Finding $P(A\cap B')$ Suppose $A$ and $B$ are events in the same sample space. Also $P(A \cap B)  = 0.5$ and $P(B') = 0.25$. Find value of $P(A\cap B')$.
My try: The solution gives $\frac{1}{6}$ but I think the information isn't enough. We know value of $P(B)$ and $P(B\cap A')$ for sure but there is no constrain on $P(A\cap B')$. 
 A: From $P(B')=0.25$ it follows that $P(B)=0.75$.  From that and $P(A\cap B)=0.5$ it follows that $P(A'\cap B)=0.25$.
It follows that $P(A\cap B')+P(A'\cap B')=0.25$ however from the given information alone this is not enough information to distinguish further.  We can say that $0\leq P(A\cap B')\leq 0.25$ but every value in that range is possible.  For simple examples of the extremes, consider the uniform distribution on $\{1,2,3,4\}$ and $B=\{2,3,4\}$ and compare the case where $A=\{1,2,3\}$ versus the case where $A=\{2,3\}$.  Both examples satisfy the given conditions of $P(A\cap B)=0.5$ and $P(B')=0.25$.

If we were given as an additional piece of information that $A$ and $B$ are independent (which I cannot stress enough is not a valid assumption to make under normal circumstances) then we could proceed further.
Given that $P(A\cap B) = 0.5$ and given that $A$ and $B$ are independent, it follows that $0.5 = P(A)\cdot P(B)$ (which is invalid if $A$ and $B$ were not independent).
From that and that $P(B)=0.75$ it follows that $P(A)=\frac{.5}{.75}=\dfrac{2}{3}$
From here it follows that $P(A\cap B') = \dfrac{2}{3}\times .25 = \dfrac{1}{6}$
Again... this followed from a flawed assumption that $A$ and $B$ are independent.  If this was not given in the problem statement then this is incorrect.
