# Calculating probability of Y<100

Consider an airline selling tickets for a flight that can hold $$100$$ passengers. The probability that a passenger shows up is $$0.95$$, and the passengers behave independently. Define with the random variable $$Y$$ the number of passengers that show up to board on the flight.

Suppose the airline sold exactly 100 tickets. What is the probability that the flight departs with empty seats: $$\Pr(Y < 100)$$?

• What’s the probability that everyone shows up?
– Joe
Apr 2, 2021 at 18:59
• The probability that a passenger shows up is 0.95.
– user909773
Apr 2, 2021 at 19:00
• Yes, the probability that any one specific person shows up is 0.95. So what’s the probability that all 100 passengers show up?
– Joe
Apr 2, 2021 at 19:01
• If you’re stuck on next steps, here’s a thought. You can rely on an independence assumption to compute that. Each passenger is independent. Apr 2, 2021 at 19:03

Let $$E$$ denote the event that the plane departs with one or more empty seats, i.e. that $$Y<100$$. Let $$F$$ denote the event that the plane departs full, i.e. $$Y=100$$. Note that $$F$$ is the complement of $$E$$, so $$P(E)+P(F)=1$$. Now, $$F$$ occurs if and only if every passenger shows up. Therefore, by independence

$$P(F) = 0.95^{100}\approx 0.006$$

and so

$$P(E) = 1 - 0.95^{100} \approx 0.994.$$

The fact that $$E$$ occurs with fairly high probability provides some justification for the practice of overbooking.

• For whoever didn't hit the link, overbooking is when airlines sell more tickets than seats are available because they know not everyone will come, which results in some people being offered money to leave the plane or forced off the plane like United Airlines if they overbook too much. Apr 2, 2021 at 19:41