Poisson distribution - stuck.

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The number of customers that arrive at the Customer’s Service counter can be modeled with a Poisson distribution with an average of 11 per hour. What is the probability that over the next 5 hours, there will be exactly 3 hours where no one shows up at the counter? All I know is that I need to compute the probability that no one shows up. But I'm not sure how to proceed.

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hint....The probability of no one turning up in a given hour is, as you say, $e^{-11}=p$

Now use this as the probabililty of "success" is a Binomial distribution, where $X$ is the number of "dead" hours, so $X$ follows a Binomial distribution with parameters $5$ and $p$. Use the Binomial formula to calculate $p(X=3)$

Can you finish this?