Question

Suppose that an airline accepted 12 reservations for a commuter plane with 10 seats.
They know that 7 reservations went to regular commuters who will show up for sure.
The other 5 passengers will show up with a 50% chance, independently of each other.
Find the probability that the
flight will be overbooked.

Answer 1

Let, X = the random variable denoting the number of non-regular passengers who will show up, X = 0, 1, 2, 3, 4, 5.

Since, the chance of each of these passengers will show up is 50% i.e. 0.5 and is independent of each other

Therefore, X ~ bin(n = 5, p = 0.5)

The p.m.f. of X is given by,

P(X = x) = $\begin{cases} \dbinom{5}{x}(0.5)^x(1-0.5)^{5-x} &\text{for } x=0,1,...,5 \\ 0 &\text{otherwise } \end{cases}$

The commuter plane has 10 seats and out of 12 reservations, airline knows that 7 reservations went to regular commuters who will show up for sure. So, there are 3 seats remaining and 5 reservations. If more than 3 passengers out of 5 reservations show up, then the flight will be overbooked.

$\therefore$ The probability that the flight will be overbooked

P(X > 3)

= P(X = 4) + P(X = 5)

= $\dbinom{5}{4}(0.5)^4(1-0.5)^{5-4}+\dbinom{5}{5}(0.5)^5(1-0.5)^{5-5}$

= 0.15625 + 0.03125

= 0.1875

Answer: The probability that the flight will be overbooked is 0.1875.