Chapter 6, Problem 77CE

A small feeder airline knows that the probability is .10 that a reservation holder will not show up for its daily 7:15 a.m. flight into a hub airport. The flight carries 10 passengers. (a) If the flight is fully booked, what is the probability that all those with reservations will show up? (b) If the airline overbooks by selling 11 seats, what is the probability that no one will have to be bumped? (c) That more than one passenger will be bumped? *(d) The airline wants to overbook the flight by enough seats to ensure a 95 percent chance that the flight will be full, even if some passengers may be bumped. How many seats would it sell?

To determine

Find the probability of all passengers with reservation will show up.

Answer to Problem 77CE

The probability of all passengers with reservation will show up is 0.3487.

Explanation of Solution

Calculation:

The given information is that the airline knows that the probability of a reservation holder will not show up its regular 7:15 a.m. flight into a hub airport is 0.10. The number of passengers carries in the flight is 10.

Binomial Distribution:

The random variable X is the number of success in n trails. The random variable X follows binomial distribution with parameters n and $\pi$.

The PDF of the distribution is,

$P\left(X=x\right)=\frac{n!}{x!\left(n-x\right)!}{\pi }^x{\left(1-\pi \right)}^{n-x}$ for X = 0, 1, 2, 3, 4, …, n.

Here, the random variable X is the number of no shows, n is 10 and the probability of no shows is 0.10.

The probability of all passengers with reservations will show up is $P\left(X=0\right)$.

Probability = BINOM.DIST (numbers, trails, probability, cumulative).

Software procedure:

Step by step procedure to obtain the probability by using EXCEL software:

• Open new EXCEL file.
•  In cell A1, enter the formula as “=BINOM.DIST (0,10,0.10,0)”.

Output using EXCEL software is given as follows:

From the output, the value of $P\left(X=0\right)$ is 0.3487.

Thus, the probability of all passengers with reservation will show up is 0.3487.

To determine

Find the probability of no one passengers will have to be bumped if selling of 11 seats.

Answer to Problem 77CE

The probability of no one passengers will have to be bumped if selling of 11 seats is 0.6862.

Explanation of Solution

Calculation:

The airline overbooks by the selling of 11 seats, that is n is 11.

The probability of no one will be bumped means more than one passenger is not show up is, $P\left(X\ge 1\right)=1-P\left(X=0\right)$.

Software procedure:

Step by step procedure to obtain the probability by using EXCEL software:

• Open new EXCEL file.
•  In cell A1, enter the formula as “=BINOM.DIST (0,11,0.10,0)”.

Output using EXCEL software is given as follows:

From the output, the value of $P\left(X=0\right)$ is 0.3138.

$\begin{array}{c}P\left(X\ge 1\right)=1-P\left(X=0\right)\\ =1-0.3138\\ =0.6862\end{array}$

Thus, the probability of no one passengers will have to be bumped if selling of 11 seats is 0.6862.

To determine

Find the probability of more than one passenger has to be bumped if selling of 11 seats.

Answer to Problem 77CE

The probability of more than one passenger has to be bumped if selling of 11 seats is 0.

Explanation of Solution

Calculation:

The airline overbooks by the selling of 11 seats, that is n is 11.

The probability of more than one passenger has to be bumped is, $P\left(X>1\right)$.

In practical situation, the number of passenger is 10 if the airline overbooks by selling 11 seats. The one person may has to be bumped but it is not possible to bumped more than one person. That is the probability of more than one passenger has to be bumped is 0.

Thus, the probability of more than one passengers will have to be bumped if selling of 11 seats is 0.

To determine

Find the number of seats would be sell to ensure that the flight will be filled by 95%.

Answer to Problem 77CE

The number of seats would be sell to ensure that the flight will be filled by 95% is 13.

Explanation of Solution

Calculation:

Assume the random variable X represents the number of passengers show up and the probability for the event is 0.90.

Calculate the number of seats need to be sell to get the 95% of the flight to be filled,$P\left(X\ge 10\right)\ge 0.95$ for various values of n.

The value of $P\left(X\ge 10\right)=1-P\left(X\le 9\right)$.

Probability = BINOM.DIST (9, n, 0.90, 1).

For $n=11$.

Software procedure:

Step by step procedure to obtain the probability by using EXCEL software:

• Open new EXCEL file.
•  In cell A1, enter the formula as “=BINOM.DIST (9,11,0.90,1)”.

Output using EXCEL software is given as follows:

From the output, the value of $P\left(X\le 9\right)$ for n as 11 is 0.3026.

$\begin{array}{c}P\left(X\ge 10\right)=1-P\left(X\le 9\right)\\ =1-0.3026\\ =0.6974\end{array}$

For $n=12$.

Software procedure:

Step by step procedure to obtain the probability by using EXCEL software:

• Open new EXCEL file.
•  In cell A1, enter the formula as “=BINOM.DIST (9,12,0.90,1)”.

Output using EXCEL software is given as follows:

From the output, the value of $P\left(X\le 9\right)$ for n as 12 is 0.1109.

$\begin{array}{c}P\left(X\ge 10\right)=1-P\left(X\le 9\right)\\ =1-0.1109\\ =0.8891\end{array}$

For $n=13$.

Software procedure:

Step by step procedure to obtain the probability by using EXCEL software:

• Open new EXCEL file.
•  In cell A1, enter the formula as “=BINOM.DIST (9,13,0.90,1)”.

Output using EXCEL software is given as follows:

From the output, the value of $P\left(X\le 9\right)$ for n as 13 is 0.0342.

$\begin{array}{c}P\left(X\ge 10\right)=1-P\left(X\le 9\right)\\ =1-0.0342\\ =0.9658\end{array}$

Thus, the number of seats would be sell to ensure that the flight will be filled by 95% is 13.