Als Alternative über die Hypergeometrische Verteilung
\(\frac{\Pi\left(\left(\begin{array}{r}M_i\\m_i\\\end{array}\right) \right) \; \left(\begin{array}{r}N - \Sigma M_i\\n - \Sigma m_i\\\end{array}\right)}{\left(\begin{array}{r}N\\n\\\end{array}\right)}\)
also
\(\left\{N= 100, n=10, M=\left\{2, 8, 20, 70\right\}, m= \left\{0, 2, 3, 5\right\}\right\}\)
\( \binom{2}{0} \cdot \binom{8}{2}\cdot \binom{20}{3}\cdot \binom{70}{5} / \binom{100}{10} = 0.02232 \)