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10-06-2016 , 10:23 AM
There are 900 stocks from which you must pick 50 at random. What is the probability that you pick two in the top 30 return stocks? If necessary to assume a return distribution we can assume normal, but I don't think it is?
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10-06-2016 , 10:53 AM
If you want to know the probability of picking exactly x stocks from the best 30, the formula is:

C(30,x)*C(870,50-x)/C(900,50)

where C(n,k) is the number of combinations (i.e. n!/(k!*(n-k)!) ). In your case x=2 and so the probability is ~ 0.2761215.

If you want to know at least 2, you use the formula above for the x=0 and x=1 case and take the complement:

1 - (C(30,1)*C(870,50-1) + C(30,2)*C(870,50-2))/C(900,50) = 0.4044247
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10-06-2016 , 10:58 AM
The hypergeometric distribution applies, though the binomial gives a very good approximation. The chance of exactly 2 top-30 stocks in a sample of 50 from a population of 900 is 27.6%. I think a more meaningful situation is the probability of at least two, and that value is 50.6%, The binomial gives 50.0%

Edit NTG has it but he has a 'math typo' for at least 2
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10-06-2016 , 01:29 PM
Ta, both are useful.
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10-06-2016 , 02:09 PM
Quote:
Originally Posted by statmanhal
NTG has it but he has a 'math typo' for at least 2
Yep, my bad. I plugged 1 and 2 instead of 0 and 1. The correct formula:

1 - (C(30,0)*C(870,50-0) + C(30,1)*C(870,50-1))/C(900,50) = 0.5056985
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