Probability circle to be in a circle We have circle with radius $R$ and inside it we choose at random 2 points $A$ and $B$.Find the probability the circle with center A and radius $AB$ to be inside the circle(It is possible both circles to be tangent)
What I've concluded : the first point A has range from $[0;R]$ and the second point B must have range from $[0;R-OA]$ .But how should I find the probability?
 A: First, I'll assume the two points are obtained with an uniform distribution in the circle.
You have a good conclusion there. Let's build from it. As you've noticed, once you fix the distance from $A$ to the center of the circle, $B$ has to fall inside the little circle: $B$ can fall in an area of $\pi (R-r)^2$, from a total of $\pi R^2$. Here $r$ is the distance from the center of the original circle and $A$.
Now, if you use polar coordinates to express your points, the probability density function of the radius of a uniformly random point $A$ is $2r/R^2$. The only thing we have left is to compute the integral, which in words would mean "Compute the product of the probability of $A$ having distance $r$ from the center multiplied by the probability of $B$ falling close enough, and sum for each possible $r$":
$$\int_0^R \Big(\frac{R-r}{R}\Big)^2 \frac{2}{R^2}r \hspace{.2cm}dr = \int_0^R \Big(1-\frac{r}{R}\Big)^2 \frac{2}{R^2}r \hspace{.2cm}dr$$
Which is just a polynomial, so we can solve it. It's a bit tedious, so I'll leave it here with the answer of $\frac{1}{6}$ from wolfram alpha.
A: I can propose a little different approach. We want to choose $2$ points $($call them $X,Y)$ inside a circle of radius $R$. Due to circle being rotation-invariant, we can always rotate so that $X$ lies on $[0,R] \times \{0\} \subset \mathbb R^2$. So that we need to find the distribution of $Z:=||X||$. We have $F_Z(t) = \frac{\pi t^2}{\pi R^2}\chi_{[0,R]}(t) + \chi_{(R,+\infty)}(t)$, so that the density is $g_Z(t) = \frac{2}{R^2}t \chi_{[0,R]}(t)$.
Let $A \in \mathcal F$ be an event - circle with center $X$ and radius $||Y-X||$ is inside our circle of radius $R$.
Then we have $\mathbb P(A | Z) = \frac{\pi (R-Z)^2}{\pi R^2} = \frac{(R-Z)^2}{R^2}$ 
And by total expectation $$\mathbb P(A) = \mathbb E[ \mathbb P(A|Z)] = \frac{1}{R^2}\mathbb E[ (R-Z)^2] = \frac{1}{R^2} \int_0^R (R-z)^2 z \frac{2}{R^2} dz = \frac{2}{R^4} \int_0^R s^2(R-s)ds = \frac{2}{R^4}(\frac{R^4}{3} - \frac{R^4}{4}) = \frac{1}{12} $$
