# Exam P Practice Problem 6

Problem 6a
An auto insurer offers collison coverage to two large groups of policyholders, Group 1 and Group 2. On the basis of historical data, the insurer has determined that the loss due to collision for a policyholder in Group 1 has an exponential distribution with mean 5. On the other hand, the loss due to collision for a policyholder in Group 2 has an exponential distribution with mean 10.

Considering the two groups as one block, about 75% of the losses are from Group 1.

1. Given a randomly selected loss in this block, what is the probability that the loss is greater than 15?
2. If a randomly selected loss is greater than 15, what is the probability that it is a from a policyholder in Group 1?

Problem 6b
An auto insurer has two groups of policyholders – those considered good risks and those considered bad risks. On the basis of historical data, the insurer has determined that the number of car accidents during a policy year for a policyholder classified as good risk follows a binomial distribution with $n=2$ and $p=\frac{1}{10}$. The number of car accidents for a policyholder classified as bad risk follows a binomial distribution with $n=2$ and $p=\frac{3}{10}$. In this block of policies, 75% are classified as good risks and 25% are classified as bad risks. A new customer, whose risk class is not yet known with certainty, has just purchased a new policy.

1. What is the probability that this new policyholder is not accident-free in the upcoming policy year?
2. By the end of the policy year, it is found that this policyholder is not accident-free, what is the probability that the policyholder is a “good risk” policyholder?

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Soultion is found below.

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Solution to Problem 6a
Let $X$ be the loss amount of a randomly selected policyholder. The conditional probabilities of a loss greater than 7.5 are:

$\displaystyle P(X>15 \lvert \text{ Group 1 Policyholder})=e^{-\frac{15}{5}}$
$\displaystyle P(X>15 \lvert \text{ Group 2 Policyholder})=e^{-\frac{15}{10}}$

By the law of total probability, the unconditional probability is:

\displaystyle \begin{aligned}P(X>15)&=P(X>15 \lvert \text{ Group 1 Policyholder}) \times P(\text{ Group 1 Policyholder}) \\&\ \ \ +P(X>15 \lvert \text{ Group 2 Policyholder}) \times P(\text{ Group 2 Policyholder}) \\&=\frac{3}{4} \times e^{-\frac{15}{5}}+\frac{1}{4} \times e^{-\frac{15}{10}} \\&=\frac{3}{4} \times e^{-3}+\frac{1}{4} \times e^{-1.5} \\&=0.0931228413 \end{aligned}

The above calculation indicates that the unconditional probability is the weighted average of the conditional probabilities. The answer to the second question is obtained by applying the Bayes’ theorem:

\displaystyle \begin{aligned}P(\text{Group 1 Policyholder } \lvert X>15)&=\frac{P[(\text{Group 1 Policyholder}) \cap (X>15)]}{P(X>15)} \\&=\frac{\frac{3}{4} \times e^{-3}}{\frac{3}{4} \times e^{-3}+\frac{1}{4} \times e^{-1.5}} \\&=0.400978973 \end{aligned}

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$\displaystyle 6b.1 \ \ \ \ \frac{108}{400}=0.27$
$\displaystyle 6b.2 \ \ \ \ \frac{57}{108}=0.5278$