[tex]\Bar{\Bar{\Omega}} = \dfrac{20!}{3!(9-3)!} = \dfrac{20*19*18*17!}{3!*17!} = \dfrac{20*19*18}{3*2*1} = \dfrac{6840}{6} = 1140\\\\\Bar{\Bar{A}} = \dfrac{9!}{3!(9-3)!} = \dfrac{9*8*7*6!}{3!*6!} = \dfrac{9*8*7}{3*2} = \dfrac{504}{6} = 84\\\\P(A) = 1-\dfrac{\Bar{\Bar{\Omega}}}{\Bar{\Bar{A}}} = 1-\dfrac{1140}{84} = 1-\dfrac{7}{95}=\dfrac{88}{95}\\\\Odp. Prawdopodobienstwo\ wynosi\ \dfrac{88}{95}.[/tex]
[tex]Zadanie\ mozna\ wykonac\ takze\ przy\ pomocy\ drzewka\ (zalacznik), wtedy\\wykluczamy\ scenariusz\ gdy\ jest\ trzech\ mezczyzn\\\\\dfrac{9}{20}*\dfrac{8}{19}*\dfrac{7}{18}=\dfrac{504}{6840} = \dfrac{7}{95}\\\\P(A) = 1-\dfrac{7}{95} = \dfrac{88}{95}\\\\Odp. Prawdopodobienstwo\ wynosi\ \dfrac{88}{95}.[/tex]
