Question

Because all airline passengers do not show up for their reserved seat, an airline sells 125 tickets for a flight that holds only 120 passengers. The probability that a passenger does not show up is 0.10, and the passengers behave independently. a. What is the probability that every passenger who shows up can take the flight?b. What is the probability that the flight departs with empty seats?

Answer 1

Answer:

a) 0.9961

b) 0.9886          

Step-by-step explanation:

We are given the following information:

We treat passenger not showing up as a success.

P(passenger not showing up) = 0.10

Then the number of passengers follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 125

a) probability that every passenger who shows up can take the flight

$P(x \geq 5) = 1- P(x = 0) - P(x = 1)-P(x = 2) - P(x = 3) - P(x = 4) \\= 1-\binom{125}{0}(0.10)^0(1-0.10)^{125} -...-\binom{125}{4}(0.10)^4(1-0.10)^{121}\\=0.9961$

b)  probability that the flight departs with empty seats

$P(x > 5) =P(x\geq 5) - P(x = 5) \\= 0.9961 -\binom{125}{5}(0.10)^5(1-0.10)^{120}\\=0.9961-0.0075\\ = 0.9886$