Question

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

Answer 1

As, the statem ent shows problem is to be related to Baye's law.
Let C,S,B,T be the events when person is going by car, scooter, bus or train, respectively.
$\therefore P(C)=\frac{1}{7},P(S)=\frac{3}{7},P(B)=\frac{2}{7},P(T)=\frac{1}{7}$
Again, $L$ be the event of the person reaching office late
$\therefore \overline{L}$ be the event of the person reaching office in time.
Then ,$P\left(\frac{\bar{L}}{C}\right)=\frac{7}{9},P\left(\frac{\bar{L}}{S}\right)=\frac{8}{9},P\left(\frac{\bar{L}}{B}\right)=\frac{5}{9}$ and $P\left(\frac{\bar{L}}{T}\right)=\frac{8}{9}$
$P\left(\frac{C}{L}\right)=\frac{P\left(\frac{\bar{L}}{C}\right).P(C)}{P\left(\frac{\bar{L}}{C}\right).P(C)+P\left(\frac{\bar{L}}{S}\right).P(S)+P\left(\frac{\bar{L}}{B}\right).P(B)+P\left(\frac{\bar{L}}{T}\right).P(T)}$
$=\frac{\frac{7}{9}\times \frac{1}{7}}{\frac{7}{9}\times \frac{1}{7}+\frac{8}{9}\times \frac{3}{7}+\frac{5}{9}\times \frac{2}{7}\frac{8}{9}\frac{1}{7}}=\frac{1}{7}$