Conditional probability in Poisson distribution The number of accidents that a person has in a given year is a Poisson random variable with mean $\lambda$. However we may suppose that the rate $\lambda$ depends on the person or proportion of the population. Suppose that $\lambda = 4$ for 65% of the population and $\lambda = 5$ for other 35% of the population. What is the conditional probability that a person will have 3 accidents in a given year, given that he/she had no accidents in the preceding year?
$\textbf{Answer:}$
Let's define two events for a person A = 3 accidents in a given year and B = No accident in the last year.
So, we want to calculate $P(A|B)$. Now, $P(\text{k accidents in 1 year}) = \frac{(rt)^ke^{-rt}}{k!}$ with $\lambda =rt $. Since, time is 1 year here, $\lambda = r$.
Now, the events in poisson distribution happen independently, i.e the number of accidents happening this year will not be affected by the number of accidents in last year, and hence   $P(A|B) = P(A)$
$$ P(A) =  0.65 \times \frac{4^3e^{-4}}{3!} + 0.35 \times \frac{5^3e^{-5}}{3!} = 0.17612$$
But, this answer is being marked incorrect, Is the solution incorrect? why? If so, then to solve this conditional probability, we will have to take the usual approach, i.e
$$ P(A|B) = \frac{P(A\cap B)}{P(B)} $$
How to calculate the $ P(A\cap B) $ then?
Thanks for the help.
 A: I think the problem is your assumption of independence is at the wrong place.
Think of the situation as a coupled experiment. First, you choose $\lambda$ randomly according to the mentioned probabilities. Then, the number of accidents in year $j$ is a $\text{Poisson}(\lambda)$-distributed random variable $X_j$.
Intuitively, if the accidents in year $1$ were $0$, then it is more likey, that $\lambda$ is $4$ not $5$.
So $A = \{X_2 = 3\}$ and $B = \{X_1 = 0\}$ are not independent, they're only independent given that $\lambda = 4$ or $\lambda = 5$.
More formally:
$$
\mathbb{P}(A \cap B) = \mathbb{P}(A \cap B \, \vert \, \lambda = 4)\mathbb{P}(\lambda = 4) + \mathbb{P}(A \cap B \, \vert \, \lambda = 5)\mathbb{P}(\lambda = 5) \\
= \mathbb{P}(A \, \vert \, \lambda = 4)\mathbb{P}(B \, \vert \, \lambda = 4)\mathbb{P}(\lambda = 4) 
+ \mathbb{P}(A \, \vert \, \lambda = 5)\mathbb{P}(B \, \vert \, \lambda = 5)\mathbb{P}(\lambda = 5) 
$$ and
$$
\mathbb{P}(B) = \mathbb{P}(B \, \vert \, \lambda = 4)\mathbb{P}(\lambda = 4) + \mathbb{P}(B \, \vert \, \lambda = 5)\mathbb{P}(\lambda = 5).
$$.
Now just compute using the Poisson distribution as you did above.
