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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:

$B$ : event that a person has an accident

$B'$ : event that a person has no accident

$A_1$ : event that a person is good risk

$A_2$ : event that a person is average risk

$A_3$ : event that a person is bad risk

Now, $P(A_1)=0.20$, $P(A_2)=0.50$, $P(A_3)=0.30$, $P(B|A_1)=0.05$, $P(B|A_1)=0.15$ and $P(B|A_3)=0.30$

\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,

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

\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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