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Question

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


Solution

Let E1,E2 and E3 be the respective 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 drivers =2000+4000+6000=12000
P(E1)=P(driver is a scooter driver)=200012000=16

P(E2)=P(driver is a car driver)=400012000=13

P(E3)=P(driver is a truck driver)=600012000=12

P(A/E1)=P(scooter driver meets with an accident)=0.01=1100

P(A/E2)=P(car driver meets with an accident)=0.03=3100

P(A/E3)=P(truck driver meets with an accident)=0.15=15100

The probability that the driver is a scooter driver, given that he met with an accident, is given by P(E1/A).
By using Baye's theorem, we obtain
P(E1/A)=P(E1).P(A/E1)P(E1).P(A/E1)+P(E2).P(A/E2)+P(E3).P(A/E3)
=16×110016×1100+13×3100+12×15100=152
152=pq
qp=521=51

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