Probability relations I found this question on probability.
Given that
$$P(a \cup b) = \frac{2}{3}$$
$$P(a \cap \neg b) = \frac{1}{3}$$
$$P(a|b) = \frac{1}{6}$$ 
I have to find $P(a)$ and $P(b)$.
I tried a lot but felt it was missing data. 
Thanks.
 A: $$P(A|B) = P(A \cap B) / P(B)$$
$$P(B) = P(A \cup B) - P(A \cap B^\prime)$$
So then if we re-arrange, we get
$$P(A|B) = P(A \cap B) / (P(A \cup B) - P(A \cap B^\prime))$$
Where
$P(A|B) = \frac{1}{6}$, $P(A \cup B) = \frac{2}{3}$, and $P(A \cap B^\prime) = \frac{1}{3}$
Now you can solve for $P(A \cap B)$ and from there you can solve for $P(B)$ and then $P(A)$
A: Draw a picture (Venn diagram) and everything will become clear. We follow tradition by using capital letters for events. If $E$ is an event, then $E^c$ denotes the complement of $E$. 
Since $\Pr(A\cap B^c)=\frac{1}{3}$, the part of $A$ outside $B$ has probability $\frac{1}{3}$. But $A\cup B$ is the disjoint union of $A\cap B^c$ and $B$. It follows that $\Pr(B)=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$.
Note that $\Pr(A| B)=\dfrac{\Pr(A\cap B)}{\Pr(B)}$. Thus $\Pr(A\cap B)=\frac{1}{6}\cdot\frac{1}{3}=\frac{1}{18}$.
Finally, we can use the formula $\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B)$ to find $\Pr(A)$. We get $\Pr(A)=\frac{2}{3}-\frac{1}{3}+\frac{1}{18}=\frac{7}{18}$.
