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
Text Solution
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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`
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