Question

A small airline has a policy of booking as many as 60 persons on an airplane that can seat only 53. (Past studies have revealed that only 77% of the booked passengers actually arrive for the flight.)

Find the probability that if Air-USA books 60 persons, not enough seats will be available.

Answer 1

Answer:

Approximately $0.83\%$, assuming that the presence of the passengers are all independently of one another.

Step-by-step explanation:

The question suggests that the event that each passenger shows up is a random event with a chance of $0.77$. Let $X_1$, $X_2$, $\dots$, $X_{60}$ represent whether each of the sixty passenger showed up, with $1$ denoting that the passenger showed up for the flight, and $0$ otherwise.

Let $Y$ denote the sum of these sixty random variables. $Y\!$ would represent the number of passengers that show up for the flight.

Each of these $X$ would be a Bernoulli random variable with probability $p = 0.77$.

Assume that whether one passenger show up does not depend on that of another passenger. In other words, assume that those sixty $X_1$, $X_2$, $\dots$, $X_{60}$ are all independent of one another.

$X_1,\, X_2\, \dots,\, X_{60} \overset{\text{iid}}{\sim} \text{Bernoulli}(0.77)$.

$Y$, the sum of these $60$ independent Bernoulli random variables of the same probability $p = 0.77$, would be a binomial random variable with sixty trials ($n = 60$) and a probability of $0.77$ at each trial ($p = 0.77\!$.) That is:

$\displaystyle \left(X_1 + X_2 + \cdots + X_{60}\right) \sim B(60,\, 0.77)$.

$\displaystyle Y \sim B(60,\, 0.77)$.

For $y \in \left\lbrace1,\, 2,\, \dots,\, 60\right\rbrace$, the probability that $Y = y$ would be:

$\begin{aligned}P(Y = y) &= \begin{pmatrix}60 \\ y\end{pmatrix}\, (0.77)^{y}\, (1 - 0.77)^{60 - y} \\ &= \frac{60!}{y!\, (60 - y)\!} \, (0.77)^{y}\, (1 - 0.77)^{60 - y} \end{aligned}$.

There would not be enough seats for those passengers whenever $54 \le Y \le 60$. Calculate the probability of that event:

$\begin{aligned}P(54 \le Y \le 60) &= P(Y = 54) + \cdots + P(Y = 60)\\ & \approx 0.0083 = 0.83\%\end{aligned}$.