A person goes to office either by car, scooter, bus or train, the probability of which being 1/7, 3/7, 2/7 and 1/7, respectively. Probability that he reaches office late, if he taks car, scooter, bus or train is 2/9, 1/9, 4/9 and 1/9 respectively. Given that he reached office in time, then what is probability that he traveled by a car.

Answer

A

Explanation

As, the statement 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)=(1)/(7),P(S)=(3)/(7),P(B)=(2)/(7),P(T)=(1)/(7)`
Again, L be the event of the person reaching office late.
`therefore barL` be the event of the person reaching office in time.
Then, `P((barL)/(C))=(7)/(9),P((barL)/(S))=(8)/(9),P((barL)/(B))=(5)/(9)`
and `P((barL)/(T))=(8)/(9)`
`therefore P((C)/(L))=(P((barL)/(C))*P(C))/(P((barL)/(C))*P(C)+P((barL)/(S))*P(S)+P((barL)/(B))*P(B)+((barL)/(T))*P(T))`
`=((7)/(9)xx(1)/(7))/((7)/(9)xx(1)/(7)+(8)/(9)xx(3)/(7)+(5)/(9)xx(2)/(7)+(8)/(9)xx(1)/(7))=(1)/(7)`