독립사건 확률

두 사건 $A$와 $B$는 서로 독립이므로 $\mathrm{P} \mathit{\left( A \cap B \right)}$$= \mathrm{P} \mathit{\left( A \right)} \mathrm{P} \mathit{\left( B \right)}$ $\displaystyle \mathrm{P} \mathit{\left( A \cap B ^{C} \right)} = \frac{4}{9}$에서 $\mathrm{P} \mathit{\left( A \cap B ^{C} \right)}$$= \mathrm{P} \mathit{\left( A \right)} - \mathrm{P} \mathit{\left( A \cap B \right)}$ $= \mathrm{P} \mathit{\left( A \right)} - \mathrm{P} \mathit{\left( A \right)} \mathrm{P} \mathit{\left( B \right)}$ $= \mathrm{P} \mathit{\left( A \right)} \times \left\{ 1 - \mathrm{P} \mathit{\left( B \right)} \right\}$ $= \left\{ \mathrm{P} \mathit{\left( A \right)} \right\} ^{2} \left( \because \mathrm{P} \mathit{\left( A \right)} + \mathrm{P} \mathit{\left( B \right)} = 1 \right)$ $\displaystyle \left\{ \mathrm{P} \mathit{\left( A \right)} \right\} ^{2} = \frac{4}{9}$이므로 $\displaystyle \mathrm{P} \mathit{\left( A \right)} = \frac{2}{3}$