Question

A person goes to office either by car, scooter, bus or train, probability of which being $\frac{1}{7},\frac{3}{7},\frac{2}{7}\text{and}\frac{1}{7}$ respectively. Probability that he reaches office late, if he takes car, scooter, bus or train is $\frac{2}{9},\frac{1}{9},\frac{4}{9}\text{and}\frac{1}{9},$ respectively. Given that he reaches office in time, the probaility 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 $E_1$, $E_2$, $E_3$,$E_4$, be the events that the person goes to office by car, scooter, bus, train respectively. Let $A$ be the event that he reaches office in time. Then $\overline{A}$ be the event he reaches office late.
$P\left(E_1\right)=\frac{1}{7},P\left(E_2\right)=\frac{3}{7},P\left(E_3\right)=\frac{2}{7}\text{and}P\left(E_4\right)=\frac{1}{7}$
Also $P\left(\overline{A}/E_1\right)=\frac{2}{9},P\left(\overline{A}/E_2\right)=\frac{1}{9},P\left(\overline{A}/E_3\right)=\frac{4}{9},P(\overline{A}/E_4=\frac{1}{9})$
$\Rightarrow P\left(A/E_1\right)=\frac{7}{9},P\left(A/E_2\right)=\frac{8}{9},P\left(A/E_3\right)=\frac{5}{9},P\left(A/E_4\right)=\frac{8}{9}$
Now, $P\left(E_1/A\right)=\frac{P\left(E_1\right).P\left(A/E_1\right)}{\sum P\left(E_i\right).P\left(A/E_i\right)}\text{where}(i=1,2,3,4)$
$=\left(\frac{(\frac{1}{7}\times \frac{7}{9})}{(\frac{1}{7}\times \frac{7}{9})+(\frac{3}{7}\times \frac{8}{9})+(\frac{2}{7}\times \frac{5}{9})+(\frac{1}{7}\times \frac{8}{9})}\right)=\left[\frac{7}{7+24+10+8}\right]=\frac{7}{49}=\frac{1}{7}$