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Question

Suppose that an insurance company classifies people into one of three classes — good risks, average risks, and bad risks. Their records indicate that the probabilities that good, average, and bad risk persons will be involved in an accident over a 1-year span are, respectively, .05, .15, and .30. If 20 percent of the population are “good risks,” 50 percent are “average risks,” and 30 percent are “bad risks,” what proportion of people have accidents in a fixed year? If policy holder A had no accidents in 1987, what is the probability that he or she is a good (average) risk?

Solution

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Define:

BB : event that a person has an accident

BB' : event that a person has no accident

A1A_1 : event that a person is good risk

A2A_2 : event that a person is average risk

A3A_3 : event that a person is bad risk

Now, P(A1)=0.20P(A_1)=0.20, P(A2)=0.50P(A_2)=0.50, P(A3)=0.30P(A_3)=0.30, P(BA1)=0.05P(B|A_1)=0.05, P(BA1)=0.15P(B|A_1)=0.15 and P(BA3)=0.30P(B|A_3)=0.30

P(B)=P(A1)P(BA1)+P(A2)P(BA2)+P(A3)P(BA3)=0.20×0.05+0.50×0.15+0.30×0.30=0.01+0.075+0.09=0.175\begin{align*} P(B)&=P(A_1)P(B|A_1)+P(A_2)P(B|A_2)+P(A_3)P(B|A_3)\\ &=0.20 \times 0.05+0.50 \times 0.15+0.30 \times 0.30\\ &=0.01+0.075+0.09\\ &=0.175 \end{align*}

So,

P(B)=1P(B)=10.175=0.825P(B')=1-P(B)=1-0.175=0.825

P(A1B)=P(A1)P(BA1)P(B)=0.20×(10.05)0.825=0.190.825=0.2303\begin{align*} P(A_1|B')&=\frac{P(A_1)P(B'|A_1)}{P(B')}\\ &=\frac{0.20 \times (1-0.05)}{0.825}\\ &=\frac{0.19}{0.825}\\ &=0.2303 \end{align*}

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