Question

Suppose a college is considering a new placement test, which has 40% of all students taking remedial arithmetic. If 8 randomly chosen students take the placement test, what is the probability that exactly 2 of them will have to take remedial arithmetic?

Answer 1

The probability is `~~ 20.9%`.

Explanation:

Each student has an independent chance of 0.4 (or 40%) of taking remedial arithmetic. Assuming the population of students is large enough that sampling without replacement does not affect that 40% probability for the remaining students, we can apply the Binomial theorem.

Let `X` be the number of students chosen out of 8 that will have to take remedial arithmetic. Then `X` is `"Bin"(n=8,p=0.4)`. So

`P(X=x)=((n),(x))p^x(1-p)^(n-x)=((8),(x))0.4^x(0.6)^(8-x)`

and so

`P(X=2)=((8),(2))0.4^2(0.6)^(6)`
`color(white)(P(X=2))=28(0.16)(0.046656)`
`color(white)(P(X=2))=0.20901888`
`color(white)(P(X=2))~~0.209"                        "=20.9%`.