\n\\displaystyle
\n\\boxed{P(x=a;N,A,n,a) = \\frac{{A \\choose a} \\cdot {N-A \\choose n-a} }{{N \\choose n}}}
\n\\]<\/span><\/p>\n
where \\(N\\)<\/span> is the total number of balls in, \\(A\\)<\/span> the number of \u201cred\u201d balls in the urn, and \\(n\\)<\/span> the sample size. The question answered by the above expression is the probability of finding \\(x=a\\)<\/span> red balls in the sample.<\/p>\n However, a different – equally worthy – viewpoint is that of a tree with conditional probabilities.<\/p>\n Here is an example for \\(N=10, A=4, n=3\\)<\/span><\/p>\n So there are 3 leafs with exactly one red ball in the sample. Following the tree we multiply the conditional probabilities of the tree egdes to get to the \u201cand\u201d probability of the leafs. It should be clear that all three probabilites are identical – but for the order of multiplication:<\/p>\n \\[ How many leafs contain exactly one red ball? Exactly \\({n \\choose a} = {3 \\choose 1} = 3\\)<\/span>. The probability for the precise event \u201ca red balls followed by n-a blue balls\u201dis:<\/p>\n \\[ The left numerator is simply \\(_AV_n = A!\/(A-a)!\\)<\/span> in analogy the right numerator \\(_{N-A}V_{n-a} = (N-A)!\/(N-A-(n-a))!\\)<\/span> So, all in all \\[ \\[ which leaves us with \\[<\/p>\n
\nP (\\mbox{one red}) = 3 \\cdot P_{leaf} = 3 \\cdot \\frac{4}{10} \\cdot \\frac{6}{9} \\cdot \\frac{5}{8} = 3 \\cdot \\frac{4 \\cdot 6 \\cdot 5}{10 \\cdot 9 \\cdot 8}
\n\\]<\/span><\/p>\nAlternative formula<\/h3>\n
\n \\frac{A}{N} \\cdot \\frac{A-1}{N-1} \\cdots \\frac{A-a+1}{N-a+1} \\cdot \\frac{N-A}{N-a} \\cdot \\frac{N-A-1}{N-a-1} \\cdots \\frac{N-A-(n-a-1)}{N-a-(n-a-1)}
\n\\]<\/span> The last denominator is simply \\(N-n+1\\)<\/span>, i.e.\u00a0the full denominator is \\(_NV_n=N!\/(N-n)!\\)<\/span><\/p>\n
\n{n \\choose a} \\cdot \\frac{_AV_n \\cdot _{N-A}V_{n-a}}{_NV_n} = {n \\choose a} \\cdot \\frac{A! \\cdot (N-A)! \\cdot (N-n)!}{(A-a)! \\cdot (N-A-(n-a))! \\cdot N!}
\n\\]<\/span><\/p>\n
\n= {n \\choose a} \\cdot \\frac{\\frac{A!}{(A-a)!} \\cdot \\frac{(N-A)!}{(N-A-(n-a))!} }{\\frac{N!}{(N-n)!}} = \\frac{{A \\choose a} \\cdot {N-A \\choose n-a} }{{N \\choose n}}
\n\\]<\/span><\/p>\n
\n\\displaystyle
\n\\boxed{P(x=a;N,A,n,a) = {n \\choose a} \\cdot \\frac{_AV_n \\cdot _{N-A}V_{n-a}}{_NV_n} }
\n\\]<\/span><\/p>\n<\/div>\n<\/div>\n