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An airline operates a flight having 50 seats. As they expect some passenger to not show up, they overbook the flight by selling 51 tickets. The probability that an individual passenger will not show up is 0.01, independent of all other tourists. Each ticket costs Rs.10 thousand, and is non-refundable if a tourist fails to show up. If a tourist shows up and a seat is not available, the airline has to pay a compensation of Rs.1lakh to that passenger. What is the expected revenue of the airline?

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Let R be a random variable indicating revenue of airlines,and P(R) be the probability of that event.

Probability of turning up=0.01 and not turning up=1-0.01=0.99

Case 1:

0 people are not showing up or 51 people are showing up ,seat of 1 person will not be available as there are 50 seats and all 51 persons showed up,and they have to pay the compensation for a single person.Let this probability be P.

Case 2:

If at least 1 person not shows up or at most 50 person shows up,in all such cases,airlines will be earning 10k for all 51 tickets sold.Let this probability be Q=1-P (As, it's either the first case or second).

Distribution table goes as follows:-

$R_{i}$ $(51\times 10^{4})- 10^{5}$ $(51\times 10^{4})$
$P(R_{i})$ $(0.01)^{0}(0.99)^{51}$ $1-(0.01)^{0}(0.99)^{51}$

$E(R)=\sum_{i=1}^{2}R_{i}\times P(R_{i})$

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P(Passenger $x$ shows up) = 0.99 for some passenger $x$

P(All passengers show up) = $(0.99)^{51}$ = 0.599

Revenue from 51 tickets = 51 * 10000, irrespective of how many passengers show up

Loss if all passengers show up = 100000

Loss occurs with probability 0.599

Therefore, net income = 51 * 10000 – 0.599*100000 = 4.5 Lakh

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