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November 23, 2014

November 23, 2014

Posted by **Stacey** on Friday, December 15, 2006 at 8:16am.

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.

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