Zadanie 1.
[tex]P(A') = 1 - P(A)\\\frac{1}{3} \cdot P(A) = 1 - P(A)\\1\frac{1}{3} \cdot P(A) = 1\\P(A) = 1 : 1\frac{1}{3} = \frac{3}{4}[/tex]
Zadanie 2.
[tex]| \Omega | = {15 \choose 4} = \frac{15!}{4!(15-4)!} = \frac{15!}{4! \cdot 11!} = \frac{11! \cdot 12 \cdot 13 \cdot 14 \cdot 15}{11! \cdot 2 \cdot 3 \cdot 4} = \frac{32760}{24} = 1365[/tex]
[tex]|A| = {9 \choose 3} \cdot {6 \choose 1} = \frac{9! \cdot 6!}{3!(9-3)! \cdot 1!(6-1)!} = \frac{9! \cdot 6!}{3! \cdot 6! \cdot 1! \cdot 5!} = \frac{5! \cdot 6 \cdot 7 \cdot 8 \cdot 9}{5! \cdot 3!} = \frac{3024}{6} = 504[/tex]
[tex]P(A) = \frac{|A|}{|\Omega|} = \frac{504}{1365} = \frac{24}{65}[/tex]
