독립사건 합사건

두 사건 $A$, $B$가 서로 독립이므로 $\displaystyle \mathrm{P} \left( \mathit{A} \cap B \right) = \frac{1}{6}$에서 $\displaystyle \mathrm{P} \left( \mathit{A} \right) \times \mathrm{P} \left( \mathit{B} \right) = \frac{1}{6}$ $\displaystyle \mathrm{P} \left( \mathit{A} \right) = \frac{2}{3}$이므로 $\displaystyle \mathrm{P} \left( \mathit{B} \right) = \frac{1}{4}$ $\therefore$ $\mathrm{P} \mathit{\left( A \cup B \right)}$$\mathrm{=} P \mathit{\left( A \right)} + \mathrm{P} \mathit{\left( B \right)} - \mathrm{P} \mathit{\left. \left( A \cap B \right) \right.}$ $\displaystyle = \frac{2}{3} + \frac{1}{4} - \frac{1}{6} = \frac{3}{4}$