Question

# Airlines sometimes overbook flights. Suppose that for a plane with 50 seats, 55 passengers have tickets....

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.

y4546474849505152535455
p(y)0.050.100.120.140.240.180.060.050.030.020.01

(a) What is the probability that the filight will accommodate all ticketed passengers who show up?

(b) What is the probability that not all ticketed passengers who show up can be accommodated?

(c) If you are the first person on the standby list (which means you will be the first one to get on the plane if there 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 an the standby list?

(a)

The known probability distribution of Y is,

 Y=y P(Y=y) 45 0.05 46 0.10 47 0.12 48 0.14 49 0.24 50 0.18 51 0.06 52 0.05 53 0.03 54 0.02 55 0.01 Total ∑P(Y=y)=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,

P(Y≤50) = P(45)+P(46)+P(47)+P(48)+P(49)+P(50) = 0.05+0.10+0.12+0.14+0.24+0.18

P(Y≤50) = 0.83

b)

P(Y>50)=1−P(Y≤50)=

1-0.83= 0.17

c)

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

P(First person on the standby list)

=P(Y≤49)

=P(45)+P(46)+P(47)+P(48)+P(49)

=0.05+0.10+0.12+0.14+0.24

=0.65

P(Third person on the standby list)=P(45)+P(46)+P(47)=0.05+0.10+0.12=0.27

Solution:-

(a) p(y<=50): 0.05+0.10+0.12+0.14+0.24+0.18 = 0.83

(b) p(y>50): 1 - P(Y <= 50) = 0.17

(c) p(y<50): 0.05+0.10+0.12+0.14+0.24 = 0.65

(d) P(y<48) = 0.05+0.10+0.12 = 0.27

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