Question

Suppose that the probability that a passenger will miss a flight is 0.0959. Airlines do not like flights with empty seats, but it is also not desirable to have overbooked flights because passengers must be "bumped" from the flight. Suppose that an airplane has a seating capacity of 5353 passengers.

(a) If 55 tickets are sold, what is the probability that 54 or 55 passengers show up for the flight resulting in an overbooked flight?

(b) Suppose that 59 tickets are sold. What is the probability that a passenger will have to be "bumped"?

(c) For a plane with seating capacity of 55 passengers, how many tickets may be sold to keep the probability of a passenger being "bumped" below 55%?

Answer #1

Ans:

a)Let x be the number of people who show up for flight.

Then x has binomial distribution with n=55 and p=1-0.0959=0.9041

P(overbooked)=P(x>53)=P(x=54)+P(x=55)

=^{55}C_{54}*0.9041^54*0.0959+0.9041^55

**=0.0267**

b)Then x has binomial distribution with n=59 and p=0.9041

P(bumped)=P(x>53)=1-P(x<=53)

=1-BINOMDIST(53,59,0.9041,TRUE)

**=0.4964**

c)

P(bumped)=P(x>55)=1-P(x<=55)

=1-BINOMDIST(55,n,0.9041,TRUE)

n | 1-BINOMDIST(55,n,0.9041,TRUE) |

56 | 0.0035 |

57 |
0.0225 |

58 | 0.0744 |

59 | 0.1705 |

Maximum number of tickets to be sold,n=57 to keep the probability of a passenger being "bumped" below 5%.

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