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I am trying to deal with a problem that says "40% of people are known to have a condition X. Find a probability that 12/14 randomly selected people will have a condition X." I thought I could do the following: Binomial coefficient 14 pick 2, times $(0.4^{12}*0.6^{2})$ since the order doesn't matter. To test this, I tried testing it for 10 people, expecting the answer equal to 40%. I did (10 combination 4) * $(0.4^{4}*0.6^{6})$ but got only ~25%. What am I doing wrong? Thanks.

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2 Answers

up vote 1 down vote accepted

You have done nothing wrong, except expecting the answer equal to 40%.

If you tried with $n=1000$, you would think the probability of exactly $400$ people having the condition to be fairly small, and so it is, at about $0.026$.

If you want 40% as the answer, then you need to ask a different question, for example working out the probabilities for $0$ to $n$ (where $n=10$ in you second example), and then working out the expected proportion to have the condition. So $$\sum_{i=0}^{n} \frac{i}{n} {n \choose i} 0.4^i 0.6^{n-i}$$ which should come out to be $0.4$.

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Your calculation looks correct, so your expectation of 40% for your test case was wrong. =)

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So I was not really calculating something that has 40% probability of happening, I think I am getting it... –  probq Jul 27 '11 at 12:37
    
@probq: The thing that has 40% probability of happening is any individual having the condition. Maybe making it coin flipping helps. The chance of one toss being heads is 50%. But that doesn't mean the chance of getting $n$ heads out of $2n$ trials is 50%. –  Ross Millikan Jul 27 '11 at 12:57
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