N=10
n=3
x=2
k=4
Hypergeometric distribution
"P(X) = \\frac{\\binom{k}{x} \\binom{N-k}{n-x}}{\\binom{N}{n}} \\\\\n\nP(2) = \\frac{\\binom{4}{2} \\binom{10-4}{3-2}}{\\binom{10}{3}} \\\\\n\nP(2) = \\frac{6 \\times 6}{120} = 0.3 \\\\\n\nP(2) = 0.3"
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