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.

## Best Answer

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.