Решебник по алгебре и начала математического анализа 11 класс Колягин Проверь себя VI. Сложный вариант

\[\mathbf{1}.\]

\[p = 0,9;\]

\[q = 1 - p = 0,1;\]

\[P = C_{2}^{1} \bullet p \bullet q = 2 \bullet 0,9 \bullet 0,1 = 0,18.\]

\[Ответ:\ \ 0,18.\]

\[\mathbf{2}\text{.\ }\]

\[p = 0,8;\]

\[q = 1 - p = 0,2;\]

\[\overline{P} = q^{3} = {0,2}^{3} = 0,008;\]

\[P = 1 - \overline{P} = 1 - 0,008 = 0,992.\]

\[Ответ:\ \ 0,992.\]

\[\mathbf{3.}\ \]

\[P = \frac{20 - 5}{20} \bullet \frac{8}{20} = \frac{15 \bullet 2}{20 \bullet 5} =\]

\[= \frac{3}{10} = 0,3.\]

\[Ответ:\ \ 0,3.\]

\[\mathbf{4}\text{.\ }\]

\[N_{1} = 2 + 3 + 4 = 9;\]

\[N_{2} = 1 + 2 + 3 = 6;\]

\[\overline{P} = \frac{2}{9} \bullet \frac{1}{6} + \frac{3}{9} \bullet \frac{2}{6} + \frac{4}{9} \bullet \frac{3}{6} =\]

\[= \frac{1}{27} + \frac{1}{9} + \frac{2}{9} = \frac{1}{27} + \frac{3}{9} =\]

\[= \frac{1}{27} + \frac{9}{27} = \frac{10}{27};\]

\[P = 1 - \overline{P} = 1 - \frac{10}{27} = \frac{17}{27}.\]

\[Ответ:\ \ \frac{17}{27}.\]