# Math - Probability

posted by
**Stacey**
.

I want to know that if 1 person in 50 has one blue eye and one green and you have 700 people living in an area in groups of four. What is the probability that three of those people would end up together?

Would it be greater or less of one of those people only had a relation who had this eye feature?

I know this is an odd one but it's been bugging me. Thanks to those who attempt to answer this.

Out of 700, you have 14 candidates.

Pr(three)=14/700 * 13/699 * 12/698 * 686/697 = 6.29380573 × 10-6 or about a six in a million chance.

Not quite bobpursley.

You have given the probability of a particular drawing sequence having exacty three "with blue-gree". That is, the sequence is (yes,yes,yes,no). There are exactly 4 different sequences that would produce three yeses (n,y,y,y), (y,n,y,y), (y,y,n,y) and (y,y,y,n).

Soooooo, in short, multiply your answer by 4

Actually, we shouldn't assume that there are 14 people with blue-green eyes. The probability is 4*(1/50)^3*(49/50)= 3.1*10^(-5)

Count Iblis is right in that, under the original question, we shouldnt assume that there are 14 people with green eyes. However, I think Count Iblis's solution applies if we were drawing from a very large population. I think that since the problem asks us to draw from a finite population of 700, this changes the solution criteria. (Or is 700 sufficiently large).

This problem is beyond me.

The more I think about it, the more I think Count Iblis is right all the way around. The problem doesnt draw from a finite population of 700 but an expected population of 700. This should remove my finite population concerns.

Thanks for all your help everyone. At least I have a rough idea of the numbers involved. I am happy to assume larger numbers. I just happen to be using 700 for the problem.

I didn't need an exact answer as I know it was an odd question but the idea of the probablility was very useful so thanks to those who answered.