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Q. 7. 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 is a scooter driver?

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Let A : scooter drivers = 2000

B : car drivers = 4000

C : truck drivers = 6000

Total drivers = 12000

$P(A)=\frac{2000}{12000}=\frac{1}{6}=0.16$

$P(B)=\frac{4000}{12000}=\frac{1}{3}=0.33$

$P(C)=\frac{6000}{12000}=\frac{1}{2}=0.5$

D : the event that person meets with an accident.

$P(D|A)= 0.01$

$P(D|B)= 0.03$

$P(D|C)= 0.15$

$P(A|D)=\frac{P(A).P(D|A)}{P(B).P(D|B)+P(A).P(D|A)+P(C).P(D|C)}$

$P(A|D)= \frac{0.16\times 0.01}{0.16\times 0.01+0.33\times 0.03+0.5\times 0.15}$

$P(A|D)= \frac{0.0016}{0.0016+0.0099+0.075}$

$P(A|D)= \frac{0.0016}{0.0865}$

$P(A|D)= 0.019$

Posted by

seema garhwal

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