Question

A person goes to office either by car,scooter , bus or train, the probability if which being 1/7,3/7,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 reaches office in time, then what is the probability that he travelled by a car.

Answer 1

Let us define the following events

C: person goes by car

S: person goes by scooter

B: person goes by bus

T: person goes by train

L: person reaches late

Then,we are given in the question

$P\left(C\right)=\frac{1}{7},P\left(S\right)=\frac{3}{7},P\left(B\right)=\frac{2}{7},P\left(T\right)=\frac{1}{7}$

$P\left(L/C\right)=\frac{2}{9},P\left(L/S\right)=\frac{1}{9},P\left(L/B\right)=\frac{4}{9},P\left(L/T\right)=\frac{1}{9}$

We have to find $P\left(C/\overline{L}\right)$ [Since reaches in time$\equiv$ not late].Using Bayes's theorem,

$P\left(C/\overline{L}\right)=\frac{P\left(\overline{L}/C\right)P\left(O\right)}{P\left(\overline{L}/C\right)P\left(C\right)+P\left(\overline{L}/S\right)P\left(S\right)+P\left(\overline{L}/B\right)+A\left(\overline{L/T}\right)P\left(T\right)}$.................(1)

Now,

$P\left(\overline{L}/C\right)=1-\frac{2}{9}=\frac{7}{9},P\left(\overline{L}/S\right)=1-\frac{1}{9}=\frac{8}{9}$

$P\left(\overline{L}/B\right)=1-\frac{4}{9}=\frac{5}{9},P\left(\overline{L}/T\right)=1-\frac{1}{9}=\frac{8}{9}$

Substituting these values in eq.1, we get

$P\left(C/\overline{L}\right)=\frac{\frac{7}{9}\times \frac{1}{7}}{\frac{7}{9}\times \frac{1}{7}+\frac{8}{9}\times \frac{3}{7}+\frac{5}{9}\times \frac{2}{7}+\frac{8}{9}\times \frac{1}{7}}$

$=\frac{7}{7+24+10+8}=\frac{7}{49}=\frac{1}{7}$