Question

Airlines sometime overbook flights. Suppose that for a planewith 50 seats, 55 passengers have tickets. Define the randomvariable Y as the number of ticketed passengers who actually showup for the flight. The probability mass function of Y appears inthe 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 flightwill accommodate all ticketed passengers who show up?

b.)    What is the probability that not allticketed passengers who show up can be accommodated?

c.)     If you are the first person on thestandby list (which means you will be the first one to get on theplane if there are any seats available afar all ticketed passengershave been accommodated), what is the probability that you will beable to take the flight? What is this probability if you are thethird person on the standby list?

Answer 1

Concepts and reason

The problem deals with the concepts of finding the simple probabilities using the discrete probability distribution of a random variable Y.

Fundamentals

For any discrete random variable Y, we have $\sum {P\left( {Y = y} \right)} = 1$ .

(a)

The known probability distribution of Y is,

$Y = y$	$P\left( {Y = y} \right)$
45	0.05
46	0.10
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
Total	$\sum {P\left( {Y = y} \right)} = 1$

Let Y be the number of ticketed passengers who actually show up for the flight.

For a plane with 50 seats, there are at most 50 ticketed passengers who actually show up for the flight.

Using the probability distribution, the probability that the flight will accommodate all ticketed passengers who show up is,

$\begin{array}{c}\\P\left( {Y \le 50} \right) = P\left( {45} \right) + P\left( {46} \right) + P\left( {47} \right) + P\left( {48} \right) + P\left( {49} \right) + P\left( {50} \right)\\\\ = 0.05 + 0.10 + 0.12 + 0.14 + 0.25 + 0.17\\\\ = 0.83\\\end{array}$

(b)

For a plane with 50 seats, there are at most 50 ticketed passengers who actually show up for the flight. Also, it is known that not all ticketed passengers who show up can be accommodated are questionable.

Using the probability distribution, the probability that not all ticketed passengers who show up can be accommodated is,

$\begin{array}{c}\\P\left( {Y > 50} \right) = 1 - P\left( {Y \le 50} \right)\\\\ = 1 - \left[ {P\left( {45} \right) + P\left( {46} \right) + P\left( {47} \right) + P\left( {48} \right) + P\left( {49} \right) + P\left( {50} \right)} \right]\\\\ = 1 - \left( {0.05 + 0.10 + 0.12 + 0.14 + 0.25 + 0.17} \right)\\\\ = 1 - 0.83\\\\ = 0.17\\\end{array}$

(c)

Using the discrete probability distribution, the probability that the first person on the standby list is,

$\begin{array}{c}\\P\left( {{\rm{First person on the standby list}}} \right) = P\left( {Y \le 49} \right)\\\\ = P\left( {45} \right) + P\left( {46} \right) + P\left( {47} \right) + P\left( {48} \right){\rm{ + }}P\left( {49} \right)\\\\ = 0.05 + 0.10 + 0.12 + 0.14 + 0.25\\\\ = 0.66\\\end{array}$

Using the discrete probability distribution, the probability that the third person on the standby list is,

$\begin{array}{c}\\P\left( {{\rm{Third person on the standby list}}} \right) = P\left( {45} \right) + P\left( {46} \right) + P\left( {47} \right)\\\\ = 0.05 + 0.10 + 0.12\\\\ = 0.27\\\end{array}$

Ans: Part a

Therefore, the probability that the flight will accommodate all ticketed passengers who show up is 0.83.

Part b

Therefore, the probability that not all ticketed passengers who show up can be accommodated is 0.17.

Part c

Therefore, the probability that the first person on the standby list is 0.66. The probability that the third person on the standby list is 0.27.