Question

Musquodoboit World Airways operates a fleet of small passenger planes. Like most major airlines it has a
problem with no-shows, people who make reservations, but don't show up for their flight or cancel at the
last minute. Their no-show rate is 20%, comparable to the industry average. One type of aircraft used by
MWA can accommodate up to 16 passengers.
a) If they accept 16 reservations, what is the probability that the plane departs with every seat filled (0 noshows)?

b) As is common in the industry, MWA overbooks its flights. For this size aircraft, it accepts up to 17
reservations. Suppose that MWA has accepted 17 reservations for a particular flight. What is the
probability that everyone who shows up will get a seat?

Answer 1

Answer on Question #51325 – Math – Statistics and Probability

Musquodoboit World Airways operates a fleet of small passenger planes. Like most major airlines it has a problem with no-shows, people who make reservations, but don't show up for their flight or cancel at the last minute. Their no-show rate is 20%, comparable to the industry average. One type of aircraft used by MWA can accommodate up to 16 passengers.

a) If they accept 16 reservations, what is the probability that the plane departs with every seat filled (0 noshows)?

b) As is common in the industry, MWA overbooks its flights. For this size aircraft, it accepts up to 17 reservations. Suppose that MWA has accepted 17 reservations for a particular flight. What is the probability that everyone who shows up will get a seat?

Solution

a) It is Binomial Distribution problem with $n = 16$ and $p(no \ show) = 0.2$; $q = 1 - p = 0.8$.

The probability that the plane departs with every seat filled (0 noshows) is

$$P(x = 0) = \frac{16!}{(16 - 0)! \ 0!} \cdot 0.2^0 \cdot 0.8^{16 - 0} = 0.8^{16} = 0.02815.$$

b) It is Binomial Distribution problem with $n = 17$ and $p(no \ show) = 0.2$; $q = 1 - p = 0.8$.

The probability that everyone who shows up will get a seat is

$$\begin{array}{l} P(\text{at least 1 noshow out of 17}) = 1 - P(0 \text{ noshows out of 17}) = 1 - \frac{17!}{(17 - 0)! \ 0!} \cdot 0.2^0 \cdot 0.8^{17 - 0} \\ = 1 - 0.8^{17} = 1 - 0.0225 = 0.9775. \end{array}$$

Answer: a) 0.02815; b) 0.9775.

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