독립사건 확률

[출제의도] 확률의 계산식에서 확률을 구할 수 있는가? $\displaystyle \mathrm{P} \left( \mathit{A} \cup B \right) = \frac{5}{6}$에서 $\displaystyle \mathrm{P} \left( \mathit{A} \right) + \mathrm{P} \left( \mathit{B} \right) - \mathrm{P} \left( \mathit{A} \cap B \right) = \frac{5}{6}$ $\cdots \cdots$ ㉠ 두 사건 $A$와 $B$가 서로 독립이므로 $\mathrm{P} \left( \mathit{A} \cap B \right) = \mathrm{P} \left( \mathit{A} \right) \mathrm{P} \left( \mathit{B} \right)$ $\cdots \cdots$ ㉡ ㉠, ㉡에서 $\displaystyle \mathrm{P} \left( \mathit{A} \right) + \mathrm{P} \left( \mathit{B} \right) - \mathrm{P} \left( \mathit{A} \right) \mathrm{P} \left( \mathit{B} \right) = \frac{5}{6}$ 이때, $\displaystyle \mathrm{P} \left( \mathit{A} \right) = \frac{2}{3}$이므로 $\displaystyle \frac{2}{3} + \mathrm{P} \left( \mathit{B} \right) - \frac{2}{3} \mathrm{P} \left( \mathit{B} \right) = \frac{5}{6}$, $\displaystyle \frac{1}{3} \mathrm{P} \left( \mathit{B} \right) = \frac{1}{6}$ 따라서 $\displaystyle \mathrm{P} \left( \mathit{B} \right) = \frac{1}{2}$