Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 577

\[1)\ 2\ раза:\]

\[P_{8}(2) = C_{8}^{2} \bullet \left( \frac{1}{2} \right)^{2} \bullet \left( 1 - \frac{1}{2} \right)^{8 - 2} =\]

\[= \frac{8!}{2!(8 - 2)!} \bullet \frac{1}{2^{2}} \bullet \frac{1}{2^{6}} =\]

\[= \frac{8 \bullet 7 \bullet 6!}{2 \bullet 6! \bullet 2^{8}} = \frac{28}{2^{8}} = \frac{7}{2^{6}} = \frac{7}{64}.\]

\[2)\ 6\ раз:\]

\[P_{8}(6) = C_{8}^{6} \bullet \left( \frac{1}{2} \right)^{6} \bullet \left( 1 - \frac{1}{2} \right)^{8 - 6} =\]

\[= \frac{8!}{6!(8 - 6)!} \bullet \frac{1}{2^{6}} \bullet \frac{1}{2^{2}} =\]

\[= \frac{8 \bullet 7 \bullet 6!}{6! \bullet 2! \bullet 2^{8}} = \frac{8 \bullet 7}{2 \bullet 2^{8}} = \frac{28}{2^{8}} =\]

\[= \frac{7}{2^{6}} = \frac{7}{64}.\]

\[3)\ Не\ менее\ 6\ раз:\]

\[P = P_{8}(6) + P_{8}(7) + P_{8}(8);\text{\ \ \ }\]

\[p = \frac{1}{2};\ \ \ q = 1 - \frac{1}{2} = \frac{1}{2};\]

\[= \frac{1}{2^{8}} \bullet \left( \frac{8 \bullet 7 \bullet 6!}{6! \bullet 2!} + 8 + 1 \right) =\]

\[= \frac{1}{2^{8}} \bullet \left( \frac{8 \bullet 7}{2} + 9 \right) = \frac{37}{2^{8}} = \frac{37}{256}.\]

\[4)\ Не\ более\ 2\ раз:\]

\[P = P_{8}(0) + P_{8}(1) + P_{8}(2);\text{\ \ \ }\]

\[p = \frac{1}{2};\ \ \ q = 1 - \frac{1}{2} = \frac{1}{2};\]

\[= \frac{1}{2^{8}} \bullet \left( 1 + 8 + \frac{8 \bullet 7 \bullet 6!}{2 \bullet 6!} \right) =\]

\[= \frac{1}{2^{8}} \bullet (9 + 28) = \frac{37}{2^{8}} = \frac{37}{256}.\]

\[Ответ:\ \ \frac{37}{256}.\]