Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 519

\[1)\ Оба\ билета\ выигрышные:\]

\[n = C_{15}^{2};\ \ \ m = C_{3}^{2};\]

\[P(A) = \frac{m}{n} = \frac{C_{3}^{2}}{C_{15}^{2}} =\]

\[= \frac{3!}{2!(3 - 2)!}\ :\frac{15!}{2!(15 - 2)!} =\]

\[= \frac{3 \bullet 2}{2 \bullet 1} \bullet \frac{2 \bullet 13!}{15 \bullet 14 \bullet 13!} = \frac{6}{210} = \frac{1}{35}.\]

\[2)\ Один\ билет\ выигрышный:\]

\[n = C_{15}^{2};\ \ \ m = C_{3}^{1} \bullet C_{12}^{1};\]

\[P(A) = \frac{m}{n} = \frac{C_{3}^{1} \bullet C_{12}^{1}}{C_{15}^{2}} =\]

\[= (3 \bullet 12)\ :\frac{15!}{2!(15 - 2)!} =\]

\[= 36 \bullet \frac{2 \bullet 13!}{15 \bullet 14 \bullet 13!} = \frac{72}{210} = \frac{12}{35}.\]

\[3)\ Выигрышного\ нет:\]

\[n = C_{15}^{2};\ \ \ m = C_{12}^{2};\]

\[P(A) = \frac{m}{n} = \frac{C_{12}^{2}}{C_{15}^{2}} =\]

\[= \frac{12!}{2!(12 - 2)!}\ :\frac{15!}{2!(15 - 2)!} =\]

\[= \frac{12 \bullet 11 \bullet 10!}{2 \bullet 10!} \bullet \frac{2 \bullet 13!}{15 \bullet 14 \bullet 13!} =\]

\[= \frac{132}{210} = \frac{22}{35}.\]