ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1131

\[\boxed{\mathbf{1131}\mathbf{.}}\]

\[n = C_{15}^{2} = \frac{15!}{(15 - 2)! \bullet 2!} =\]

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

\[= 15 \bullet 7 = 105.\]

\[1)\ обе\ выбранные\ лампочки\ \]

\[испорчены:\]

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

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

\[P = \frac{m}{n} = \frac{6}{105} = \frac{2}{35}.\]

\[2)\ одна\ лампочка\ исправная,\ \]

\[а\ другая\ нет:\]

\[m = C_{11}^{1} \bullet C_{4}^{1} =\]

\[= \frac{11!}{(11 - 1)! \bullet 1!} \bullet \frac{4!}{(4 - 1)! \bullet 1!} =\]

\[= \frac{11!}{10!} \bullet \frac{4!}{3!} = 11 \bullet 4 = 44;\]

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

\[3)\ обе\ выбранные\ лампочки\ \]

\[исправны:\]

\[m = C_{11}^{2} = \frac{11!}{(11 - 2)! \bullet 2!} =\]

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

\[= 55;\]

\[P = \frac{m}{n} = \frac{55}{105} = \frac{11}{21}\text{.\ \ }\]