1. ## Binomials

We're talking about jury duty. There are 16000 men who are available for jury duty and whom only 26% are black.

What is the probability that at least one is African American?

My figure of thought was finding what the probability that all men on jury are black and then subtracting that answer from 1. It doesnt seem to be right since the answer I get is so high. Which...now that I think about it, couldent be possible. How can this be solved?

2. Originally Posted by l0v3n
We're talking about jury duty. There are 16000 men who are available for jury duty and whom only 26% are black.

What is the probability that at least one is African American?

My figure of thought was finding what the probability that all men on jury are black and then subtracting that answer from 1. It doesnt seem to be right since the answer I get is so high. Which...now that I think about it, couldent be possible. How can this be solved?

I think you have the right idea, but what you need is the prob that it is all non black.

Prob no black is

$\frac{\binom{11840}{12} \binom{4160}{0}}{\binom{16000}{12}} \approx 0.02692$

so the prob of at least one is

$1 - 0.02692=0.97308$

3. ## Jury duty

This is a combination problem i think. If there are 12 people in the jury, then the denominator should be 16000 combination 12, which is a really large number. After finding that number, you have to find the number of cases with a black person among the 12 members. Since 26% are black, that'd mean that 4160 men are black. So the possibly that there are:

12 black - 4160 C 12
11 black - 4160 C 11
10 black - 4160 C 10 etc...

add these numbers together to get the numerator of the fraction. The denominator is 16000 C 12. So that should be it. I hope T.T