Find the value of $P(Y|X)$

Question

If $P(X)=\frac{1}{4},$ $P(Y)=\frac{1}{3},$ and $P(X \cap Y) =\frac{1}{2}$ then find $$P(Y | X)= \frac{P(X \cap Y)}{P(X)}.$$

But by using this formula, I got an incorrect answer ($1/3$ is the right answer).

Answer 1

Suppose that $A$ and $B$ are any two events. If $A\subset B$, then $P(A)\le P(B)$. We have that $X\cap Y\subset X$, but $1/2=P(X\cap Y)>P(X)=1/4$. Clearly, there is a mistake somewhere.

I guess that the real statement of the exercise is as follows. If $P(X)=1/4$, $P(Y)=1/3$ and $P(X\cup Y)=1/2$ ($\cup$ denotes the union ("or") and $\cap$ denotes the intersection ("and")), find $P(Y\mid X)$. We have that $$ P(Y\mid X)=\frac{P(X\cap Y)}{P(X)}=\frac{P(X)+P(Y)-P(X\cup Y)}{P(X)}=\frac{1/4+1/3-1/2}{1/4}=\frac13. $$