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A printer receives a number of jobs in an hour, which is poisson distributed with parameter $\lambda$. Every job is recognized with a probability $p$ such that the job is faulty and wont be printed.

(1) What is the distribution of printed jobs?

(2) If we assume that we printed $k$ jobs, what is the probability that $n$ jobs arrived.

For (1) I did the following: Let $a$ be the number of jobs in an hour, then $\mathbb P(n=a)=\frac{\lambda^n}{n!}e^{-\lambda}$. Let $X$ be the number of printed jobs and $F$ the set of faulty jobs, then $\mathbb P(F)=p$. What we want is $\mathbb P(k=X)$, correct? I guess it is possion again, but why?

For (2) I guess we need $\mathbb P(n=k | X=k )$ or something similar?

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1 Answer 1

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Let $X$ denote the number of printed jobs and $Y$ the number of jobs received, thus $Y\geqslant X$. First, one knows that, for every $n\geqslant0$, $q(n)=P[Y=n]$ is $$q(n)=\mathrm e^{-\lambda}\frac{\lambda^n}{n!}. $$ Second, the fact that each job happens to be faulty with probability $p$ independently of the others means that, for every $0\leqslant k\leqslant n$, $P[X=k\mid Y=n]=p(k,n)$ where $$ p(k,n)={n\choose k}p^{n-k}(1-p)^k. $$ Thus:

  • For every $k\leqslant n$, $$ P[X=k,Y=n]=P[X=k\mid Y=n]\cdot P[Y=n]=p(k,n)q(n). $$
  • For every $k$, $P[X=k]=r(k)$ with $$ r(k)=\sum_nP[X=k,Y=n]=\sum_{n\geqslant k}p(k,n)q(n). $$
  • For every $k\leqslant n$, $P[Y=n\mid X=k]=s(n,k)$ with $$ s(n,k)=\frac{P[X=k,Y=n]}{P[X=k]}=\frac{p(k,n)q(n)}{r(k)}. $$

Surely you can identify $r(k)$ and $s(n,k)$... Hint: for each $k$, the probability distribution $s(\ ,k)$ is almost Poisson.

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