Conditional Probability Situation 
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?
 A: 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
A: 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.
$$\mathsf P(C_2\mid C_1) {= \dfrac{\mathsf P(C_1,C_2)}{\mathsf P(C_1)}\\~~\vdots} $$
