An automobile insurance company divides customers into three categories, good risks, medium risks, and poor risks. Assume that 70% of the customers are good risks, 20% are medium risks, and 10% are poor risks. Assume that during the course of a year, a good risk customer has probability 0.005 of filing an accident claim, a medium risk customer has probability 0.01, and a poor risk customer has probability 0.025. A customer is chosen at random.
Given that the customer has filed a claim, what is the probability that the customer is a good risk?
I think the total probability that the customer has filed a claim is (0.7*0.005+0.2*0.01+0.1*0.025) So the probability that the customer is a good risk is (0.7*0.005)/(0.7*0.005+0.2*0.01+0.1*0.025).
Am I right?
$\endgroup$2 Answers
$\begingroup$You are correct, this is a direct application of Bayes' theorem. You just need to finish the math:
$$\begin{split} P(Good|Claim) &= \frac{P(Good \cap Claim)}{P(Claim)} \\ &= \frac{0.7 \cdot 0.005}{0.7 \cdot 0.005 + 0.2 \cdot 0.01 + 0.1 \cdot 0.025} \\ &= \frac{0.0035}{0.008} \\ &= 0.4375 \end{split}$$
$\endgroup$ $\begingroup$We note that we can apply Bayes' Theorem here:
$$P(A \mid B)=\frac{P(B \mid A)P(A)}{P(B)}$$
So in our problem, we wish to find:
$$P(\text{good} \mid \text{claimed})=\frac{P(\text{claimed} \mid \text{good})P(\text{good})}{P(\text{claimed})}$$
So we note that the question gives us two of these quantities right from the start. We have: $P(\text{claimed}\mid \text{good})= 0.005$ and $P(\text{good})= 0.7$. We can now calculate the probability that a claim is made:
$$P(\text{claimed}) = 0.7\times 0.005 + 0.2\times 0.01 + 0.1 \times 0.025 = 0.008$$
Thus:
$$P(\text{good}\mid \text{claimed})=\frac{0.005 \times 0.7}{0.008} = 0.4375$$
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