Question

On average, 3 traffic accidents per month intersection. What is the probability that in any given month at this intersection between 3 and 4 accidents occur?

Answer 1

Let $X$ be the random variable representing the number of accidents in a month: $X\sim Po(\lambda t)$

$$P(X=x)=\dfrac{e^{-\lambda t}(\lambda t)^x}{x!}$$

Average number of accidents per month, $\lambda=3, t=1.$

Then $\lambda t=3(1)=3$ and

$$P(X=x)=\dfrac{e^{-3}(3)^x}{x!}$$

$$P(3\leq X\leq4)=P(X=3)+P(X=4)=$$

$$=\dfrac{e^{-3}(3)^3}{3!}+\dfrac{e^{-3}(3)^4}{4!}=\dfrac{63e^{-3}}{8}\approx0.392073$$

The probability that in any given month at this intersection between 3 and 4 accidents occur is $0.392073.$