Answer:
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
