Question

Repeat Problem 7.68 when the proportion of the population having a value of $\lambda$less than $x$is equal to $1-e^{-x}$.

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. We are required to find $P(N=0).$

b. We are required to find $P(N=3)$.

c. Define $M$as the number of accidents in a preceding year. As likely as $N$we are require to find.

Short answer

a. $P(N=0)$using the law of the total probability value found to be$\frac{1}{2}$.

b. $P(N=3)$using the law of the total probability value found to be$\frac{1}{16}$.

c. The number of accidents in a preceding year required to value found to be$\frac{2}{81}$.

Step-by-step solution

Given Information (Part a)

Define$N$as the number of accidents in a year we are required to find the$P(N=0)$.

Explanation (Part a)

Define $N$as the number of accidents in a year. We know that

$N\mid \lambda ~Pois\left(\lambda \right)$

Where $\lambda ~Expo\left(1\right)$which implies $f_{\lambda }\left(s\right)=e^{-s}$for $s>0$,

We are required to find $P(N=0)$.

Explanation (Part a)

Using the law of the total probability, we have that,

$P(N=0)={\int }_0^{\infty }$

$P(N=0\mid \lambda =s)f_{\lambda }\left(s\right)ds$

Substitute the value,

$={\int }_0^{\infty }\frac{s^0}{0!}{e}^{-s}·e^{-s}ds$

${\int }_0^{\infty }{e}^{-2s}ds=\frac{1}{2}$.

Final answer (Part a)

The $P(N=0)$using the law of the total probability value found to be$\frac{1}{2}$.

Given Information (Part b) 

We are required to find$P(N=3)$. Using the law of the total probability.

Explanation (Part b)

We are required to find $P(N=3)$

$P(N=3)={\int }_0^{\infty }$

$P(N=3\mid \lambda =s)f_{\lambda }\left(s\right)ds$

Substitute,

$={\int }_0^{\infty }\frac{s^3}{3!}{e}^{-s}·e^{-s}ds$

$=\frac{1}{3!}{\int }_0^{\infty }{s}^3{e}^{-2s}ds$.

Explanation (Part b) 

In order to solve this integral, rewrite $2s=u$which implies $S=\frac{u}{2}$

Hence,

$\frac{1}{3!}{\int }_0^{\infty }{s}^3{e}^{-2s}ds=\frac{1}{3!}\times \frac{1}{16}$

Divide the value,

${\int }_0^{\infty }{u}^3{e}^{-u}du=\frac{T\left(4\right)}{3!·16}$

$=\frac{1}{16}$.

Final answer (Part b)

The$P(N=3)$using the law of the total probability value found to be$\frac{1}{16}$.

Given Information (Part c)

Define$M$as the number of accidents in a preceding year. As likely as$N$ we have that.

Explanation (Part c)

Define $M$as the number of accidents in a preceding year.

$M\mid \lambda ~Pois\left(\lambda \right)$

But observe that $M$and $N$are not independent random variables, since both of them depend on $\lambda .$

But, they are independent if we fix parameter$\lambda .$We are required to find.

$P(N=3\mid M=0)=\frac{P(N=3,M=0)}{P(M=0)}$.

Explanation (Part c) 

We know that $P(M=0)=P(N=0)$

$=\frac{1}{2}$

Also, we have that

$P(N=3,M=0)={\int }_0^{\infty }$$P(N=3,M=0\mid \lambda =s)f_{\lambda }\left(s\right)ds$

Simplify

$={\int }_0^{\infty }\frac{s^3}{3!}{e}^{-s}·e^{-s}·e^{-s}ds$

$=\frac{1}{3!}{\int }_0^{\infty }{s}^3{e}^{-3s}ds$.

Explanation (Part c) 

In order to solve this integral, rewrite $3s=u$which implies$s=\frac{u}{3}$

$\frac{1}{3!}{\int }_0^{\infty }{s}^3{e}^{-3s}ds=\frac{1}{3!}\times \frac{1}{3^4}$

Simplify

${\int }_0^{\infty }{u}^3{e}^{-u}du=\frac{T\left(4\right)}{3!\times 3^4}$

Divide the value,

$=\frac{1}{81}$.

Finally we have that,

Substitute,

$P(N=3\mid M=0)=\frac{P(N=3,M=0)}{P(M=0)}$

$=\frac{\frac{1}{81}}{\frac{1}{2}}$

Divide the value,

$=\frac{2}{81}$.

Final answer (Part c)

The number of accidents in a preceding year required to value found to be$\frac{2}{81}$.