an airline has a policy of booking as many as 24 person on an airplane that can seat only 23 (past studies have revealed that only 94 % of booked passengers actually arrive for the flight) find the probability that if the airline booked 24 persons, not enough seats will be available
P(x = 24 = 24C24
binompdf(24, .94, 24) =?
24C24(.94)^24(.06)^0 =?
To find the probability that if the airline booked 24 persons, not enough seats will be available, we need to calculate the probability of more than 23 passengers showing up for the flight.
Given that past studies have revealed that only 94% of booked passengers actually arrive for the flight, we can assume that the probability of a passenger showing up for the flight is 0.94.
Let's use the binomial probability formula to calculate the probability:
P(X > 23) = sum of P(X = 24) + P(X = 25) + ... + P(X = 49)
Where P(X) is the probability of X passengers showing up for the flight, and X ranges from 24 to 49 (since we are considering the possibility of all 24 passengers showing up plus additional passengers).
To calculate each individual probability, we can use the binomial probability formula:
P(X = k) = nCk * p^k * (1-p)^(n-k)
Where n is the total number of trials (number of passengers booked), k is the number of successful trials (number of passengers showing up), p is the probability of success (probability of a passenger showing up), and (1-p) is the probability of failure (probability of a passenger not showing up).
Now, let's calculate the probability P(X > 23):
P(X > 23) = P(X = 24) + P(X = 25) + ... + P(X = 49)
P(X > 23) = (24C24 * 0.94^24 * (1-0.94)^(24-24)) + (24C25 * 0.94^25 * (1-0.94)^(24-25)) + ... + (24C49 * 0.94^49 * (1-0.94)^(24-49))
You can plug in these values into a calculator or use a statistical software program to calculate the exact value of P(X > 23).
Please note that this calculation assumes that each passenger's arrival is independent of others, which may not be a perfect representation of reality. Additionally, other factors such as overbooking policies and passenger preferences can affect the actual probabilities in practice.