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Chapter 3: Q12E (page 107)

Airlines sometimes overbook flights. Suppose that for a plane with 50 seats, 55 passengers have tickets. Define the random variable Y as the number of ticketed passengers who actually show up for the flight. The probability mass function of Y appears in the accompanying table.

y

45

46

47

48

49

50

51

52

53

54

55

p(y)

.05

.10

.12

.14

.25

.17

.06

.05

.03

.02

.01

a. What is the probability that the flight will accommodateall ticketed passengers who show up?

b. What is the probability that not all ticketed passengerswho show up can be accommodated?

c. If you are the first person on the standby list (whichmeans you will be the first one to get on the plane ifthere are any seats available after all ticketed passengers have been accommodated), what is the probability that you will be able to take the flight? What is this probability if you are the third person on the standby list?

Short Answer

Expert verified

a. The probability that the flight will accommodate all the ticketed passengers who shows up is 0.83

b.The probability that not all ticketed passengers who show up can be accommodated is0.17.

c.The required probability is 0.66. The probability that you are the third person on the standby list is 0.27

Step by step solution

01

Given information

The data are provided that consists of pmf of a random variable Y:

Y

P(Y)

45

0.05

46

0.1

47

0.12

48

0.14

49

0.25

50

0.17

51

0.06

52

0.05

53

0.03

54

0.02

55

0.01

02

Compute the probability that the flight will accommodateall ticketed passengers who show up

a.

Here, the probability thatthe flight will accommodate all ticketed passengers who show up is obtained.

The required probability is\(P\left( {Y \le 50} \right)\). It can be computed as:

\(P\left( {Y \le 50} \right) = \left( \begin{array}{l}P\left( {X = 45} \right) + P\left( {X = 46} \right) + P\left( {X = 47} \right)\\ + P\left( {X = 48} \right) + P\left( {X = 49} \right) + P\left( {X = 50} \right)\end{array} \right)\)

Substitute the values of\(P\left( {X = 45} \right) = 0.05\),\(P\left( {X = 46} \right) = 0.1\),\(P\left( {X = 47} \right) = 0.12\),\(P\left( {X = 48} \right) = 0.14\),\(P\left( {X = 49} \right) = 0.25\)and\(P\left( {X = 50} \right) = 0.17\)in the above formula,

\(\begin{array}{c}P\left( {X \le 50} \right) = 0.05 + 0.1 + 0.12 + 0.14 + 0.25 + 0.17\\ = 0.83\end{array}\)

Thus, the required probability is 0.83.

03

Compute the probability that not all ticketed passengers the flight who show up can be accommodated in the flight

b.

Here, the probability thatnot all ticketed passengers who show up can be accommodated is obtained.

So, the required probability is\(1 - P\left( {Y \le 50} \right)\). It can be computed as:

\(1 - P\left( {Y \le 50} \right) = 1 - \left( \begin{array}{l}P\left( {X = 45} \right) + P\left( {X = 46} \right) + P\left( {X = 47} \right)\\ + P\left( {X = 48} \right) + P\left( {X = 49} \right) + P\left( {X = 50} \right)\end{array} \right)\)

Substitute the values of\(P\left( {X = 45} \right) = 0.05\),\(P\left( {X = 46} \right) = 0.1\),\(P\left( {X = 47} \right) = 0.12\),\(P\left( {X = 48} \right) = 0.14\),\(P\left( {X = 49} \right) = 0.25\)and\(P\left( {X = 50} \right) = 0.17\)in the above formula,

\(\begin{array}{c}1 - P\left( {X \le 50} \right) = 1 - \left( {0.05 + 0.1 + 0.12 + 0.14 + 0.25 + 0.17} \right)\\ = 1 - 0.83\\ = 0.17\end{array}\)

Thus, the required probability is 0.17.

04

Compute the probability that if you are the first person on standby list then you will be able to take the flight whenall ticketed passengersshow up

c.

In this case, if a person is the first person on the stand list then the probability that a person will able to take the flight is when there is one seat left out of 50 seats or at most 49 ticketed passengers would shows up in a plane.

Therequired probability is\(P\left( {Y \le 49} \right)\). It can be computed as:

\(P\left( {Y \le 49} \right) = \left( \begin{array}{l}P\left( {X = 45} \right) + P\left( {X = 46} \right) + P\left( {X = 47} \right)\\ + P\left( {X = 48} \right) + P\left( {X = 49} \right)\end{array} \right)\)

Substitute the values of\(P\left( {X = 45} \right) = 0.05\),\(P\left( {X = 46} \right) = 0.1\),\(P\left( {X = 47} \right) = 0.12\),\(P\left( {X = 48} \right) = 0.14\),\(P\left( {X = 49} \right) = 0.25\)and\(P\left( {X = 50} \right) = 0.17\)in the above formula,

\(\begin{array}{c}P\left( {X \le 49} \right) = \left( {0.05 + 0.1 + 0.12 + 0.14 + 0.25} \right)\\ = 0.66\end{array}\)

Thus, the required probability is 0.66.

05

Compute the probability that if you are the third person on standby list then you will be able to take the flight when all ticketed passengers show up

In this case, if a person is the third person on the stand list then the probability that person will able to take the flight is when there is three seats left out of 50 seats or at most 47 ticketed passengers would shows up in a plane.

Therequired probability is\(P\left( {Y \le 47} \right)\). It can be computed as:

\(P\left( {Y \le 47} \right) = P\left( {X = 45} \right) + P\left( {X = 46} \right) + P\left( {X = 47} \right)\)

Substitute the values of\(P\left( {X = 45} \right) = 0.05\),\(P\left( {X = 46} \right) = 0.1\), and\(P\left( {X = 47} \right) = 0.12\),in the above formula,

\(\begin{array}{c}P\left( {X \le 47} \right) = \left( {0.05 + 0.1 + 0.12} \right)\\ = 0.27\end{array}\)

Thus, the required probability is 0.27.

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