Question

An airline knows that 5 percent of the people making reservations on a certain flight will not show up. Consequently, their policy is to sell 52 tickets for a flight that can hold only 50 passengers. What is the probability that there will be a seat available for every passenger who shows up?

Answer 1

By condition, the probability that the passenger will arrive on the flight is $p = 1 - 0.05=0.95 \Rightarrow q = 0.05$ .

Using the Bernoulli formula, we find the probabilities that 51 and 52 passengers will arrive on the flight, respectively:

$P(51) = C_{52}^{51}{p^{51}}q = 52 \cdot {0.95^{51}} \cdot 0.05 = {\rm{0}}{\rm{.19005408893348192330750711636671}}$

$P(52) = {p^{52}} = {0.95^{52}} = {\rm{0}}{\rm{.06944284018723377967005067713399}}$

Then the probability that more than 50 passengers will arrive on the flight is

$P(x > 50) = P(51) + P(52) = 0.2594969291207157029775577935007$

Then the wanted probability is

$P = 1 - P(x > 50) = 0.7405030708792842970224422064993$

Answer: $P = 0.7405030708792842970224422064993$