An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 track drivers. The probabilities of an accident involving a scooter driver, car driver and a truck driver are 0.01, 0.03 and 0.15 respectively. One of the insured person meets with an accident. What is the probability that he is a scooter driver?

Explanation

Let `E_(1),E_(2), and E_(3)` be the respectively events that the driver is a scooter driver, a car driver, and a truck driver.
Let A be the event that the person meets with an accident.
There are 2000 scooter drivers, 4000 car drivers, and 6000 truck drivers.
Total number of drovers `=2000+4000+6000=12000`
`P(E_(1))=P` (drive is a scooter driver) `=2000/12000=1/6`
`P(E_(2))=P` (driver is a car driver) `4000/12000=1/3`
`P(E_(3))=P` (drivers is a car driver) `=6000/12000=1/2`
`P(A//E_(1))=P` (sector driver met with an accident) `=0.01=1/100`
`P(A//E_(3))=P` (truck driver met with an acciden) `=0.15=15/100`
The probability that the driver is a scooter driver, given that he met with an accident, is given by `P(E_(1)//A).`
By using Bayes' theorem, we obtain
`P(E_(1)//A)=(P(E_(1)).P(A//E_(1)))/(P(E_(1)).P(A//E_(1))+P(E_(2)).P(A//E_(2)))`
`+P(E_(3)).P(A//E_(3))`
`(1/6xx1/100)/(1/6xx1/100+1/3xx3/100+1/2xx15/100)`
`=(1/6xx1/100)/(1/100(1/6+1+15/2))=1/52`