독립사건의 확률

$\displaystyle \mathrm{P} \mathit{\left( B \right)} = \mathrm{P} \mathit{\left( A \cap B \right)} + \mathrm{P} \mathit{\left( A ^{C} \cap B \right)} = \frac{3}{4}$ $\mathrm{P} \mathit{\left( A \right)} = k$라 하면 $\mathrm{P} \mathit{\left( A \cap B \right)} = \mathrm{P} \mathit{\left( A \right)} \mathrm{P} \mathit{\left( B \right)}$이므로 $\displaystyle \frac{1}{2} = k \times \frac{3}{4}$, $\displaystyle k = \frac{2}{3}$ 따라서 $\displaystyle \mathrm{P} \mathit{\left( A \right)} = \frac{2}{3}$ [다른 풀이] 두 사건 $A$, $B$가 서로 독립이므로 두 사건 $A ^{C}$, $B$도 서로 독립이다. $\displaystyle \mathrm{P} \mathit{\left( A \cap B \right)} = \mathrm{P} \mathit{\left( A \right)} \mathrm{P} \mathit{\left( B \right)} = \frac{1}{2}$ $\mathrm{P} \mathit{\left( A ^{C} \cap B \right)}$$= \mathrm{P} \mathit{\left( A ^{C} \right)} \mathrm{P} \mathit{\left( B \right)}$ $= \left\{ 1 - \mathrm{P} \mathit{\left( A \right)} \right\} \mathrm{P} \mathit{\left( B \right)}$ $= \mathrm{P} \mathit{\left( B \right)} - \mathrm{P} \mathit{\left( A \right)} \mathrm{P} \mathit{\left( B \right)}$ $\displaystyle = \mathrm{P} \mathit{\left( B \right)} - \frac{1}{2} = \frac{1}{4}$ $\displaystyle \mathrm{P} \mathit{\left( B \right)} = \frac{3}{4}$ 따라서 $\displaystyle \mathrm{P} \mathit{\left( A \right)} = \frac{1}{2} \times \frac{4}{3} = \frac{2}{3}$