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:

Since we have given that

Probability that does not show up = 0.10

Probability that show up = 0.90

Here, we use "Binomial distribution":

n = 125 and p = 0.90

Number of passengers that hold in a flight = 120

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

$P(X\leq 120)=\sum_{x=0}^{120}^{125}C_x(0.90)^x(0.10)^{125-x}=0.9961$

(b) What is the probability that the flight departs with empty seats?

$P(X\leq 119)=\sum _{x=0}^{119}^{125}C_x(0.90)^x)(0.10)^{125-x}=0.9886$

Hence, a) 0.9961, b) 0.9886