Question

Airlines and hotels often grant reservations in excess of capacity to minimize losses due to no-shows. Suppose the records of a hotel show that, on the average, \(10 \%\) of their prospective guests will not claim their reservation. If the hotel accepts 215 reservations and there are only 200 rooms in the hotel, what is the probability that all guests who arrive to claim a room will receive one?

Short answer

Answer: The probability that all guests who arrive at the hotel will have a room available is approximately 96.76%.

Step-by-step solution

Identify the binomial distribution parameters

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)

Calculate the probability

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.

Find the probability that all guests who arrive will receive a room

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.