Conditional Probability Situation

Question

An auto insurance compnay divides its customers into 2 types, $A$ and $B$. Type $A$ customers have a probability of $1/3$ of making a claim in a given year, while type $B$ have a probability of $1/8$ of making a claim in a given year.

The probability that a random customer is type $A$ is $0.25$, which implies that the probability of a random customer being type $B$ is $0.75$.

Suppose a customer makes a claim in 2001. Find the probability that he will make a claim in 2002.

This seems like it should be relatively straighforward, but I'm getting stuck. Essentially what I'm interested in finding is this:

$P($customer makes claim in 2002$|$customer made claim in 2001$)=\frac{P(2002\cap2001)}{P(2001)}$

where I just used the years as shorthand. But I'm confused as to how to calculate the numerator of this expression. How would you do this?

Answer 1

Graham makes a good point. Assume that the customer stays the same type each year and we have independence between a customer making a claim between years.

Then we have

$$\begin{align*} P(2002|2001) &=\frac{P(2002\cap2001)}{P(2001)}\\\\ &=\frac{\left(\frac{1}{4}\cdot\left(\frac{1}{3}\right)^2\right)+\left(\frac{3}{4}\cdot\left(\frac{1}{8}\right)^2\right)}{\left(\frac{1}{4}\cdot\frac{1}{3}\right)+\left(\frac{3}{4}\cdot\frac{1}{8}\right)}\\\\ &\approx0.223 \end{align*}$$

where the numerator comes from the probability of them being a given type and then the probability of making a claim in two consecutive years

Answer 2

Use $C_1, C_2$ for the event of a claim by the same customer in 2001, 2002. Assume the probability of making claims is independent over the years, when given the customer type. Let $A,B$ the complementary events of the customer types.

You have $\mathsf P(A,C_1)=\mathsf P(A)\;\mathsf P(C_1\mid A) =\tfrac 14\cdot\tfrac 13,\\ \mathsf P(A,C_1,C_2)=\mathsf P(A)\;\mathsf P(C_1\mid A)\;\mathsf P(C_2\mid A)= =\tfrac 14\cdot(\tfrac 13)^2, \\\mathsf P(B,C_1)=\mathsf P(B)\;\mathsf P(C_1\mid B)=\tfrac 34\cdot\tfrac 18\\ \mathsf P(B,C_1,C_2)=\mathsf P(B)\;\mathsf P(C_1\mid B)\;\mathsf P(C_2\mid B)=\tfrac 34\cdot(\tfrac 18)^2$

Now find $\mathsf P(C_2\mid C_1)$ using the Definition of Conditional Probability and the Law of Total Probability.

$$\mathsf P(C_2\mid C_1) {= \dfrac{\mathsf P(C_1,C_2)}{\mathsf P(C_1)}\\~~\vdots} $$