An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he i

Options

A. `(1)/(52)`
B. `(3)/(52)`
C. `(15)/(52)`
D. `(19)/(52)`

Answer

A

Explanation

Let `E_(1), E_(2), E_(3)` & A be events defined as follows.
`E_(1)`=person chosen is a scooter driver
` E_(2)`= person chosen is a car driver.
`E_(3)`= person chosen is a truck driver & A=person meets with an accident
`P(E_(1))=(2000)/(12000)=(1)/(6), P(E_(2))& =(1)/(3)P(E_(3))=(1)/(2)`
Probability that a person meet with an accident given that he is a scooter driver =`p(A/E_(3))` =0.01
`P(A/E_(2))=0.03 & P(A/E_(3))=0.15`
the person meets with an accident.
the probability that he was a scooter driver
`=P(E_(1)/A)=((1)/(6)xx0.01)/(((1)/(6xx0.01))+((1)/(3)xx0.03)+((1)/(2)xx0.015))=(1)/(52)`