Probability, binominal distribution

A hotel manager knows that approximately 5 % of the reservations are so called “no-shows” – people who have made a reservation, but do not arrive without cancelling their room. The hotel has 200 rooms and accepts a certain rate of overbooking. If 215 reservations are accepted, what is the probability that some
guests arriving will not have a room though they have a reservation?

I figured that It would go something like this:

5%*215= 10.75
215-10.75= 204.25
4.25/215= 0.019767 = 1.98%

but then I thought that it can't be that simple, my thinking must be wrong.

teacher gave a hint and said that we should calculate P(x>200), how do I do this?
Thanks for the hint! I used the binomial distribution formula and n=215, p=.05 x=14. I got 0.068228. Am I on the right track here?
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This is the probability of "exactly" 14 no-shows. It doesn't answer your question. You can either use the cumulative function in EXCEL or sum each case individually to get the answer you are looking for.
I changed values to n=215, p=.95 x=200/x=201 in order to take advantage of my teachers hint. With cumulative probability I got
x=200, P(x>200) = 0.87727
x=201, P(x>=201) = 0.87727
Thanks for the help everyone!

For future reference, I used Excel and Binomdist function.

For example in Excel:

A1 = Number of Trials (n), 215
A2 = Number of successes, 201-215
A3 = Probability (1-p), 1-0,05=0,95

I used number of successes from 201 to 215 and the summed up the probabilities.