Question

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

Answer 1

Answer:

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:

$X \sim Bin (n = 50, p =0.02)$

Thus;

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

$= (1 - 0.02)^{50}$

= 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)

= $=(^{50}_1) 0.02(1-0.02)^{50-1} +(^{50}_2) 0.02(1-0.02)^{50-2} + (^{50}_3) 0.02(1-0.02)^{50-3} + ... + (^{50}_5) 0.02(1-0.02)^{50-5}$

$=(^{50}_1) 0.02(0.98)^{49} +(^{50}_2) 0.02(0.98)^{48} + (^{50}_3) 0.02(0.98)^{47} + ... + (^{50}_5) 0.02(0.98)^{45}$

= 0.6354

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