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

f is the midpoint of