The number of accidents that a person has in a given year is a Poisson random variable with mean $\lambda$̣ However, suppose that the value of $\lambda$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 number of accidents that a person has in a given year is a Poisson random variable with mean$\lambda$.

Let $X$is a number of accidents that a person has in a given year is a Poisson random variable with mean $\lambda$.

$\lambda =2$with $P\{\lambda =2\}=0.6$and $\lambda =3$

With $P\{\lambda =3\}=0.4$

$P\{X=0\}=e^{-\lambda }$.

Substitute

$=0.6·e^{-2}+0.4·e^{-3}$

Add the value,

$=0.10112$.

A number of $0$accidents value found to be$P\{X=0\}=0.10112$.

Exactly $3$accidents in a certain year and given that he had no accidents the preceding year

Exactly$3$accidents in a certain year

Substitute,

$=0.6\frac{2^3{e}^{-2}}{3!}+0.4\frac{3^{3e^{-3}}}{3!}$

$=0.6\frac{8e^{-2}}{6}+0.4\frac{27e^{-3}}{6}$

Simplify,

$=\frac{1}{6}(0.645+0.54)$

$=0.1979$.

$P\{X=3\mid$he had no accidents the preceding year$\}$

$=\frac{P\{X=3,X=0\}}{P\{X=0\}}$

Substitute,

$=\frac{0.6\frac{2^3{e}^{-2}}{3!}{e}^{-2}+0.4\frac{3^{3e^{-3}}}{3!}{e}^{-3}}{0.10112}$

Divide,

$=\frac{0.019114265}{0.10112}$

$=0.189$

The exact $3$accidents in a certain year value are $P\{X=3\}=0.1979$and $P\{X=31$he had no accidents the preceding year$\}=0.189$.