91Ó°ÊÓ

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

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

$Nâˆ\pounds \mathrm{λ}~Pois\left(\mathrm{λ}\right)$

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

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

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

$P(N=0)={∫}_0^{∞}$

$P(N=0âˆ\pounds \mathrm{λ}=s)f_{\mathrm{λ}}\left(s\right)ds$

Substitute the value,

$={∫}_0^{∞}\frac{s^0}{0!}{e}^{-s}·e^{-s}ds$

${∫}_0^{∞}{e}^{-2s}ds=\frac{1}{2}$.

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

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

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

$P(N=3)={∫}_0^{∞}$

$P(N=3âˆ\pounds \mathrm{λ}=s)f_{\mathrm{λ}}\left(s\right)ds$

Substitute,

$={∫}_0^{∞}\frac{s^3}{3!}{e}^{-s}·e^{-s}ds$

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

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

Hence,

$\frac{1}{3!}{∫}_0^{∞}{s}^3{e}^{-2s}ds=\frac{1}{3!}×\frac{1}{16}$

Divide the value,

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

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

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

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

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

$Mâˆ\pounds \mathrm{λ}~Pois\left(\mathrm{λ}\right)$

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

But, they are independent if we fix parameter$\mathrm{λ}.$We are required to find.

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

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

$=\frac{1}{2}$

Also, we have that

$P(N=3,M=0)={∫}_0^{∞}$$P(N=3,M=0âˆ\pounds \mathrm{λ}=s)f_{\mathrm{λ}}\left(s\right)ds$

Simplify

$={∫}_0^{∞}\frac{s^3}{3!}{e}^{-s}·e^{-s}·e^{-s}ds$

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

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

$\frac{1}{3!}{∫}_0^{∞}{s}^3{e}^{-3s}ds=\frac{1}{3!}×\frac{1}{3^4}$

Simplify

${∫}_0^{∞}{u}^3{e}^{-u}du=\frac{T\left(4\right)}{3!×3^4}$

Divide the value,

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

Finally we have that,

Substitute,

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

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

Divide the value,

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

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