Probability and independence of 2 gaussians Let X and Z two independent normal random variables centered reduced. 
I want to calculate  $ P(X+Z<0,Z>0) $, so i have done : 
 $$ 
P(X+Z<0,Z>0)=P(|Z|<|X|,Z>0,X<0)
 $$ 
And I am blocked here. 
But the correction says only that it is equal to  $ 1/8 $ because the r.vs are independants and centered (and no more details). 
However my question is : Could we split like that 
 $$ 
P(|Z|<|X|,Z>0,X<0)=P(|Z|<|X|)P(Z>0)P(X<0)
 $$ 
 ? And if yes, why  ? 
 A: Maybe a simulation will help you visualize the relationships among variables.
I simulated 100,000 realizations of $X \sim Norm(0,1)$ and independently the
same number of realizations of $Z \sim Norm(0,1)$ in R statistical software. 
Then I plotted the
points with $X + Z < 0$ in orange. The points of interest to you
are the orange ones above the x-axis. (Of course, you can draw a
similar sketch without any simulation, if you understand the
symmetry of the bivariate uncorrelated standard normal distribution.)
m = 10^5;  x = rnorm(m);  z = rnorm(m)
plot(x, z, pch=".")
  cond = (x + z < 0)
  points(x[cond], z[cond], pch=".", col="orange")
  abline(h = 0, col="green", lwd=2)
  abline(v = 0, col="green", lwd=2)
mean(z > 0);  mean(x + z < 0)
## 0.49889  # aprx P(Z > 0) = 1/2
## 0.49951  # aprx P(X + Z < 0) = 1/2
mean(x + z < 0 & z > 0); 1/8 
## 0.12254  # aprx P(X + Z = 0, Z > 0) = 1/8
## 0.125


A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{\sigma > 0}$, the answer is given by the following expression:
\begin{align}
&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\bracks{{1 \over \root{2\pi}\sigma}\,
\exp\pars{-\,{x^{2} \over 2\sigma^{2}}}}
\bracks{{1 \over \root{2\pi}\sigma}\,
\exp\pars{-\,{z^{2} \over 2\sigma^{2}}}}\bracks{x + z < 0}\bracks{z > 0}
\dd x\,\dd z
\\[5mm] = &\
{1 \over 2\pi\sigma^{2}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\exp\pars{-\,{{x^{2} + z^{2} \over 2\sigma^{2}}}}
\bracks{0 < z < -x}\dd x\,\dd z
\\[5mm] = &\
{1 \over \pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\expo{-x^{2}\ -\ z^{2}}\,\,\bracks{0 < \root{2}\sigma z < -\root{2}\sigma x}
\dd x\,\dd z
\\[5mm] = &\
{1 \over \pi}\int_{0}^{2\pi}\int_{0}^{\infty}
\expo{-r^{2}}\,\,\bracks{0 < r\sin\pars{\theta} < -r\cos\pars{\theta}}r
\,\dd r\,\dd\theta
\\[5mm] = &\
{1 \over \pi}\int_{0}^{2\pi}\bracks{0 < \sin\pars{\theta} < -\cos\pars{\theta}}\
\underbrace{\int_{0}^{\infty}\expo{-r^{2}}r\,\dd r}_{\ds{1 \over 2}}\
\dd\theta =
{1 \over 2\pi}\int_{0}^{\pi}
\bracks{\sin\pars{\theta} < -\cos\pars{\theta}}\,\dd\theta
\\[5mm] = &\
{1 \over 2\pi}\int_{3\pi/4}^{\pi}\,\dd\theta =\
\bbox[#ffe,5px,border:1px dotted navy]{\ds{1 \over 8}}
\end{align}
