조건부확률 역산

$\mathrm{P} ( \mathrm{A} ) = 0.3 , \mathrm{P} ( \mathrm{B} ) = 0.4 , \mathrm{P} ( \mathrm{B} | \mathrm{A} ) = 0.8$에서 $\displaystyle \mathrm{P} ( \mathrm{B} | \mathrm{A} ) = \frac{\mathrm{P} ( \mathrm{A} \cap \mathrm{B} )}{\mathrm{P} ( \mathrm{A} )} = 0.8 \therefore \mathrm{P} ( \mathrm{A} \cap \mathrm{B} ) = 0.24$ $\displaystyle \mathrm{P} ( \mathrm{A} | \mathrm{B} ) = \frac{\mathrm{P} ( \mathrm{A} \cap \mathrm{B} )}{\mathrm{P} ( \mathrm{B} )} = \frac{0.24}{0.4} = 0.6$