How to find the probability of such events? I solve the following problem. Let $\xi_1, \xi_2, \xi_3, \xi_4$ - independent random variables having a normal distribution with a parameter $(0, 1)$. This means that values $\xi_i$ have a p.d.f. $p_{\xi_j} (x) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{x^2}{2}}$. Need to find the probability $\mathbb{P} \{ (\xi_1 + \xi_2)^2 < (\xi_3 - \xi_4)^2 \}$.
How to start to solve it? I have two ideas:
1)We introduce two random variables $f = (\xi_1 + \xi_2)^2$ and $g = (\xi_3 - \xi_4)^2$. Next, find the density $p_f (x)$, $p_g (x)$ and find the probability of the event $\mathbb{P}(f < g) = \iint\limits_{u < v} p_f (u) p_g (v) {d}u {d}v$. But I think that would be difficult to find density $p_f (x)$ and $p_g (x)$.
2) $\mathbb{P} \{ (\xi_1 + \xi_2)^2 < (\xi_3 - \xi_4)^2 \} = \iint\limits_{(u_1 + u_2)^2 < (u_3 - u_4)^2} p_{\xi_1} (u_1) p_{\xi_2} (u_2) p_{\xi_3} (u_3) p_{\xi_4} (u_4) {d}u_1 \ldots {d}u_4$. But I'm afraid that this integral cannot be calculated.
Are there any other ideas for a solution?

$x_1=\xi_1+\xi_2$ has distribution ${\cal N}(0,2)$. $x_2=\xi_3-\xi_4$ has distribution ${\cal N}(0,2)$. You want to find $P(|x_1| < |x_2|)$. Geometric and symmetry  considerations give the answer ${1 \over 2}$.
