# Binomial Probability Distribution

## Homework Statement

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. Consider a random sample of 25 students who have recently taken the test.

a.) What is the probability that exactly 1 received a special accommodation?
b.) What is the probability that at least 1 received a special accommodation?

## Homework Equations

$$\begin{pmatrix} n\\ x\\ \end{pmatrix}p^x(1-p)^{(n-x)}$$

## The Attempt at a Solution

a.) $$\begin{pmatrix} 25\\ 1\\ \end{pmatrix}.02^x(1-.02)^{(25-1)} = .3079$$

30% seems kind of high. If 40,000 students out of 2,000,000 were {success} and chose 25 of the 2,000,000, there's a 30% chance I'd get 1 of the 40,000? That, just seems too high for some reason.

b.) $$\sum_{x=1}^{25} \begin{pmatrix} 25\\ x\\ \end{pmatrix}.02^x(.98)^{(25-x)}$$

Does this look right? How can I calculate this? I don't believe I have to simplify it, per instructor's orders...but I'm not sure if I have it right to begin with.

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## Homework Statement

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. Consider a random sample of 25 students who have recently taken the test.

a.) What is the probability that exactly 1 received a special accommodation?
b.) What is the probability that at least 1 received a special accommodation?

## Homework Equations

$$\begin{pmatrix} n\\ x\\ \end{pmatrix}p^x(1-p)^{(n-x)}$$

## The Attempt at a Solution

a.) $$\begin{pmatrix} 25\\ 1\\ \end{pmatrix}.02^x(1-.02)^{(25-1)} = .3079$$

30% seems kind of high. If 40,000 students out of 2,000,000 were {success} and chose 25 of the 2,000,000, there's a 30% chance I'd get 1 of the 40,000? That, just seems too high for some reason.

b.) $$\sum_{x=1}^{25} \begin{pmatrix} 25\\ x\\ \end{pmatrix}.02^x(.98)^{(25-x)}$$

Does this look right? How can I calculate this? I don't believe I have to simplify it, per instructor's orders...but I'm not sure if I have it right to begin with.
As far as A is concerned, it looks OK to me... wait for other opinions :\

as far as B is concerned, you can find the probability that zero out of the 25 students get the special accommodation and calculate
1-(zero out of 25)= at least one

they are the complements of each other.

(none)+(at least 1)= 1

That is exactly what I thought. however, how can you calculate that 0 of 25 are {success}? It would come out as this:

$$1-[{ \begin{pmatrix} 25\\ 0\\ \end{pmatrix}.02^0(1-.02)^{(25-0)}}]$$

Do I ignore the 25C0 at the beginning? If I do, I get:

$$1-[{(1-.02)^{(25)}}] = 1 - .603 = .397$$

25 choose 0 is 1. Intuitively there is only one way to choose 0 things (namely by choosing nothing). But 0! is defined to be 1. Also your part a) is correct. It's only a sample after all.

that is exactly what i thought. However, how can you calculate that 0 of 25 are {success}? It would come out as this:

$$1-[{ \begin{pmatrix} 25\\ 0\\ \end{pmatrix}.02^0(1-.02)^{(25-0)}}]$$

do i ignore the 25c0 at the beginning? If i do, i get:

$$1-[{(1-.02)^{(25)}}] = 1 - .603 = .397$$
c(25,0)=1

Ooh. Right. 0!=1. Somehow, that slipped my mind.

Thanks a lot.

vela
Staff Emeritus