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A person goes to office by car, scooter, bus and 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
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The correct answer is:
$1 / 7$
Let the event of person goes to office by a car, scooter, bus or train be $A, B, C$ and $D,$ respectively. We have, $P(A)=\frac{1}{7}, P(B)=\frac{3}{7}, P(C)=\frac{2}{7}$
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 $P\left(\frac{\bar{E}}{D}\right)=\frac{1}{9}$
$\begin{aligned} P\left(\frac{A}{E}\right)=& P(A) \cdot P\left(\frac{E}{A}\right) \\ &+P(C) \cdot P\left(\frac{E}{C}\right)+P(D) \cdot P\left(\frac{E}{D}\right) \\ &=\frac{\frac{1}{7} \cdot \frac{7}{9}}{P(A) \cdot P\left(\frac{E}{A}\right)+P(B) \cdot P\left(\frac{E}{B}\right)} \\ &=\frac{1}{7} \cdot \frac{7}{9}+\frac{3}{7} \cdot \frac{8}{9}+\frac{2}{7} \cdot \frac{5}{9}+\frac{1}{7} \cdot \frac{8}{9} \\=& \frac{7}{7+24+10+8}=\frac{7}{49}=\frac{1}{7} \end{aligned}$
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 $P\left(\frac{\bar{E}}{D}\right)=\frac{1}{9}$
$\begin{aligned} P\left(\frac{A}{E}\right)=& P(A) \cdot P\left(\frac{E}{A}\right) \\ &+P(C) \cdot P\left(\frac{E}{C}\right)+P(D) \cdot P\left(\frac{E}{D}\right) \\ &=\frac{\frac{1}{7} \cdot \frac{7}{9}}{P(A) \cdot P\left(\frac{E}{A}\right)+P(B) \cdot P\left(\frac{E}{B}\right)} \\ &=\frac{1}{7} \cdot \frac{7}{9}+\frac{3}{7} \cdot \frac{8}{9}+\frac{2}{7} \cdot \frac{5}{9}+\frac{1}{7} \cdot \frac{8}{9} \\=& \frac{7}{7+24+10+8}=\frac{7}{49}=\frac{1}{7} \end{aligned}$
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