Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 578

\[1)\ 2\ раза:\]

\[P_{5}(2) = C_{5}^{2} \bullet \left( \frac{1}{6} \right)^{2} \bullet \left( 1 - \frac{1}{6} \right)^{5 - 2} =\]

\[= \frac{5!}{2!(5 - 2)!} \bullet \frac{1}{6^{2}} \bullet \frac{5^{3}}{6^{3}} =\]

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

\[= \frac{1250}{7776} = \frac{625}{3888}.\]

\[2)\ 3\ раза:\]

\[P_{5}(3) = C_{5}^{3} \bullet \left( \frac{1}{6} \right)^{3} \bullet \left( 1 - \frac{1}{6} \right)^{5 - 3} =\]

\[= \frac{5!}{3!(5 - 3)!} \bullet \frac{1}{6^{3}} \bullet \frac{5^{2}}{6^{2}} =\]

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

\[= \frac{250}{7776} = \frac{125}{3888}.\]

\[3)\ Не\ более\ 2\ раз:\]

\[P = P_{5}(0) + P_{5}(1) + P_{5}(2);\text{\ \ \ }\]

\[p = \frac{1}{6};\ \ \ q = 1 - \frac{1}{6} = \frac{5}{6};\]

\[= \frac{5^{3}}{6^{5}} \bullet \left( 5^{2} + 5 \bullet 5 + \frac{5 \bullet 4 \bullet 3!}{2 \bullet 3!} \right) =\]

\[= \frac{5^{3}}{6^{5}} \bullet 60 = \frac{125 \bullet 10}{1296} = \frac{625}{648}.\]

\[4)\ Не\ менее\ 4\ раз:\]

\[P = P_{5}(4) + P_{5}(5);\text{\ \ \ }\]

\[p = \frac{1}{6};\ \ \ q = 1 - \frac{1}{6} = \frac{5}{6};\]

\[= 5 \bullet \frac{1}{6^{5}} \bullet \frac{5}{6} + 1 \bullet \frac{1}{6^{5}} \bullet \frac{5^{0}}{6^{0}} = \frac{25}{6^{5}} + \frac{1}{6^{5}} =\]

\[= \frac{26}{6^{5}} = \frac{26}{7776} = \frac{13}{3888}.\]