The number of accidents that a person has in a given year is a Poisson random variable with mean $. However, suppose that the value of $ changes from person to person, being equal to 2 for 60 percent of the population and 3 for the other 40 percent. If a person is chosen at random, what is the probability that he will have (a) 0 accidents and (b) exactly 3 accidents in a certain year? What is the conditional probability that he will have 3 accidents in a given year, given that he had no accidents the preceding year?

a) 0.10112

b) 0.1979

c) 0.189

Step-by-step explanation:

Let X be number of accidents a person has in a year.

λ = 2

Pr (λ = 2) = 60%

Pr (λ = 2) = 0.6

λ = 3

Pr (λ = 3) = 40%

Pr (λ = 3) = 0.4

a) Let the probability that there are 0 accidents be Pr (X=0)

Pr (X) = (λ^X) (e^ - λ) / X!

Pr (X=0) = (λ^0) (e^ - λ) / 0!

Pr (X=0) = e^ - λ

Pr (X=0) = Pr[X=0| λ = 2] + Pr[X=0| λ = 3]

= 0.6e^ - 2 + 0.4e^ - 3

= 0.10112

b) Let the probability that exactly 3 accidents in a certain year = Pr (X=3)

Pr (X=3) = (λ^3) (e^ - λ) / 3!

= (λ^3) (e^ - λ) / 6

Pr (X=3) = Pr (X=3| λ=2) + Pr (X=3| λ=3)

= 0.6 [ (2^3) (e^ - 2) / 6] + 0.4 [ (3^3) (e^ - 3) / 6]

= 0.6[ (8e^-2) / 6] + 0.4[ (27e^-3) / 6]

= 0.1979

c) Let the conditional probability that he will have 3 accidents in a given year, given that he had no accidents the preceding year =

Pr[X=3|he had no accident the preceding year]

= Pr (X=3, X = 0) / Pr (X=0)

= [0.6 [ (8e^ - 2) / 6) e^-2 ] + 0.4 [ (27e^ - 3) / 6) e^-3]] / 0.10112

= 0.019114/0.10112

= 0.189