P(A B)=16, P( A B)...

Question

$P(A\cup B)=\dfrac16$, $P(\overline A\cap B)$... (given $P(A\cup B)$, $P(A\cap B)$, $P(\overline A)$ as $\dfrac16,\dfrac14,\dfrac14$). Determine the relationship between $A$ and $B$.

Accepted Answer

$P(\overline{A\cup B})=\dfrac16\Rightarrow P(A\cup B)=\dfrac56$; $P(\overline A)=\dfrac14\Rightarrow P(A)=\dfrac34$; $P(A\cap B)=\dfrac14$.

$$P(B) = P(A\cup B)-P(A)+P(A\cap B) = \frac56-\frac34+\frac14=\frac13$$

Since $P(A)\ne P(B)$, not equally likely; but $P(A)P(B)=\dfrac34\times\dfrac13=\dfrac14=P(A\cap B)$, so independent.

Solutions · 1