Question

A recent survey of students at John Tukey High School revealed that

18% of the students are in favor of changing the dress code. If you ran-
domly select 15 students from this school, what is the probability that
at least three of these students are in favor of the proposal to change the
dress code?

Answer 1

0.5234 = 52.34% probability that at least three of these students are in favor of the proposal to change the dress code.

Explanation:

For each student, there are only two possible outcomes. Either they are in favor, or they are not. Students are independent. This means that 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.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

18% of the students are in favor of changing the dress code.

This means that
`p = 0.18`

You randomly select 15 students

This means that
`n = 15`

What is the probability that at least three of these students are in favor of the proposal to change the dress code?

This is

`P(X \geq 3) = 1 - P(X < 3)`

In which

`P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)`

In which

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(15,0).(0.18)^(0).(0.82)^(15) = 0.051`

`P(X = 1) = C_(15,1).(0.18)^(1).(0.82)^(14) = 0.1678`

`P(X = 2) = C_(15,2).(0.18)^(2).(0.82)^(13) = 0.2578`

`P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.051 + 0.1678 + 0.2578 = 0.4766`

`P(X \geq 3) = 1 - P(X < 3) = 1 - 0.4766 = 0.5234`

0.5234 = 52.34% probability that at least three of these students are in favor of the proposal to change the dress code.