Compare the Poisson approximation with the correct binomial probability for the following cases:

$\left(a\right)$$P\{X=2\}$when$n=8,p=.1;$

$\left(b\right)$$P\{X=9\}$when$n=10,p=.95;$

$\left(c\right)$$P\{X=0\}$when$n=10,p=.1;$

$\left(d\right)$ $P\{X=4\}$when$n=9,p=.2.$

$P\{X=2\}$when$n=8,p=.1$.

We are to compare the Poisson approximation with the correct binomial probability.

In the $\left(a\right)$part we have role="math" localid="1646463415641" $\mathrm{\mathbb{P}}(X=2)$and $n=0,p=0.1$.

We know that:

$\mathbb{E}\left[X\right]=np=8·0.1=0.8$

Therefore the Poisson approximation is:

$\mathrm{\mathbb{P}}(X=2)=\frac{e^{-0.8}·0.8^2}{2!}=\frac{0.288}{2}=0.144$

On the other hand binomial is:

$\mathrm{\mathbb{P}}(Y=2)=\left(\begin{array}{l}8\\ 2\end{array}\right)0.1^2(1-0.1{)}^6=28·0.005=0.148$

The Poisson approximation with the correct binomial probability is found to be$0.144$and$0.148$.

$P\{X=9\}$ when$n=10,p=.95$.

For the$\left(b\right)$part we have $\mathrm{\mathbb{P}}(X=9)$and $n=10,p=0.95$.

So the calculation is:

$\mathrm{\mathbb{P}}(X=9)=\frac{e^{-9.5}·9.5^9}{9!}=0.13$

$\mathrm{\mathbb{P}}(Y=9)=\left(\begin{array}{c}10\\ 9\end{array}\right)·0.{95}^9·0.05=0.315$

The Poisson approximation with the correct binomial probability is found to be$0.13$and$0.315$.

$P\{X=0\}$ when$n=10,p=.1$.

In the $\left(c\right)$part we have $\mathrm{\mathbb{P}}(X=0)$and $n=10,p=0.1.$Obviously $\lambda =np=1$.

Therefore we have:

$\mathrm{\mathbb{P}}(X=0)=\frac{e^{-1}·1^0}{0!}=e^{-1}=0.37$

$\mathrm{\mathbb{P}}(Y=0)=\left(\begin{array}{c}10\\ 0\end{array}\right)·0.1^0·0.9^{10}=0.35$.

The Poisson approximation with the correct binomial probability is found to be$0.37$and$0.35$.

$P\{X=4\}$ when $n=9,p=.2$.

Finally, in the $\left(d\right)$part we have $n=9,p=0.2⟹\lambda =9·\frac{2}{10}=1.8$.

Now it follows:

$\mathrm{\mathbb{P}}(X=4)=\frac{e^{-1.8}·1.8^4}{4!}=0.07$

$\mathrm{\mathbb{P}}(Y=4)=\left(\begin{array}{l}9\\ 4\end{array}\right)·0.2^4·0.8^5=0.066$

The Poisson approximation with the correct binomial probability is found to be$0.07$and$0.066$.

All four part of the assignment we have $X~P\left(np\right),Y~B(n,p)$.

Hence, we are done.