91Ó°ÊÓ

Define $X~Pois\left(\mathrm{λ}\right)$. Define $Y$as the random variable that counts the events that are accepted. We are required to show that $Y~Pois\left(p\mathrm{λ}\right)$. Take any $k∈{\mathrm{ℕ}}_0$and using LOTP we have that

$P(Y=k)=\underset{n=k}{\overset{∞}{∑}}P(Y=kâˆ\pounds X=n)P(X=n)$

Observe that if we are given $X=n,Y$has binomial distribution since every of $n$events are independently accepted with the probability $p$.

$P(Y=k)=\underset{n=k}{\overset{∞}{∑}}\left(\begin{array}{l}n\\ k\end{array}\right)p^k(1-p{)}^{n-k}·\frac{{\mathrm{λ}}^n}{n!}{e}^{-\mathrm{λ}}$

$=\underset{n=k}{\overset{∞}{∑}}\frac{n!}{k!(n-k)!}{p}^k(1-p{)}^{n-k}·\frac{{\mathrm{λ}}^n}{n!}{e}^{-\mathrm{λ}}$

$=\frac{p^k}{k!}{e}^{-\mathrm{λ}}\underset{n=k}{\overset{∞}{∑}}\frac{1}{(n-k)!}(1-p{)}^{n-k}·{\mathrm{λ}}^n$

$=\frac{p^k}{k!}{e}^{-\mathrm{λ}}\underset{n=0}{\overset{∞}{∑}}\frac{1}{n!}(1-p{)}^n·{\mathrm{λ}}^{n+k}$

$=\frac{p^k}{k!}{e}^{-\mathrm{λ}}{\mathrm{λ}}^k\underset{n=0}{\overset{∞}{∑}}\frac{1}{n!}(1-p{)}^n·{\mathrm{λ}}^n$

$=\frac{p^k}{k!}{e}^{-\mathrm{λ}}{\mathrm{λ}}^k\underset{n=0}{\overset{∞}{∑}}\frac{\left(\mathrm{λ}\right(1-p){)}^n}{n!}$

$=\frac{p^k}{k!}{e}^{-\mathrm{λ}}{\mathrm{λ}}^k·e^{\mathrm{λ}(1-p)}$

$=\frac{(\mathrm{λ}p{)}^k}{k!}{e}^{-\mathrm{λ}p}$

Since it holds for every $k$, we have proved that $Y~Pois\left(\mathrm{λ}p\right)$. Observe that it is consistent with our intuition: the number of accepted events also behaves like Poisson process since the original distribution is Poisson and the average rate is $p\mathrm{λ}$. For the last question, we are given that $\mathrm{λ}p=10·\frac{1}{50}=0.2$.

(a) $P(Y=1)=0.2·e^{-0.2}$

(b) $P(Yâ‰\yen 1)=1-P(Y=0)=1-e^{-0.2}$

(c)$P(Y≤1)=P(Y=0)+P(Y=1)=e^{-0.2}+0.2·e^{-0.2}$