If 20 reservations are confirmed for a flight, what is the probability that no passengers will be bumped and there will be no empty seats on the plane? (Note: “Bumped” means that a passenger holding a reservation will not have a seat.)
The correct answer and explanation is:
To answer this question, some important details are needed but are not provided explicitly: the number of seats on the plane and the probability that each passenger actually shows up for the flight. Usually, problems like this assume a plane has a fixed number of seats and each passenger independently shows up with some probability.
Let’s assume the plane has exactly 20 seats (since the question mentions 20 reservations). The goal is to find the probability that exactly 20 passengers show up out of 20 reservations, so that no one is bumped and there are no empty seats.
If each passenger shows up independently with probability pp, the number of passengers who show up follows a Binomial distribution: X∼Binomial(n=20,p)X \sim \text{Binomial}(n=20, p)
The probability that exactly 20 passengers show up is: P(X=20)=(2020)p20(1−p)0=p20P(X = 20) = \binom{20}{20} p^{20} (1-p)^0 = p^{20}
This means that no one is bumped (because the number who show up does not exceed 20 seats) and no seats are empty (because all 20 show up).
Summary: The probability that no passengers will be bumped and no seats will be empty is simply the probability that all 20 reserved passengers show up. If each shows up independently with probability pp, the answer is: p20\boxed{p^{20}}
Explanation
This problem involves understanding the relationship between the number of reservations, the number of seats, and the actual number of passengers who show up. “Bumped” means there are more passengers than seats, so to avoid this, the number who show up must be at most 20. To have no empty seats, the number who show up must be exactly 20.
Each passenger can either show up or not, with some independent probability pp. The total number who show up follows a binomial distribution because it is a sum of independent yes/no events.
The probability of exactly kk passengers showing up out of nn reservations is given by the binomial formula: P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
Here, n=20n=20, and the case k=20k=20 represents everyone showing up.
Because the problem asks for no one to be bumped (so X≤20X \leq 20) and no empty seats (so X=20X = 20), the only value of interest is exactly 20.
If the question provided or implied a value for pp, you would plug it into p20p^{20} to get a numerical answer. Without pp, this is the simplest general form.
If the number of seats is different from 20, or the problem involves other probabilities, the answer would be adjusted accordingly. But based on the information given, this is the correct approach.