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Worked solution

Reverse an overbooking event into a conditional no-show count

An airline company sold 105 tickets for a particular flight. Based on past experience, 8% of passengers do not turn up for the flight. Let XX be the number of passengers who do not turn up for that flight.

(i) State, in context, two assumptions needed for XX to be well modelled by a binomial distribution. marks

(ii) Find the probability that fewer than 5 do not turn up for the flight. marks

“Overbooking” occurs when there are more passengers who turn up than the number of seats available for the flight. When this happens, the airline will re-accommodate the excess passengers, that is, arrangements will be made for them to take another flight.

(iii) It is given that the particular flight has only 100 seats. Find the probability that at least 3 passengers have to be re-accommodated given that “overbooking” has occurred. marks

[2017/MJC/MYE/I/9]

CJC · Review Problems · Question 18

Answer

(ii) 0.07060.0706; (iii) 0.1150.115.

(i)

Assume each passenger has the same no-show probability .08 and that whether one passenger turns up is independent of whether any other turns up.

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(ii)

Let X∼B(105,.08)X\sim B(105,.08). Then P(X<5)=P(X≤4)≈.07062938348P(X<5)=P(X\leq4)\approx.07062938348, printed as .070629 (5 s.f.) and .0706 (3 s.f.).

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(iii)

The number turning up is 105−X105-X. Overbooking means 105−X>100105-X>100, or X≤4X\leq4. At least three excess passengers means 105−X−100≥3105-X-100\geq3, or X≤2X\leq2. Hence P(X≤2∣X≤4)=P(X≤2)P(X≤4)≈.008106234001.07062938348=.1147714110=.115P(X\leq2\mid X\leq4)=\dfrac{P(X\leq2)}{P(X\leq4)}\approx\dfrac{.008106234001}{.07062938348}=.1147714110=.115 (3 s.f.). The printed rounded numerator .0081062 and denominator .070629 give the same final result.

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Common mistakes

More excess passengers means fewer no-shows; overbooking excludes X=5.

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