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The life insurance company issues standard and preferred policies.
Of all policyholders:
60% have standard policy and probability of 0.01 of dying next year.
40% have preferred policy and probabilty of 0.08 of dying next year.
A policyholder dies in the next year.
What is the conditional probability that a deceased having a preferred policy?

I know this could be solved via Bayes Theorem, but I am completely confused where to plug all the number at. Thank you!

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The Bayes Theorem states that:

$P(A|B)=\frac{P(B|A)*P(A)}{P(B)}$

Now for your case let's define $A$ as the event of a person having a preferred policy.

And let's define $B$ as the event of a person dying next year.

From the Bayes Theorem we have:

$P(A|B)=\frac{P(B|A)*P(A)}{P(B)}=\frac{0.08*0.4}{0.08*0.4+0.01*0.6}=\frac{32}{32+6}\simeq 0.84$

Here is a way(not the most mathematical perhaps) to think of it if it confuses you:

You have 1000 people.

600 of them have standard policy. 400 have preferred policy. The ones that are going to die next year are 600*0.01 from the ones with standard policy and 400*0.08 from the ones with preferred policy. So a total of $600*0.01+400*0.08$ people are going to die next year. So given that someone died, the probability of him/her being from the ones with preferred policy is

$\frac{400*0.08}{600*0.01+400*0.08}$

That is what the Bayes Theorem actually says here.

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First define your events, and use them to express what you have been told and seek in symbols.

Let $S$ be the event of a (specified) policy holder having a standard polity, and $D$ the event of that policy holder dying. You are given:

$$\begin{align}\mathsf P(S)&=0.60\\\mathsf P(D\mid S)&=0.01\\\mathsf P(S^\complement)&=0.49\\\mathsf P(D\mid S^\complement)&=0.08\end{align}$$

You seek $\mathsf P(S\mid D)$ and may now find it with the above using Bayes' Rule, and the Law of Total Probability.

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