$A,B$ are events with $P(A)=P(B)=0.5$ and $P(A\cap B)=0.25$. What is probability that... $A,B$ are events with $P(A)=P(B)=\frac{1}{2}$ and $P(A\cap B)=\frac{1}{4}$. What is probability that exactly one of this events occur?
No sure how you do it good. There is $3$ event, two of them have same probability $0.5$ and the other have lower probability of $0.25$.
That why I do this $$\frac{\frac{1}{2}+\frac{1}{2}+\frac{1}{4}}{3}= \frac{\frac{5}{4}}{3}= \frac{5}{12}= 0.41\bar{6}= 41.\bar{6}\text{%}$$
This is probability that exactly one of the event occur. Is my solution good or all wrong?
 A: The two answers above are correct, but I would like to give some intuition.  Suppose that the rectangle below represents the sample space, with two events $A$ and $B$ shown.

If we assume that the rectangle has a total area of 1 square unit, then, in a way that can be made mathematically rigorous, the probability of an event is the area of the region covered by that event.  Thus
$$ P(A) = P(B) = 0.5 $$
indicates that each of the ellipses has area 0.5 square units, and
$$ P(A\cap B) = 0.25 $$
indicates that the area of the overlap is 0.25 square units.
The event "either $A$ or $B$, but not both" is the area shaded in the lighter green color.  Observe that we can write that area as
$$ P(A\setminus B) + P(B\setminus A), \tag{1} $$
i.e. all of the stuff in $A$ that is not in $B$ (the light green area associated to $A$ on the left part of the figure), and all of the stuff in $B$ but not $A$ (the region on the right).
But note that
$$ P(A) = P(A \setminus B) + P(A \cap B). $$
That is, we can write $A$ as the light green region (the region $A\setminus B$) plus the area where $A$ and $B$ overlap (the region $A\cap B$).  But we know the areas of two of those regions from the statement of the problem, and so we have
$$
0.5 = P(A \setminus B) + 0.25
 \implies P(A \setminus B) = 0.25. \tag{2}
$$
By similar reasoning,
$$
P(B \setminus A) = 0.25. \tag{3}
$$
Substituting the results in (2) and (3) back into (1), we obtain
$$
P(A\setminus B) + P(B\setminus A) = 0.25 + 0.25 = 0.5.
$$
A: $A$ and $B$ are independently and you are interested in probability of $A'\cap B+A\cap B'$. So  the result is $$P(A'\cap B+A\cap B') = P(A'\cap B)+P(A\cap B')=0,5$$
A: Or, more formally,
$P(A) + P(B) = P(A \cap B) + P(A \cap B') + P(A \cap B) + P(A' \cap B)$
$\Rightarrow P(A \cap B') + P(A' \cap B) = P(A) + P(B) - 2P(A \cap B)$
