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A restaurant has two waiters. suppose that the number of customers serviced daily by each waiter can be viewed as independent Poisson random variables with parameters $\lambda$ and $\mu$. whats the probability that the two waiters together handle less than $6$ customers during the day?

I was thinking along the lines of... $$ P(X\leq 6,Y\leq 6) = P(X\leq6)\cdot P(Y\leq 6), $$ but for the probability mass function of Poisson, what would $\lambda$ be?? We are not given any data regarding the average number of customers served.

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    $\begingroup$ $X \sim P(\lambda),\ Y \sim P(\mu)$. If X and Y are independent, then $X+Y \sim P(\lambda+\mu)$ $\endgroup$
    – Coiacy
    Feb 21 '13 at 11:12
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Let's let $X$ be the number of customers served by the first waiter, and $Y$ the number served by the second waiter.

Note that $P(X \le 6, Y \le 6)$ is not what you want: it's the probability that each waiter serves at most 6 customers. But it includes the event that, say, the first waiter serves 5 and the second waiter serves 4, in which case the total is more than 6.

What you want is $P(X + Y \le 6)$. It's a fact, which you should be able to find in your textbook, that the sum of independent Poisson random variables is another Poisson random variable, whose mean is the sum of the means (here, $\lambda + \mu$).

So you are looking for the probability for the Poisson random variable $X+Y$, whose mean is known to you, to be less than or equal to 6. I expect you can take it from here.

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  • $\begingroup$ Ah I see. So Z=X+Y and m = λ+μ. which leads to p(Z<=6) = sigma(i=0to6) e^-m(m^i/i!) Thank you so much! $\endgroup$
    – user63303
    Feb 21 '13 at 11:36

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