Respuesta :
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.
[tex]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.
10% of his customers buy a magazine
This means that [tex]P = 0.1[/tex]
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
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 5) = C_{13,5}.(0.1)^{5}.(0.9)^{8} = 0.0055[/tex]
0.55% probability that exactly 5 out of the first 13 customers buy a magazine
Answer:
[tex] P(X=5) = (13C5) (0.1)^5 (1-0.1)^{13-5}= 0.00554[/tex]
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:
[tex]X \sim Binom(n=13, p=0.1)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
We want to find this probability:
[tex] P(X=5)[/tex]
And using the probability mass function we got:
[tex] P(X=5) = (13C5) (0.1)^5 (1-0.1)^{13-5}= 0.00554[/tex]
So then he probability that exactly 5 out of the first 13 customers buy a magazine is 0.0554