Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1180

\[\boxed{\mathbf{1180}\mathbf{.}}\]

\[n = C_{5 + 7}^{3} = C_{12}^{3} = \frac{12!}{(12 - 3)! \bullet 3!} =\]

\[= \frac{12 \bullet 11 \bullet 10 \bullet 9!}{9! \bullet 3 \bullet 2} = 4 \bullet 11 \bullet 5 =\]

\[= 220.\]

\[1)\ все\ шары\ белого\ цвета:\]

\[m = C_{5}^{3} = \frac{5!}{(5 - 3)! \bullet 3!} = \frac{5!}{2! \bullet 3!} =\]

\[= \frac{5 \bullet 4 \bullet 3!}{2 \bullet 3!} = 5 \bullet 2 = 10;\ \]

\[P = \frac{m}{n} = \frac{10}{220} = \frac{1}{22}.\]

\[2)\ все\ шары\ черного\ цвета:\]

\[m = C_{7}^{3} = \frac{7!}{(7 - 3)! \bullet 3!} = \frac{7!}{4! \bullet 3!} =\]

\[= \frac{7 \bullet 6 \bullet 5 \bullet 4!}{4! \bullet 3 \bullet 2} = 7 \bullet 5 = 35;\]

\[P = \frac{m}{n} = \frac{35}{220} = \frac{7}{44}.\]

\[3)\ один\ шар\ белый,\ \]

\[а\ два\ других\ черные:\]

\[m = C_{5}^{1} \bullet C_{7}^{2} = 5 \bullet \frac{7!}{(7 - 2)! \bullet 2!} =\]

\[= \frac{5 \bullet 7 \bullet 6 \bullet 5!}{5! \bullet 2} = 5 \bullet 7 \bullet 3 = 105;\]

\[P = \frac{m}{n} = \frac{105}{220} = \frac{21}{44}.\]

\[4)\ один\ шар\ черный,\ \]

\[а\ два\ других\ белые:\]

\[m = C_{5}^{2} \bullet C_{7}^{1} = \frac{5!}{(5 - 2)! \bullet 2!} \bullet 7 =\]

\[= \frac{5 \bullet 4 \bullet 3! \bullet 7}{3! \bullet 2} = 5 \bullet 2 \bullet 7 = 70;\]

\[P = \frac{m}{n} = \frac{70}{220} = \frac{7}{22}.\]