Binomial and Normal Distribution Probability Problems: Airport Limousine

Published on June 11, 2026

Statistics

Solve probability problems involving binomial distributions for passenger no-shows and normal distributions for daily profit, calculating expected values and probabilities.

Question

Solve

Answer

Based on the problem description, we have the following parameters:
  • Maximum capacity of the limousine: passengers.
  • Maximum reservations accepted: .
  • Probability that a passenger with a reservation does not show up: .
  • Probability that a passenger with a reservation shows up: .
Let be the random variable representing the number of passengers who show up for a trip. Since each reservation is independent, follows a binomial distribution: .

(a) Probability that at least one individual cannot be accommodated

An individual cannot be accommodated if the number of passengers who show up exceeds the capacity of the limousine ( ).
We need to find .
Using the binomial formula :
Summing these probabilities:
Answer (a): The probability is .

(b) Expected number of available places when the limousine departs

The number of available places is the capacity minus the number of passengers who actually board. Since the limousine can only take up to 4 people, the number of passengers boarding is .
The number of available places is .
  • If
  • If
  • If
  • If
  • If
We calculate the probabilities for :
The expected value is:
Answer (b): The expected number of available places is .

© Probability of recording a daily profit two standard deviations above the mean

The problem states that the distribution of daily profit can be approximated by a normal distribution. Let be the standard normal variable , where is the daily profit, is the mean, and is the standard deviation.
We are asked for the probability of recording a profit “two standard deviations above the mean”, which means .
This is equivalent to finding .
Using the standard normal distribution table:
Answer ©: The probability is approximately .