The airline is selling tickets for $100 for a plane with 6 seats. Each ticket holder independently has the probability of 0.15 of not turning up to the flight. Suppose 7 people want tickets. The airline has a choice of two strategies:
X: sell 6 tickets
Y: sell 7 tickets, but if everyone turns up the airline has to pay $300 in compensation.
Let X and Y be the random variables denoting the money made by following strategy X and Y respectively.
TRUE or FALSE: E(Y)>E(X)?
My approach:
Probability of showing up of a passenger = 1 – 0.15 = 0.85
Profit earned in case of X = {600}
Probable profit earned in case of Y = {700,400} (700 earned when 1 passenger doesnt show up and 400= 700-300 in compensation when all passengers show up)
P(X)= 0.85*0.85*0.85*0.85*0.85*0.85
E(X)= P(X) * 600 = 226.2897
P(Y) = 7C6 * 0.15 * (0.85)^6 + 7C7 * (0.85)^7
E(Y) = 7C6 * 700 * 0.15 * (0.85)^6 + 7C7 * (0.85)^7 * 400 = 405.436
Hence my solution is TRUE.
Is this approach and my answer correct?
Best Answer
Presumably the passengers pay regardless of whether they show up or not? If so, $E(X) = 600$. If, as you were attempting to show, they only pay if they show up, your approach is not correct. Each passenger is independent of one another, so letting $X_i$ be a Bernoulli variable that is $1$ indicating arrival with $p = 0.85$. \begin{equation} E(X) = \sum_{i=0}^6 E(100X_i) = 100\sum_{i=0}^6 P(X_i) = 510 \end{equation}
Again for the Expectation of $Y$ you have the problem that it is unclear whether passengers only pay if they show up/not. In your calculations here, you have now made the assumption that they pay regardless of whether they show up, while you made the opposite assumption. Your answer here fails to account for the cases that only $1,2,\ldots,5$ passengers show. I will assume that everyone who got a ticket pays, in which case you were close.
Let $N$ indicate the number of passengers who show up, $N \sim Binom(7, 0.85)$:
\begin{equation} E(Y) = 400P\{N=7\} + 700(1-P\{N=7\}) = 400\cdot0.85^7 + 700(1-0.85^7) \end{equation} So $E(Y) = 603.83\ldots$ which is marginally better than $E(X)=600$ under the same assumption. If only people who show up pay, you will need to calculate $E(Y)$ given all of the cases of N.