Probability: there are 8 women and 2 men. What is the probability that at least one man is present in a randomly picked group of 3 people?
From the hypergeometric distribution: $$\frac{\binom{2}{1}\binom{8}{2}+\binom{2}{2}\binom{8}{1}}{\binom{10}{3}}=\frac{2\cdot28+1\cdot8}{120}=\frac{8}{15}$$
Alternatively, use the complement: $$1-\frac{\binom{2}{0}\binom{8}{3}}{\binom{10}{3}}=1-\frac{1\cdot56}{120}=1-\frac{7}{15}=\frac{8}{15}$$