Question

Suppose that the probability that a passenger will miss a flight is 0.09070 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 54 passengers.

(a) If 56 tickets are sold, what is the probability that 55 or 56 passengers show up for the flight resulting in an overbooked flight?

(b) Suppose that 60 tickets are sold. What is the probability that a passenger will have to be "bumped"?

(c) For a plane with seating capacity of 53 passengers, how many tickets may be sold to keep the probability of a passenger being "bumped" below 10%?

Answer #1

Ans:

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

Then x has binomial distribution with n=56 and p=0.9093

P(overbooked)=P(x>54)

=P(x=55)+P(x=56)

=^{56}C_{55}*0.9093^55*0.0907+0.9093^56

**=0.0321**

b)Now,x has binomial distribution with n=60 and p=0.9093

P(bumped)=P(overbooked)=P(x>54)

=1-P(x<=54)

=1-binomdist(54,60,0.9093,true)

**=0.5353**

c)

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

=1-binomdist(53,n,0.9093,true)

n | P(bumped) |

54 | 0.0059 |

55 |
0.0347 |

56 | 0.1067 |

57 | 0.2286 |

58 | 0.3860 |

So,maximum number of ticket sold to keep probability of being bumped will be 55.

So,**n=55**

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