A homework problem on probability theory I just can't figure how to approach this:
If A and B are two events such that $P(A \cup B)$= 5/6, $P(A\cap B)$ =1/3, then which one of the following is correct?
(a) A and B are independent, (b) A and B' are independent, (c) A' and B are independent or (d) A and B are dependent.The answer says: d(that they are dependent). 
 A: Let $P(A\cap B^c)=x,\ P(B\cap A^c)=y.$ since
$$
\frac{5}{6}=P(A\cup B)=P(A\cap B)+P(A\cap B^c)+P(B\cap A^c)=\frac{1}{3}+x+y,
$$
we obtain $x+y=1/2$, and hence $y=1/2-x$. 
Hence we can rewrite $P(A)=x+1/3$ and $P(B)=5/6-x.$ We observe that
$$
P(A)P(B)=(x+\frac{1}{3})(\frac{5}{6}-x)=-(x-\frac{1}{4})^2+\frac{19}{72} \leq \frac{19}{72}<\frac{24}{72}=\frac{1}{3}=P(A\cap B).
$$
Therefore, $A$ and $B$ is not independent. 
A: Hint: You're looking to show that $P(A | B) = P(A)$. You know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, and $P(A \cap B) = P(B) P(A|B)$.
A: a = P(A)
b = P(B)
P(A U B) = P(A) + P(B) - P(A n B)
P(A) + P(B) = P(A U B) + P(A n B) = 5/6 + 1/3 = 7/6
a + b = 7/6
assumption of independence
P(A n B) = P(A)P(B) = ab = 1/3
we have 2 equations
a + b = 7/6
ab = 1/3
a(7/6 - a) = 1/3
3a^2 - 7a / 2 + 1 = 0 
a = 2/3 
b = 1/2   (or other way around, no loss of generality with that0
P(A) = 2/3
P(B) = 1/2 
here with the values found
P(A|B) = P(A n B) / P(B) = (1 / 3 ) / (1 / 2) = 2 / 3 = P(A)
I don't see why the info given means they have to be dependent
