독립사건의 확률

$\displaystyle \mathrm{P} \mathit{\left( B \right)} = \mathrm{P} \mathit{\left( A \cap B \right)} + \mathrm{P} \mathit{\left( A ^{C} \cap B \right)} = \frac{1}{15} + \frac{1}{10} = \frac{1}{6}$ 두 사건 $A$, $B$는 서로 독립이므로 $\displaystyle \mathrm{P} \mathit{\left( A \cap B \right)} = \mathrm{P} \mathit{\left( A \right)} \times \mathrm{P} \mathit{\left( B \right)} = \mathrm{P} \mathit{\left( A \right)} \times \frac{1}{6}$ $\displaystyle \mathrm{P} \mathit{\left( A \right)} = 6 \times \mathrm{P} \mathit{\left( A \cap B \right)} = 6 \times \frac{1}{15} = \frac{2}{5}$