On average, 3 traffic accidents per month occur at a certain intersection. What is the probability that in any given month at this intersection (a) exactly 5 accidents will occur? (b) fewer than 3 accidents will occur? (c) at least 2 accidents will occur?

We will assume that the number of accidents at the intersection follows a Poisson distribution with a mean (\mu) of 3 per month

$a)P[X=5]=p(5)=\frac{e^{-\mu }{\mu }^x}{x!}=\frac{e^{-3}{3}^5}{5!}=0.10082\approx 0.10$

$b)P[X<3]=F(2)=\sum _{x=0}^2\frac{e^{-\mu }{\mu }^x}{x!}=\sum _{x=0}^2\frac{e^{-3}{3}^x}{x!}=0.4232$

$c)P[X\ge 2]=1-P[X<2]=1-P[X\le 1]=1-F[1]=1-(\frac{e^{-3}{3}^0}{0!}+\frac{e^{-3}{3}^1}{1!})=0.800852\approx 0.8$

a) 

λ=3, t=1, e=2.71828, X=5

=(e^(-3))*(3^(5)) ÷ 5!

=12.0982576134 ÷ 120

=0.10081881344 

b)

X=0,1,2 as X<3

p(0;3)+p(1;3)+p(2;3)

p(0;3) = (e^(-3))*(3^(0)) ÷ 0! = 0.04978706836 / 1 = 0.04978706836

p(1;3) = (e^(-3))*(3^(1)) ÷ 1! = 0.1493612051 

p(2;3) = (e^(-3))*(3^(2)) ÷ 2! = 0.44808361531 / 2 = 0.22404180765 

p(0;3) + p(1;3) + p(2;3) = 0.04978706836 + 0.1493612051 + 0.22404180765 = 0.42319008111 

c)

X>=2

[1- ( p(0;3) + p(1;3) + p(2;3) ) ] 

As I already derived the values for p(0;3), p(1;3) & p(2;3)

I can directly solve it :

1 - 0.42319008111 = 0.57680991889