Question :
A travel agent company organizes a tour with ticket price $\$50$ and the ticket is non-refundable. The company has a bus with $20$ seats. The company knows that the participant might not attend the tour so the company sells $21$ ticket bus. The probability of participant not attending the tour is $0.02$, independent with the other. If the participant attends the tour and there is no vacant seat, the company will pay $\$100$. What is the expected revenue of the company?
My thought :
It is obvious the company will get $21\times\$50=\$1,050$ from the ticket selling and the company will pay if there are 21 participants attending the tour. Let $X$ be the random variable for the number of participant attending the tour. We use binomial distribution here with $p=0.02$ and $q=1-0.02=0.98$. So, the expected revenue of the company is $$\$1,050-\$100\cdot P[X>20]=\$1,050-\$100\cdot 0.02^{0}\cdot 0.98^{21}\approx\$984.57$$Is this correct? If not, what kind approach should I use to answer the problem? Any idea? Any help would be appreciated. Thanks in advance.
Best Answer
Let $X$ be the random revenue of the company. Let $Y \sim \mathrm{Binomial}(n = 21, p = 0.02)$ be the number of participants that fail to show. Then the company's revenue is a discrete (location-scale Bernoulli) random variable with probability mass function as follows: $$\begin{align*} \Pr[X = 950] &= \Pr[Y = 0], \\ \Pr[X = 1050] &= \Pr[Y \ge 1] = 1 - \Pr[Y = 0]. \end{align*}$$ The expected value of $X$ is therefore $$\mathrm{E}[X] = 950 \Pr[Y = 0] + 1050 (1 - \Pr[Y = 0]).$$ Since $\Pr[Y = 0] = (1-0.02)^{21}$, it is a straightforward calculation to obtain $\mathrm{E}[X]$.
Note that the use of explicit parametric models allows us to generalize the problem; we can compute, for example, the expected revenue of the company if they book $n = 25$ seats and for each attendee that shows up in excess of the $20$ seats on the bus, they refund \$100. In such a case, the relationship between the number of no-shows $Y$ and the revenue $X$ can be described by $$X = \begin{cases} 1050 - 100(5-Y), & 0 \le Y \le 4, \\ 1050, & 5 \le Y \le 25. \end{cases}$$ This permits us to express $\mathrm{E}[X]$ in the following manner: $$\mathrm{E}[X] = 1050 \Pr[Y \ge 5] + \sum_{y = 0}^4 (1050 - 100(5-y))\Pr[Y = y],$$ where in this case $Y \sim \mathrm{Binomial}(n = 25, p = 0.02)$ if the same individual probability of a no-show is assumed.