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How to find $P(X_1 + X_2 + X_3 + X_4 \geq 3)$ for uniformly distributed independent random variables $X_1$, $X_2$, $X_3$, $X_4\sim U(0,1)$?

It follows from independence that their cumulative density function is 1, but I'm struggling with integration space.

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    $\begingroup$ (Big) Hint: By symmetry, this is equal to $$P(X_1+X_2+X_3+X_4\le1)$$ Now, compute the latter... $\endgroup$ – Did Dec 4 '18 at 15:23
  • $\begingroup$ @Did I can't figure that out. Looks like it gets us to equation $$F(1) = 1 - F(3)$$ but I can't recall any formal symmetry connected with that, $\endgroup$ – DeuzharNickens Dec 4 '18 at 17:07
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    $\begingroup$ Sub-hint: $(1-X_1,1-X_2,1-X_3,1-X_4)$ is distributed like $(X_1,X_2,X_3,X_4)$. $\endgroup$ – Did Dec 4 '18 at 17:36
  • $\begingroup$ @Did Oh I see, thanks! $\endgroup$ – DeuzharNickens Dec 4 '18 at 17:54
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Once we get to the step proposed by @Did, we can obtain the solution easily using geometric probability. Our probability here would be the hyper-volume covered by $$x_1+x_2+x_3+x_4 \leq 1 \text{ and } 0\leq x_i \leq 1$$ is exactly the hyper-volume of an $4$-dimensional simplex, which is $\dfrac{1}{4!} = \dfrac{1}{24}$.

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