The College Board reports that 2% of the 2 million high school students who take the SAT each year receive special accommodations because of documented disabilities (Los Angeles Times, July 16, 2002). Consider a random sample of 25 students who have recently taken the test.
What is the probability that exactly 1 received a special accommodation?

For each student who took the SAT, there are only two possible outcomes. Either they received special acommodations, or they did not. So 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.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

In this problem we have that:

2% of the 2 million high school students who take the SAT each year receive special accommodations because of documented disabilities (Los Angeles Times, July 16, 2002). This means that [tex]p = 0.02[/tex]

Consider a random sample of 25 students who have recently taken the test. What is the probability that exactly 1 received a special accommodation?

This is [tex]P(X = 1)[/tex] when [tex]n = 25[/tex]. So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 1) = C_{25,1}.(0.02)^{1}.(0.98)^{24} = 0.3079[/tex]

There is a 30.79% probability that exactly 1 received a special accommodation.