Answer:
We want to find the following probability:
[tex] P(X=3)[/tex]
And using the probability mass function we got
[tex]P(X=3)=(3C3)(0.95)^3 (1-0.95)^{3-3}=0.857[/tex]
Step-by-step explanation:
Let X the random variable of interest "number of orders on time", on this case we now that:
[tex]X \sim Binom(n=3, p=1-0.05=0.95)[/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 the following probability:
[tex] P(X=3)[/tex]
And using the probability mass function we got
[tex]P(X=3)=(3C3)(0.95)^3 (1-0.95)^{3-3}=0.857[/tex]