Application of law of total probability to binomial random variables An actuary has done an analysis of all policies that cover two cars. $70 \%$ of the policies are of type A for both cars, and $30 \%$ of the policies are of type B for both cars. The number of claims on different cars across all policies are mutually independent. The distributions of the number of claims on a car are given in the following table.

Four policies are selected at random. Calculate the probability that exactly one of the four policies has the same number of claims on both covered cars.
Attempt:
Let $N$ be the number of claims. Let $X$ be the event that exactly one of the 4 policies has same number of claims. We want to find $P(X)$. Notice we have
$$ P(X) = P(X|A)P(A) + P(X|B)P(B)$$
We know $P(A) = 0.7$ and $P(B) = 0.3$. Now, we need to find $P(X|A)$ and $P(X|B)$. Now, we find $P(X|A)$. We have
$$ P(X|A) = \sum_{i=0}^4 P([X|A] | N=i) P(N=i) $$
And $[X|A]|N$ is $binomial$ with parameters $n=3$ and $p = P(N=i)$. so for example,
$$ P([X|A]|N=0) = {3 \choose 0} 0.4^0 0.6^3 \approx 0.216 $$
$$ P([X|A]|N=1) = {3 \choose 1} 0.4^1 0.6^2 \approx 0.432 $$
$$P([X|A]|N=2) = {3 \choose 2} 0.4^2 0.6^1 \approx 0.288$$
$$P([X|A]|N=3) = {3 \choose 3} 0.4^3 0.6^0 \approx 0.064$$
Now, we use same procedure to find $P(X|B)$.
Qs:is this a correct approach to solve this problem?
 A: Well, the notation should be $\mathsf P(X\mid A, N=k)$.   There is only ever one "pipe" dividing the events list from the conditions list.   It is not a set operator.   It doesn't make sense to talk about "an event when given a condtion when given another condition."   It is just "an event when given a junction of conditions."
But you need not use any such thing at all.   There is no conditional binomial distribution involved.
Next you should notice that the event $X$ is the event that the selected policy has the same number of claims on both cars covered; take note: each policy covers two cars.   Look in the box; you are given the distribution of the number of claims on a car for a type of policy.   Since the number of claims on each car is independent for a given policy...
$$\begin{split}\mathsf P(X\mid A)&=\sum_{k=0}^4\mathsf P(N_1=k\mid A)~\mathsf P(N_2=k\mid A) \\ &= 0.4^2+0.3^2+0.2^2+0.1^2\end{split}$$
And so on.
A: Let $p$ be probability that one policy has the same number of claims on both covered cars. Then the probability that exactly one of the four policies has the same number of claims on both covered cars is ${4 \choose 1}p(1-p)^3.$
Then $p$ is given by
$p = 0.7(0.4^2+0.3^2+0.2^2+0.1^2)+0.3(0.25^2\cdot4)=0.285.$
So the answer must be $\approx 0.417$.
