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 $P\left(\frac{\bar{E}}{D}\right)=\frac{1}{9}$
$\therefore P\left(\frac{A}{E}\right)=\frac{P(A)\cdot P\left(\frac{E}{A}\right)}{P(A)\cdot P\left(\frac{E}{A}\right)+P(B)\cdot P\left(\frac{E}{B}\right)}$
$+P(C)\cdot \bar{P}\left(\frac{E}{C}\right)+P(D)\cdot P\left(\frac{E}{D}\right)$
$=\frac{\frac{1}{7}\cdot \frac{7}{9}}{\frac{1}{7}\cdot \frac{7}{9}+\frac{3}{7}\cdot \frac{8}{9}+\frac{2}{7}\cdot \frac{5}{9}+\frac{1}{7}\cdot \frac{8}{9}}$
$=\frac{7}{7+24+10+8}$
$=\frac{7}{49}=\frac{1}{7}$