Question

Four friends decide to go to McDonald’s. It was reported that the location they visited filled approximately 90.9% of the orders correctly. What is the probability that none of the orders will be filled correctly? Which probability model should you use to calculate the answer?

Answer 1

Answer:

0.0069% probability that none of the orders will be filled correctly, using the binomial probability distribution.

Step-by-step explanation:

For each order, there are only two possible outcomes. Either it is filed correctly, or it is not. The probability of an order being filed correctly is independent from other orders. 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.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[\tex]In which [tex]C_{n,x}[\tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[\tex]And p is the probability of X happening.It was reported that the location they visited filled approximately 90.9% of the orders correctly.This means that [tex]p = 0.909$

Four friends decide to go to McDonald’s.

This means that $n = 4$

What is the probability that none of the orders will be filled correctly? Which probability model should you use to calculate the answer?

This is P(X = 0).

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[\tex]

[tex]P(X = 0) = C_{4,0}.(0.909)^{0}.(0.091)^{4} = 0.000069[\tex]

0.0069% probability that none of the orders will be filled correctly, using the binomial probability distribution.