Let the events E1, E2, E3, E4, and A be defined as follows:
E1: Event that the person goes to the office by car
E2: Event that the person goes to the office by scooter
E3: Event that the person goes to the office by bus
E4: Event that the person goes to the office by train.
A: Event that the person reaches the office in time
Then, P(E1) = \(\frac{1}{7}\) ,P(E2) = \(\frac{3}{7}\) , P(E3) = \(\frac{2}{7}\) , P(E4) = \(\frac{1}{7}\)
P(A/E1) = P(Person reaches office in time if he goes by car)
= 1 – P(Person reaches office late if he goes by car)
= \(1- \frac{2}{9}=\frac{7}{9}\)
= P(A/E2) = P(Person reaches office in time if he goes by scooter)
= 1 – P(Person reaches office late if he goes by scooter) =
\(1- \frac{1}{9} =\frac{8}{9}\)
P(A/E3) = P(Person reaches office in time if he goes by bus)
= 1 – P(Person reaches office late if he goes by bus)
\(1-\frac{4}{9}=\frac{5}{9}\)
P(A/E4) = P(person reaches office in time if he goes by train)
= 1 – P(person reaches office late if he goes by train)
= \(1-\frac{1}{9}=\frac{8}{9}\)
Now, P(person travelled by car if he reached office in time)
P(E1/A) = \(\frac {P(E_1)\times P(A/E_1)}{P(E_1)\times P(A/E_1)+ P(E_2)\times P(A/E_2)+P(E_3)\times P(A/E_3)+P(E_4)\times P(A/E_4)}\) .... Baye's Theorem
\(\frac{\frac{1}{7}\times \frac{7}{9}}{\frac{1}{7}\times \frac{7}{9}+ \frac{3}{7}\times \frac{8}{9}+ \frac{2}{7}\times \frac{5}{9}+ \frac{1}{7}\times \frac{8}{9}}\) = \(\frac{\frac{7}{63}}{\frac{7}{63}+\frac{24}{63}+\frac{10}{63}+\frac{8}{63}}\) = \(\frac{\frac{7}{63}}{\frac{49}{63}}\) = \(\frac{7}{49} =\frac{1}{7}\)