Question
The number of envelopes arriving at a mailroom in an hour is Poisson with mean 8. After all arrivals, each envelope is independently labelled for Department X with probability 1/4 (and otherwise sent elsewhere). Labels are independent of the total arrival count.
What is the probability that Department X receives no envelopes in an hour? More generally, what is the distribution of the number X sent to that department? Show why a fixed 8-envelope binomial model is not the right one.
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Hint 1
Condition first on the random total \(N=n\). Given \(N\), the number labelled \(X\) is \(\operatorname{Binomial}(n,\frac14)\).
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Solution
Let \(N\sim\operatorname{Poisson}(8)\) be the total count. Conditional on \(N=n\), \(X\) is \(\operatorname{Binomial}(n,p)\) with \(p=\frac14\) because each of the n independently sampled envelopes is labelled for X with that probability.
For any nonnegative integer k, mixing over the Poisson total gives
\[\begin{aligned}P(X=k)&=\sum_{n=k}^{\infty}[e^{-8}8^n/n!]\cdot C(n,k)p^k(1-p)^{n-k}\\&=e^{-8}(8p)^k/k!\cdot\sum_{m=0}^{\infty}[8(1-p)]^m/m!\\&=e^{-8}e^{[8(1-p)]}(8p)^k/k!\\&=\mathbf{e^{-2}2^k/k!}.\end{aligned}\]
Thus \(X\sim\operatorname{Poisson}(8p)=\) \(\operatorname{Poisson}(2)\) and the zero-count probability is \(e^{-2}\approx0.135335\). For a quick independent check,
\(\Pr(X=0)=\mathbb E[(\frac34)^N]=e^{8((\frac34)-1)}=e^{-2}\).
A \(\operatorname{Binomial}(8,\frac14)\) distribution would incorrectly treat the mean arrival count of 8 as a fixed count of exactly 8 envelopes. The total itself is random, so its thinning produces a Poisson distribution instead.
Assumptions
The total number of arrivals in one hour follows Poisson(8).
Each arrival receives an independent \(\operatorname{Bernoulli}(\frac14)\) Department X label after the total has been realized.
Arrival counts and label probabilities are stationary for the hour.
Final Answer
\(X\sim\operatorname{Poisson}(2)\); \(\Pr(X=0)=e^{-2}\approx0.135335\).