Question

At a certain airport, 75% of the flights arrive on time. A sample of 10 flights is studied. Assume each flight is independent of the others. Find the probability that eight or more of the flights will arrive on time.

Answer 1

Answer:

0.5256

Step-by-step explanation:

X ~ B(10,0.75)

q = 1 - 0.75 = 0.25

Eight or more:

P(X = 8,9,10)

10C8×0.75⁸×0.25² +

10C9×0.75⁹×0.25 +

10C10×0.75¹⁰×0.25⁰

= 0.525592804

= 0.5256

Answer 2

Answer:

52.56% probability that eight or more of the flights will arrive on time.

Step-by-step explanation:

For each flight, there are only two possible outcomes. Either it is on time, or it is not. The probability of a flight being on time is independent from other flights. So we use the binomial probability distribution to solve this question.]

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

At a certain airport, 75% of the flights arrive on time.

This means that $p = 0.75$

A sample of 10 flights is studied.

This means that $n = 10$

Find the probability that eight or more of the flights will arrive on time.

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{10,8}.(0.75)^{8}.(0.25)^{2} = 0.2816$

$P(X = 9) = C_{10,9}.(0.75)^{9}.(0.25)^{1} = 0.1877$

$P(X = 10) = C_{10,10}.(0.75)^{10}.(0.25)^{0} = 0.0563$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.2816 + 0.1877 + 0.0563 = 0.5256$

52.56% probability that eight or more of the flights will arrive on time.