1.

An insurance company insured 2000 scooter drivers. 4000, car drivers and 6000 truck driver. The probability of an accident involving a scooter, a car and a truck is 1/00,3/100and 3/20 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver ?

Answer»

SOLUTION :Total number of persons insured `=(2000+4000+6000)=12000`.
Let `E_1,E_2andE-3` be the events of choosing a scooter driver, a car driver and truck driver respectively. Then,
`P(E_1)=2000/12000=1/6,P(E_2)=4000/12000=1/3andP(E_3)=6000/12000=1/2`.
Let E be the event of an insured PERSON meeting with an accident.
Then ,
`P(E//E_1)` = probability that an insured person meets with an accident, given that he is a scooter driver
`=1/100`
Similarly, `P(E//E_2)=3/100andP(E//E_3)=3/20`.
REQUIRED probability
`=P(E_1//E)` [by BAYES's theorem]
= probability of choosing a scooter driver, given that he meets with an accident
`=(P(E//E_1).P(E_1))/(P(E//E_1).P(E_1)+P(E//E_2).P(E_2)+P(E//E_3).P(E_3))`
`=((1/100xx1/6))/((1/100xx1/6)+(3/100xx1/3)+(3/20xx1/2))=1/52`.
Hence, the required probability is `1/52`.


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