Question

Suppose that 60 percent of customers at a McDonalds purchase a hamburger, and assume that the purchases of different customers are independent. a) What is the probability that at most 7 of the next 10 customers will order a hamburger? (3 POINTS). b) What is the probability that exactly 6 of the next 10 customers will order a hamburger? (3 POINTS)

Answer 1

Answer:

a) 83.87% probability that at most 7 of the next 10 customers will order a hamburger.

b) 25.08% probability that exactly 6 of the next 10 customers will order a hamburger

Step-by-step explanation:

For each customer at McDonalds, there are only two possible outcomes. Either they purchase a hamburguer, or they do not. The purchases of different customers are independent. 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.

60 percent of customers at a McDonalds purchase a hamburger

This means that $p = 0.6$

a) What is the probability that at most 7 of the next 10 customers will order a hamburger? (3 POINTS).

This is $P(X \leq 7)$ when n = 10.

We know that either at most 7 of the next customers will order a hamburguer, or more than 7 will. The sum of these probabilities is decimal 1. So

$P(X \leq 7) + P(X > 7) = 1$

$P(X \leq 7) = 1 - P(X > 7)$

In which

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10)$

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

$P(X = 8) = C_{10,8}.(0.6)^{8}.(0.4)^{2} = 0.1209$

$P(X = 9) = C_{10,9}.(0.6)^{9}.(0.4)^{1} = 0.0403$

$P(X = 10) = C_{10,10}.(0.6)^{10}.(0.4)^{0} = 0.0001$

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) = 0.1209 + 0.0403 + 0.0001 = 0.1613$

$P(X \leq 7) = 1 - P(X > 7) = 1 - 0.1613 = 0.8387$

83.87% probability that at most 7 of the next 10 customers will order a hamburger.

b) What is the probability that exactly 6 of the next 10 customers will order a hamburger? (3 POINTS)

This is P(X = 6).

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

$P(X = 6) = C_{10,6}.(0.6)^{6}.(0.4)^{4} = 0.2508$

25.08% probability that exactly 6 of the next 10 customers will order a hamburger