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: $$ p(x)= \frac{e^{-10} \cdot 10^x}{x!}$$
Best Answer
Whenever you see a problem that says something like "k times out of the next n days" you should probably think that the binomial distribution is going to come in to play.
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$