P(B A)=25, P(A B)=17, P(A B)=19.

Question

$P(B\mid A)=\dfrac25$, $P(A\mid B)=\dfrac17$, $P(A\cap B)=\dfrac19$. Consider (S1) $P(A'\cup B)=\dfrac56$, (S2) $P(A'\cap B')=\dfrac{1}{18}$.

Accepted Answer

$P(B)=\dfrac{P(A\cap B)}{P(A\mid B)}=\dfrac{1/9}{1/7}=\dfrac79$; $P(A)=\dfrac{P(A\cap B)}{P(B\mid A)}=\dfrac{1/9}{2/5}=\dfrac{5}{18}$.

$P(A'\cup B) = 1-P(A)+P(A\cap B) = \dfrac{13}{18}+\dfrac19=\dfrac56$ (via $P(A\cap B)+P(A\cap B')... $) --- confirmed $\dfrac56$.

$P(A'\cap B') = 1-P(A\cup B) = 1-\left[P(A)+P(B)-P(A\cap B)\right] = 1-\left[\dfrac{5}{18}+\dfrac79-\dfrac19\right]=\dfrac{1}{18}$.

Both statements true.

Solutions · 1