find the probability about sum of random variables Let $X_1, X_2, X_3, Y_1, Y_2, Y_3, Z_1, Z_2, Z_3$ be random variables which have uniform distribution between 0 and 1. It means, the average of $X_1 = 0.5$
Let: $X=X_1 + X_2 + X_3,$
$Y=Y_1 + Y_2 + Y_3$, $Z=Z_1 + Z_2 + Z_3$
In this case, the probability of $\{X$ is bigger than $Y$ and $Z$ both$\}$ would be $\dfrac{1}{3}$.
My question is:
What is the probability of "$c+X$ is bigger than $Y$ and $Z$" when $c$ is a constant"?
For example: what is $\mathbb{P}\left[0.2+X >\max\{Y,Z\}\right]$?
 A: HINT


*

*Let $X = X_1+X_2+X_3$. What is the distribution of $X$? Well, a direct approach would be to find $$F_X(x) = \mathbb{P}[X_1+X_2+X_3 < x] = \iiint_{[0,1]^3} \mathbb{I}_{[a+b+c<x]} dadbdc,$$ which can be translated to a regular volume if you restrict the region of integration so the indicator is always 1.

*Then, $Y = Y_1+Y_2+Y_3$ and $Z = Z_1+Z_2+Z_3$ are defined analogously and have the same distribution with pdf $f(x) = F'(x)$. It's easy to see $f(x)$ only has support for $x \in [0,3]$.

*You want 
$$
\begin{split}
\mathbb{P}\left[c+X >\max\{Y,Z\}\right]
 &= \iiint_{[0,3]^3} \mathbb{I}_{[c+x > \max\{y,z\}]} f(x)f(y)f(z) dxdydz \\
 &= \iiint_{[0,3]^3} \mathbb{I}_{[c+x > y]}
                          \mathbb{I}_{[c+x > z]}
                          f(x)f(y)f(z)dxdydz
\end{split}
$$ which can be similarly manipulated...

A: Thank you @gt6989b
At first, the answer of $F_X(x)$ is
$F(x)=\dfrac{1}{6}x^3 $ when $0\leq x<1$
$F(x)=-\dfrac{1}{3}x^3+\dfrac{3}{2}x^2-\dfrac{3}{2}x+\dfrac{1}{2} $ when $1 \leq x <2$
$F(x)=1-\dfrac{1}{6}(3-x)^3$ when $2\leq x<3$
and secondly,
$f(x)=\dfrac{1}{2}x^2$ when $0\leq x<1$
$f(x)=-x^2 +3x-\dfrac{3}{2}$ when $1\leq x <2$
$f(x)=\dfrac{1}{2}(3-x)^2$ when $2\leq x<3$
And $f(y)$ and $f(z)$ follow same way.
Hence, 
$$
\begin{split}
\mathbb{P}\left[X >\max\{Y,Z\}\right]
 &= \iiint_{[0,3]^3} \mathbb{I}_{[x > \max\{y,z\}]} f(x)f(y)f(z) dxdydz \\
 &=  \dfrac{1}{3}\
\end{split}
$$
And I think I can go further. Thank you again.
