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Harvey the clumsy waiter is in trouble at work because he has been breaking too many dishes. Suppose that the number of dishes he breaks follows a Poisson distribution with a rate of 0.11 per hour. Harvey works an 8-hour shift today, and his boss has warned him that if he breaks any dishes during the shift, he will be red. What is the probability that Harvey will be red today?

Wouldn't lambda be lambda*t = 0.11 x 8 = 0.88? Or did I get that twisted?

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Yes, you have the appropriate $\lambda$. – André Nicolas Dec 7 '12 at 6:17
I'm not getting the correct answer though. The way I set it up is that the probability that he breaks one dish i.e P(X=1) = (.88^1 * e^-.88)/1! is wrong- I'm lost when it comes to setting up the equation. – Jack Dec 7 '12 at 6:33
Probability he will not be "red" (??) is $e^{-\lambda}$. No dishes broken. Probability of red is $1-e^{-\lambda}$. That's the probability of breaking $1$ or more dishes. You were calculating probability of breaking exactly $1$. Please report success. – André Nicolas Dec 7 '12 at 6:42
That gave me the correct answer, but I don't understand how it's 1−e^−λ – Jack Dec 7 '12 at 7:01
The term $e^{-\lambda}$ is the probability that $X$, the number of broken dishes, is $0$. So $1-e^{-\lambda}$ is the probability that more than $0$ dishes are broken. If you want to do it the hard way, find the probability number of broken dishes is $1$ (you had done that), $2$, $3$, $4$, and add up. You will get the same correct number, with a lot more work. – André Nicolas Dec 7 '12 at 7:05

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