독립사건의 확률

조건에서 ⅰ) $\displaystyle \mathrm{P} \mathit{(} A | B ) = \frac{\mathrm{P} \mathit{(} A \cap B )}{\mathrm{P} \mathit{(} B )} = \frac{\mathrm{P} \mathit{(} A ) \cdot \mathrm{P} \mathit{(} B )}{\mathrm{P} \mathit{(} B )} = \frac{3}{8}$ ($\because A , B$는 독립) $\displaystyle \therefore {\mathrm{P}} ( A ) = \frac{3}{8}$ ⅱ) $\displaystyle \begin{aligned} {\mathrm{P}} ( A \cup B ) & = {\mathrm{P}} ( A ) + {\mathrm{P}} ( B ) - {\mathrm{P}} ( A \cap B ) \\ & = {\mathrm{P}} ( A ) + {\mathrm{P}} ( B ) - {\mathrm{P}} ( A ) \cdot {\mathrm{P}} ( B ) = \frac{1}{2} \end{aligned}$ ($\because A , B$는 독립) $\displaystyle \therefore \frac{3}{8} + {\mathrm{P}} ( B ) - \frac{3}{8} {\mathrm{P}} ( B ) \frac{1}{2} , \frac{5}{8} {\mathrm{P}} ( B ) = \frac{1}{8}$ $\displaystyle \therefore {\mathrm{P}} ( B ) = \frac{1}{5}$ $\begin{aligned} {\mathrm{P}} ( A \cap B ^{c} ) & = {\mathrm{P}} ( A ) \cdot {\mathrm{P}} ( B ^{c} ) = {\mathrm{P}} ( A ) \cdot ( 1 - {\mathrm{P}} ( B ) ) \end{aligned}$ $\displaystyle = \frac{3}{8} \times \left( 1 - \frac{1}{5} \right)$ ($\because$ (ⅰ), (ⅱ)에 의하여) $\displaystyle = \frac{3}{10}$