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Thread: Probability dilemma?

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    Probability dilemma?




    Hey everybody..

    So my friend and I (high school seniors) had a bit of a disagreement. To get to the point, my contention is that, the more colleges a student applies to, the greater that student's probability of getting accepted into *at least one* of those colleges.

    Take the Ivy League schools for example. The acceptance rate at Harvard is 5.9%.. so the average applicant has a 5.9% chance of getting accepted (obviously better qualified applicants have better chances of getting accepted, but in this case we are assuming that the applicant is perfectly average in every way). Assuming I'm average, I have a 5.9% shot of getting into Harvard - but say I also apply to the 7 other Ivy League schools? Doesn't my chance of getting into at least 1 - so 1 or more - of those schools go up? By the way, the events are completely independent - one college's decision does not impact another college's decision.

    I looked around and I found a similar example: say you have 8 bags of marbles, each one containing 10 marbles - 9 white and 1 green. The bags can symbolize colleges, the white marbles can symbolize rejected applicants, and the green marbles symbolize accepted applicants. If I draw one marble out of each bag, my chance of getting 1 green marble *each draw* is 0.1, but my chance of getting 1 or more green marbles out of all 8 draws is [1 - (9/10)^8], or 0.57.

    The only problem with applying this formula to college acceptances is that each college has a different acceptance rate. The acceptance rates of the 8 Ivy League schools are 5.9%, 8.6%, 7.0%, 14.0%, 11.5%, 9.9%, 7.3%, and 6.3%. I applied to all 8 Ivy's. Assuming I'm a completely average applicant, how do I calculate my chance of getting into at least 1 Ivy?

    Thanks!

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    Re: Probability dilemma?

    You still have a similar formula. The probability of getting none of these happen is the product of probabilities of each not happening, and thus you get

    \left(1 - \frac {1} {10}\right) \times \left(1 - \frac {1} {10}\right) \times \ldots 
\times \left(1 - \frac {1} {10}\right) = \left(\frac {9} {10}\right)^8

    in the formula you posted. Now in this variation although the probabilities are not identical, the principle is the same - you just need to replace those \frac {1} {10} with the corresponding probabilities.

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    Re: Probability dilemma?

    Quote Originally Posted by BGM View Post
    You still have a similar formula. The probability of getting none of these happen is the product of probabilities of each not happening, and thus you get

    \left(1 - \frac {1} {10}\right) \times \left(1 - \frac {1} {10}\right) \times \ldots \times \left(1 - \frac {1} {10}\right) = \left(\frac {9} {10}\right)^8

    in the formula you posted. Now in this variation although the probabilities are not identical, the principle is the same - you just need to replace those \frac {1} {10} with the corresponding probabilities.
    Ahhh of course, how did I not see that? Well I sure feel silly now. Thanks! Could you please tell me what this concept/formula/probability (probability of at least one of independent events coming out positive) is formally called?

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    Re: Probability dilemma?


    Note that the assumption of independence probably doesn't hold in this case so that formula probably won't give completely accurate results.
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