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?

Short answer

The probability that there will be a seat available for every passenger who shows up can be calculated using the binomial distribution formula: \(P(X \le 50) = \sum_{k=0}^{50} {52 \choose k} (0.95)^k (0.05)^{52-k}\). By calculating each term of the sum from k=0 to k=50 and summing them, we find the cumulative probability \(P(X \le 50)\), which represents the likelihood of having a seat for everyone who shows up on the flight.

Step-by-step solution

Define the variables

Let X be the number of passengers that show up for the flight. The probability of a passenger showing up \(p=0.95\) and the probability of a passenger not showing up \(q=0.05\). There are a total of \(n=52\) tickets sold and a seating capacity of 50 passengers.

Calculate the cumulative probability using binomial distribution

We will use the binomial distribution formula to calculate the probability of up to 50 people showing up to the flight: \(P(X \le 50) = \sum_{k=0}^{50} {n \choose k} p^k q^{n-k}\) Where \(n=52, p=0.95\), and \(q=0.05\).

Calculate the binomial coefficients

For each term in the sum, calculate the binomial coefficient {n \choose k}: \({n \choose k} = \frac{n!}{k!(n-k)!}\)

Calculate each term of the sum

Calculate each term of the sum for k=0 to k=50 using the values of n, p, and q, and their binomial coefficients: \(P(X=k)={n \choose k} p^k q^{n-k}\)

Calculate the cumulative probability

Now we sum the individual probabilities from \(k=0\) to \(k=50\) to find the cumulative probability: \(P(X \le 50) = P(X=0)+P(X=1)+...+P(X=50)\)

Interpret the result

The sum calculated in the previous step represents the probability that there will be a seat available for every passenger who shows up. The result will indicate the likelihood (in percentage form) that the airline's policy of overselling the flight by 2 tickets will not result in an overfull flight.