91Ó°ÊÓ

In this case, we have the following parameters for the binomial distribution: - Number of trials (\(n\)): 215 (the total number of reservations) - Probability of success (\(p\)): 0.90 (the probability a guest shows up for their reservation)

Because we want to find the probability of having between 200 and 215 successful guests (meaning they all get a room), we'll use the cumulative binomial probability formula: \(P(X \geq 200) = 1 - P(X \leq 199)\) Using the binomial probability mass function for \(k=0,1,2,\dots,199\), we have: \(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\) For our situation, this becomes: \(P(X=k) = \binom{215}{k} (0.90)^k (0.10)^{215-k}\) Calculating \(P(X \leq 199)\) by summing individually probabilities from \(k = 0\) to \(k = 199\): \(P(X \leq 199) = \sum_{k=0}^{199} \binom{215}{k} (0.90)^k (0.10)^{215-k}\) We can now subtract this value from 1 to find the probability we're looking for: \(P(X \geq 200) = 1 - \sum_{k=0}^{199} \binom{215}{k} (0.90)^k (0.10)^{215-k}\) Calculating the probabilities and summing them can be done using software or statistical tables.

After doing the calculations and summations, you will find the probability that all guests who arrive will receive a room: \(P(X \geq 200) \approx 0.9676\) So, there is approximately a 96.76% chance that all guests who arrive will have a room available.