Using the binomial distribution, it is found that there is a 0.7941 = 79.41% probability that at least one of them is named Joe.
For each student, there are only two possible outcomes, either they are named Joe, or they are not. The probability of a student being named Joe is independent of any other student, hence, the binomial distribution is used to solve this question.
[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 that at least one of them is named Joe is:
[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_{15,0}.(0.1)^{0}.(0.9)^{15} = 0.2059[/tex]
Then:
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.2059 = 0.7941[/tex]
0.7941 = 79.41% probability that at least one of them is named Joe.
To learn more about the binomial distribution, you can take a look at https://brainly.com/question/24863377