Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1178

\[\boxed{\mathbf{1178}\mathbf{.}}\]

\[n = C_{5 + 6}^{2} = C_{11}^{2} =\]

\[= \frac{11!}{(11 - 2)! \bullet 2!} = \frac{11!}{9! \bullet 2} =\]

\[= \frac{11 \bullet 10 \bullet 9!}{9! \bullet 2} = 11 \bullet 5 = 55.\]

\[1)\ оба\ шара\ белого\ цвета:\]

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

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

\[P = \frac{m}{n} = \frac{10}{55} = \frac{2}{11}.\]

\[2)\ оба\ шара\ черного\ цвета:\]

\[m = C_{6}^{2} = \frac{6!}{(6 - 2)! \bullet 2!} = \frac{6!}{4! \bullet 2} =\]

\[= \frac{6 \bullet 5 \bullet 4!}{4! \bullet 2} = 3 \bullet 5 = 15;\]

\[P = \frac{m}{n} = \frac{15}{55} = \frac{3}{11}.\]

\[3)\ один\ шар\ белый,\ \]

\[другой\ черный:\]

\[m = C_{5}^{1} \bullet C_{6}^{1} =\]

\[= \frac{5!}{(5 - 1)! \bullet 1!} \bullet \frac{6!}{(6 - 1)! \bullet 1!} =\]

\[= \frac{5 \bullet 4!}{4!} \bullet \frac{6 \bullet 5!}{5!} = 5 \bullet 6 = 30;\]

\[P = \frac{m}{n} = \frac{30}{55} = \frac{6}{11}.\]

\[4)\ по\ крайней\ мере\ один\ шар\ \]

\[белый:\]

\[A - оба\ шара\ черные;\]

\[\overline{A} - искомое\ событие:\]

\[m = C_{6}^{2} = \frac{6!}{(6 - 2)! \bullet 2!} = \frac{6!}{4! \bullet 2} =\]

\[= \frac{6 \bullet 5 \bullet 4!}{4! \bullet 2} = 3 \bullet 5 = 15;\]

\[P\left( \overline{A} \right) = 1 - P(A) = 1 - \frac{m}{n} =\]

\[= 1 - \frac{15}{55} = 1 - \frac{3}{11} = \frac{8}{11}.\]

\[5)\ по\ крайней\ мере\ один\ шар\ \]

\[черный:\]

\[A - оба\ шара\ белые;\ \]

\[\overline{A} - искомое\ событие:\]

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

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

\[P\left( \overline{A} \right) = 1 - P = 1 - \frac{m}{n} =\]

\[= 1 - \frac{10}{55} = 1 - \frac{2}{11} = \frac{9}{11}.\]