I have done the part of two circles. Suppose that there are two intersecting circles with radius R. And let the distance between the center of two circles is D.(0$\le$D$\le$2R)
The intersection of two circles $A_{intersection}$ is given by $2R^2cos^{-1}\left(\frac{D}{2R}\right)-\frac{D}{2}\sqrt{4R^2-D^2}$.
Therefore, the union $A_{union}$ is given by $2R^2\pi-\frac{A_{intersection}}{2}$
And the probability P that the distance between the center of two circles is D is given by $\frac{2\pi D}{(2R)^2\pi}$.
The expected area of the union of two circles is $\int_{0}^{2R} P\times A_{union} dD$
I am wondering that is it correct? And how do i calculate the union of three circles?
If $B,C,D\subseteq\mathbb R^2$ have well defined area then: $$A_{B\cup C}=A_B+A_C-A_{B\cap C}$$
(not $\dots-\frac{A_{B\cap C}}{2}$ as you seem to think)
and (concerning your question about $3$ circles):
$$A_{B\cup C\cup D}=A_B+A_C+A_D-A_{B\cap C}-A_{C\cap D}-A_{B\cap D}+A_{B\cap C\cap D}$$
You are probably are dealing with a random variable $D$ that is uniformly distributed over interval $(0,2R)$.
What you (incorrectly) call "probability" is the PDF (Probability Distribution Function) of $D$: a function prescribed by $x\mapsto\frac{2\pi x}{(2R^2)\pi}=\frac{x}{2R^2}$ on $(0,2R)$ and $0$ elsewhere.