Let P(ξ≤x,η≤y)=F(x,y). [F(x,y)]²≤ F(x). F(y). Does it imply that ξ has d.f F(x) and η has d.f F(y) and are they independent? The two-dimensional distribution of the random variables $\xi$ and $\eta$ is specified by the distribution function 
$$
\mathbb{P}(\xi \leq x,\; \eta \leq y)  = F(x,y) = \begin{cases} 0 & \textrm{ if } \min(x,y) <0 \\ \min(x,y) & \textrm{ if } 0 \leq \min(x,y) \leq 1 \\ 1 & \textrm{ if } \min(x,y) > 1 \end{cases}.$$
Find $\mathbb{P}[ (\xi-\frac{1}{2})² + (\eta -\frac{1}{2})² \leq \frac{1}{4}]$.
My target is to find the required probability. But I am stuck in the middle way.
In order to find the probability this I did 
$$ 0≤min(x,y) ≤1$$
$$0≤F(x,y)≤1$$
Let us denote the events $$ξ≤x$$ as $$A$$ and $$η≤y$$ as $$B$$
Now $AB⊆ A$ and $AB⊆ B$
Therefore $P(AB)≤P(A) , P(AB)≤P(B)$
Therefore $[P(AB)]² ≤P(A). P(B)$
Therefore $[P(ξ≤x, η≤y)]² ≤ P(ξ≤x). P(η≤y)$
Therefore $[F(x,y)]² ≤ F(x). F(y)$  
Now if $ξ$ has the distribution function $F(x)$ and $η$ has the
distribution function $F(y)$ and if they are independent then I
can proceed to find the required probability. 
By the way these $F(x), F(y)~Uniform(0,1)$ with mean $1/2$
Then $T = ξ-1/2$ ~ $N(0,1/12)$
and $W = η-1/2$ ~ $N(0,1/12)$  
Then $P[12(T²+W²) ≤3]$ where $12(T²+W²)$ ~ Chi-Square Distribution with d.f 2 
Is it the correct way? 
If the independence does not work should I use this formula
$F(x,y)+F(0,0)-F(x,0)-F(0,y)$
Please Help Me.
 A: *

*The marginal distributions of $\xi$ and $\eta$ are indeed $\text{Unif}(0,1)$. But they are not independent: if they were, then $F(x,y) = xy$ for $x,y \in [0,1]$.

*Regarding your attempts to show independence: In general, independence of $\xi$ and $\eta$ is equivalent to $F_{\xi,\eta}(x,y) = F_\xi(x) F_\eta(y)$.

*I have no idea where you got the chi-squared distribution to appear.


Hint: the biggest step you can take here is to identify in plain words what the joint distribution of $(\xi,\eta)$ is.
Big, big hint:

 Your CDF is the CDF of the following distribution: $\xi \sim U(0,1)$ and $\eta = \xi$. Note that this does not mean "$\eta$ is another $U(0,1)$ random variable independent of $\xi$", but rather "$\mathbb{P}(\xi=\eta)=1$."

Check that you understand how to verify the above hint. From there, you should be able to complete the computation easily.

 Thus, $\mathbb{P}\{(\xi-\frac{1}{2})^2 + (\eta - \frac{1}{2})^2 \le \frac{1}{4}\} = \mathbb{P}\{(\xi - \frac{1}{2})^2 \le \frac{1}{8}\} = \mathbb{P}\{\frac{1}{2} - \frac{1}{\sqrt{8}} \le \xi \le \frac{1}{2} + \frac{1}{\sqrt{8}}\} = \frac{1}{\sqrt{2}}$.

