Overbooking is the practice of selling more items than are currently available. Overbooking is common in the travel industry; it
allows a vehicle (airline, train, bus, cruise ship, hotel, and so forth) to operate at or near capacity, despite cancellations, no-shows, or late arrivals.
Overselling is when more confirmed customers show up to use the vehicle than there is space available. When this happens, at least one customer will be denied the service that they paid for, either voluntarily (sometimes with an incentive provided by the supplier) or involuntarily. This is called getting "bumped."
Suppose that for a particular flight, an airline believes that 1% of ticket holders do not make the flight. The jet making the trip holds 188 passengers. If the airline sells 191 tickets, what is the probability that the flight will be oversold and they will have to bump a passenger? Assume that cancellations are independent.
Calculate the probabilities that one, two, and three people will be bumped, and then use those values to determine the probability that at least one passenger will be bumped. Give each answer to four decimal places. Avoid rounding within calculations.
P(one person is bumped)=
P(two people are bumped)=
P(three people are bumped)=
P(at least one person is bumped)=
Overselling is when more confirmed customers show up to use the vehicle than there is space available. When this happens, at least one customer will be denied the service that they paid for, either voluntarily (sometimes with an incentive provided by the supplier) or involuntarily. This is called getting "bumped."
Suppose that for a particular flight, an airline believes that 1% of ticket holders do not make the flight. The jet making the trip holds 188 passengers. If the airline sells 191 tickets, what is the probability that the flight will be oversold and they will have to bump a passenger? Assume that cancellations are independent.
Calculate the probabilities that one, two, and three people will be bumped, and then use those values to determine the probability that at least one passenger will be bumped. Give each answer to four decimal places. Avoid rounding within calculations.
P(one person is bumped)=
P(two people are bumped)=
P(three people are bumped)=
P(at least one person is bumped)=
1 answer:
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