1}{9}$
$\therefore P\left(\frac{A}{E}\right)=\frac{P(A) \cdot P\left(\frac{E}{A}\right)}{P(A) \cdot P\left(\frac{E}{A}\right)+P(B) \cdot P\left(\frac{E}{B}\right)}$
$+P(C) \cdot \bar{P}\left(\frac{E}{C}\right)+P(D) \cdot P\left(\frac{E}{D}\right)$
$=\frac{\frac{1}{7} \cdot \frac{7}{9}}{\frac{1}{7} \cdot \frac{7}{9}+\frac{3}{7} \cdot \frac{8}{9}+\frac{2}{7} \cdot \frac{5}{9}+\frac{1}{7} \cdot \frac{8}{9}}$
$=\frac{7}{7+24+10+8}$
$=\frac{7}{49}=\frac{1}{7}$