Question

the name Joe is very common at a school in one out of every ten students go by the name. If there are 15 students in one class, what is the probability that at least one of them is named Joe

Answer 1

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.

Binomial probability distribution

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

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

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• One in ten students are named Joe, hence $p = \frac{1}{10} = 0.1$.

• There are 15 students in the class, hence $n = 15$.

The probability that at least one of them is named Joe is:

$P(X \geq 1) = 1 - P(X = 0)$

In which:

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

$P(X = 0) = C_{15,0}.(0.1)^{0}.(0.9)^{15} = 0.2059$

Then:

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.2059 = 0.7941$

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 brainly.com/question/24863377