Wikipedia:Reference desk/Archives/Mathematics/2007 January 1

= January 1 =

Probability Formula
I am looking for formula that would solve this question and ones like it:

If I have a deck of 52 standard cards (26 black, 26 red). If I draw 20 cards, what is the chance that 15, or more, are red? --155.144.251.120 23:50, 1 January 2007 (UTC)


 * I don't think there's a simple formula (although there would be simplifications available in certain special cases) but the first step is calculating the probability of getting exactly 15 red, which you do as follows: the total number of ways to draw 20 cards is 52C20. The number of ways to draw 15 red (and hence 5 black) is 26C15 * 26C5 (i.e. the number of ways to draw 15 from the group of red AND 5 from the group of black), so the probability is 26C15 * 26C5 / 52C20 = 0.00403 according to both Windows and Google calcuators. Then you just have to do that for drawing 16 red, 17 red, ... 20 red, and add them up. There may be an easier way but sadly I can't think of it. Confusing Manifestation 08:48, 2 January 2007 (UTC)
 * (I have added subscripting.) The notation “52C20” denotes the number of combinations of 52 distinct items taken 20 at a time, with numeric value given as a binomial coefficient,
 * $$ {}_{n}C_{k} = C(n,k) = n?k = {n \choose k} = \frac{n!}{k! (n-k)!}. $$
 * Here n! denotes, as usual, the factorial,
 * $$ n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1 . \,\!$$
 * For example, C(26,5) is 65780. We might be able to use Vandermonde’s convolution,
 * $$ \sum_{k \in \mathbb{Z}} {r \choose {m+k}} {s \choose {n-k}} = {{r+s} \choose {m+n}}, $$
 * to simplify the proposed sum. (See Graham, Knuth, Patashnik, Concrete Mathematics.) --KSmrqT 03:20, 3 January 2007 (UTC)