ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1144

\[\boxed{\mathbf{1144}\mathbf{.}}\]

\[n = C_{24}^{3} = \frac{24!}{(24 - 3)! \bullet 3!} =\]

\[= \frac{24!}{21! \bullet 3!} = \frac{24 \bullet 23 \bullet 22 \bullet 21!}{21! \bullet 3 \bullet 2} =\]

\[= 2\ 024.\]

\[1)\ \overline{C} - выбраны\ только\ \]

\[юноши;\ \]

\[C - искомое:\]

\[m = C_{18}^{3} = \frac{18!}{(18 - 3)! \bullet 3!} =\]

\[= \frac{18!}{15! \bullet 3!} = \frac{18 \bullet 17 \bullet 16 \bullet 15!}{15! \bullet 3 \bullet 2} =\]

\[= 6 \bullet 17 \bullet 8 = 816;\]

\[P\left( \overline{C} \right) = \frac{816}{2\ 024} = \frac{102}{253};\]

\[P(C) = 1 - P\left( \overline{C} \right) = 1 - \frac{102}{253} =\]

\[= \frac{151}{253}.\]

\[Ответ:\ \ \frac{151}{253}.\]

\[2)\ \overline{C} - выбраны\ только\ \]

\[девушки;\ \]

\[C - \ искомое:\]

\[m = C_{6}^{3} = \frac{6!}{(6 - 3)! \bullet 3!} =\]

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

\[= 20;\]

\[P\left( \overline{C} \right) = \frac{20}{2\ 024} = \frac{5}{506};\]

\[P(C) = 1 - P\left( \overline{C} \right) = 1 - \frac{5}{506} =\]

\[= \frac{501}{506}.\]

\[Ответ:\ \ \frac{501}{506}.\]