# The number of traffic accidents on successive days are independent Poisson random variables with mean 2.

The number of traffic accidents on successive days are independent Poisson random variables with mean 2.

Find the probability that 3 of the next 5 days have two accidents.

Would I treat this the same as if it asked the probability that 6 accidents occur over the next 5 days? And if not, how should I approach this probability?

This is what I have so far: $$p(x)= \frac{e^{-10} \cdot 10^x}{x!}$$

• What this problem is asking for is the probability that there will be exactly two accidents on 3 of the next 5 days. This is not the same as saying 6 over the next five days. Saying that there is going to be 2 accidents on three of the next five days does not mean that only 6 accidents will occur, it just means that on three of the next five days 2 accidents must occur and for the other two days anything can happen. For example days 1, 2, and 3 could have 2 accidents but days 5 and 6 could each have 45 accidents. – alpastor Sep 7 '17 at 4:17
• Thanks, I appreciate the clarification. Any idea how to calculate specifically two accidents occurring on 3 out of the next 5 days? – varcharvi Sep 7 '17 at 4:24

First let $Y \text{~Poi}(2)$, which is the random variable that represents the number of accidents per day, and let $X \text{~Bin}(5,p)$ where $X$ is the random variable that represents the number of days out of the next 5 days that we have two accidents in one day.
We need to find $p$ for $X$ which is the probability of success. A success is that there is 2 accidents in one day so this must mean that $p = P(Y=2) = \dfrac{e^{-2}\cdot2^2}{2!}$.
Now we want to find $P(X=3) = {5 \choose 3}p^3(1-p)^2$