Using the binomial distribution, it is found that the relative frequency of rolling a 3 at least once is of 0.72.
For each roll, there are only two possible outcomes, either the result is a 3, or it is not. The probability of a roll resulting in a 3 is independent of any other roll, hence the binomial distribution is used to solve this question.
The formula is:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
In this problem:
The probability of rolling a 3 at least once is given by:
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{7,0}.(0.1667)^{0}.(0.8333)^{7} = 0.28[/tex]
Then:
[tex]P(X \geq 1) = 1 - P(X = 0) = 0.72[/tex]
More can be learned about the binomial distribution at https://brainly.com/question/14424710