Question

A small airline overbooks flights on the assumption that several passengers will not show up. Suppose that the probability that a passenger shows up is 0.91. What is the probability that a 20-seat flight with 22 tickets sold will be able to seat all passengers who arrive?

Answer 1

We have 22 tickets sold, and 20 seats. This means that at least 2 passengers must not show up (otherwise, at least 21 passengers will be present, and there wouldn't be space for them).

Considering each passenger as independent, you can think of this experiment. Suppose you toss a coin for each passenger. If the coin lands on heads, the passenger shows up. If it lands on tails, the passenger doesn't show up.

But the coin is unfair: it has a 0.91 probability of landing on heads, and thus 0.09 probability of landing on tails.

This implies that the probability of having exactly $k$ tails is

$\binom{22}{k} \cdot 0.09^k \cdot 0.91^{22-k}$

We already concluded that at least two passengers must not show up. So, if our coins lands on tails less than twice, we've lost. So, the losing probability is

$\displaystyle\binom{22}{0}\cdot 0.09^0 \cdot 0.91^{22} + \binom{22}{1}\cdot 0.09^1 \cdot 0.91^{21} \approx 0.39$

Finally, remember the rule to negate events:

$P(E) = 1-P(\lnot E)$

So, if we lose with probability 0.39, we win with probability

$1-0.39 = 0.61$