Let the events E_{1}, E_{2}, E_{3}, E_{4,} and A be defined as follows:

E_{1}: Event that the person goes to the office by car

E_{2}: Event that the person goes to the office by scooter

E_{3}: Event that the person goes to the office by bus

E_{4}: Event that the person goes to the office by train.

A: Event that the person reaches the office in time

Then, **P(E**_{1}) = \(\frac{1}{7}\) ,P(E_{2}) = \(\frac{3}{7}\) , P(E_{3}) = \(\frac{2}{7}\) , P(E_{4}) = \(\frac{1}{7}\)

P(A/E_{1}) = 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/E_{2}) = 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/E_{3}) = 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/E_{4}) = 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(E_{1}/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**

**\(\frac{\frac{1}{7}\times \frac{7}{9}}{\frac{1}{7}\times \frac{7}{9}+ \frac{3}{7}\times \frac{8}{9}+ \frac{2}{7}\times \frac{5}{9}+ \frac{1}{7}\times \frac{8}{9}}\)** = **\(\frac{\frac{7}{63}}{\frac{7}{63}+\frac{24}{63}+\frac{10}{63}+\frac{8}{63}}\)** = **\(\frac{\frac{7}{63}}{\frac{49}{63}}\)** = **\(\frac{7}{49} =\frac{1}{7}\)**