A major airline knows that about 2 percent of passengers do not manage to show up for their flight. Consider a small aircraft having 50 seats, and suppose that an airline sells precisely 50 tickets for a specific flight. Assuming that all passsengers are independent and have the same probability of not showing up, compute the probability that

a. The probability that the aircraft is fully occupied = 0.3642

b. The probability that between 1 and 5 (including 1 and  5) passengers do not show up is  = 0.6354

Step-by-step explanation:

From the given information:

We are to compute the probability that:

a. The aircraft is fully occupied( i.e. all the passengers show up)

b. Between 1 and 5 (including 1 and  5) passengers do not show up

Given that:

Percentage of passengers that do not show up  = 2% = 0.02

And the population of the passengers = 50; since 50 tickets are being sold for a specific flight.

Let assume X be the random variable that follows a binomial distribution, thus, the number of passengers from the aircraft with 50 passengers can be represented as:

Thus;

The probability that the aircraft is fully occupied i.e. all the passengers show up is:

= 0.98⁵⁰

= 0.3642

Hence, The probability that the aircraft is fully occupied = 0.3642

b.  The probability that between 1 and 5 (including 1 and  5) passengers do not show up is:

= P(X = 1) + P(X =2) + P(X = 3) + P(X = 4) + P(X = 5)

=

= 0.6354

Hence, The probability that between 1 and 5 (including 1 and  5) passengers do not show up is  = 0.6354

36 is the final answer! : )