Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 537

\[Первый\ способ:\]

\[P(A) = \frac{2}{100} = \frac{1}{50};\text{\ \ \ }\]

\[P\left( B\text{/}A \right) = \frac{100 - 2}{100 - 1} = \frac{98}{99};\]

\[P\left( \text{AB} \right) = P(A) \bullet P\left( B\text{/}A \right) =\]

\[= \frac{1}{50} \bullet \frac{98}{99} = \frac{49}{25 \bullet 99} = \frac{49}{2475}.\]

\[2)\ Второй\ способ:\]

\[P(B) = \frac{100 - 2}{100} = \frac{98}{100} = \frac{49}{50};\text{\ \ \ }\]

\[P\left( A\text{/}B \right) = \frac{2}{100 - 1} = \frac{2}{99};\]

\[P\left( \text{AB} \right) = P(B) \bullet P\left( A\text{/}B \right) =\]

\[= \frac{49}{50} \bullet \frac{2}{99} = \frac{49}{25 \bullet 99} = \frac{49}{2475}.\]

\[Ответ:\ \ \frac{49}{2475}.\]