Question

The owner of a small deli is trying to decide whether to discontinue selling magazines. He suspects that only 10% of his customers buy a magazine and he thinks that he might be able to use the display space to sell something more profitable. Before making a final decision, he decides that for one day he will keep track of the number of customers that buy a magazine. Assuming his suspicion that 10% of his customers buy a magazine is correct, what is the probability that exactly 5 out of the first 13 customers buy a magazine?

Answer 1

Answer:

$P(X=5) = (13C5) (0.1)^5 (1-0.1)^{13-5}= 0.00554$

So then  he probability that exactly 5 out of the first 13 customers buy a magazine is 0.0554

Step-by-step explanation:

Let X the random variable of interest "number of customers that buy a magazine", on this case we can model the variable of interest with this distribution:

$X \sim Binom(n=13, p=0.1)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

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

We want to find this probability:

$P(X=5)$

And using the probability mass function we got:

$P(X=5) = (13C5) (0.1)^5 (1-0.1)^{13-5}= 0.00554$

So then  he probability that exactly 5 out of the first 13 customers buy a magazine is 0.0554

Answer 2

Answer:

0.55% probability that exactly 5 out of the first 13 customers buy a magazine

Step-by-step explanation:

For each customer, there are only two possible outcomes. Either they buy a magazine, or they do not. The probability of a customer buying a magazine is independent of other customers. 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.

10% of his customers buy a magazine

This means that $P = 0.1$

What is the probability that exactly 5 out of the first 13 customers buy a magazine?

This is P(X = 5) when n = 13. So

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

$P(X = 5) = C_{13,5}.(0.1)^{5}.(0.9)^{8} = 0.0055$

0.55% probability that exactly 5 out of the first 13 customers buy a magazine