독립사건 확률 보기

[출제의도] 독립사건에서 확률의 성질 추론하기 공사건이 아닌 두 사건 $A$와 $B$가 서로 독립이므로 $\mathrm{P} ( \mathit{A} \cap B ) = \mathrm{P} ( \mathit{A} ) \cdot \mathrm{P} ( \mathit{B} )$이다. ㄱ. $\displaystyle \mathrm{P} ( \mathit{A} ^{C} | B ) = \frac{\mathrm{P} ( \mathit{A} ^{C} \cap B )}{\mathrm{P} ( \mathit{B} )}$$\displaystyle = \frac{\mathrm{P} ( \mathit{A} ^{C} ) \mathrm{P} ( \mathit{B} )}{\mathrm{P} ( \mathit{B} )}$ $= \mathrm{P} ( \mathit{A} ^{C} )$$= 1 - \mathrm{P} ( \mathit{A} )$ (참) ㄴ. $\mathrm{P} ( \mathit{A} \cup B ) = \mathrm{P} ( \mathit{A} ) + \mathrm{P} ( \mathit{B} ) - \mathrm{P} ( \mathit{A} \cap B )$ $\ne \mathrm{P} ( \mathit{A} ) + \mathrm{P} ( \mathit{B} )$ (거짓) ㄷ. (좌변)$= \mathrm{P} ( \mathit{A} \cap B ) + \mathrm{P} ( \mathit{A} ^{C} \cap B )$ $= \mathrm{P} ( \mathit{A} ) \cdot \mathrm{P} ( \mathit{B} ) + \mathrm{P} ( \mathit{A} ^{C} ) \cdot \mathrm{P} ( \mathit{B} )$ (참)