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You are given four numbers $x_1, x_2, x_3$ and $x_4$ in $[0, 1]$. What is the probability that the sum of the squares of these four numbers is less then one.

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    $\begingroup$ The sum of the numbers or the sum of the squares? $\endgroup$
    – Raffaele
    Commented Dec 18, 2020 at 10:00
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    $\begingroup$ What does the "volume", that this condition creates, look like? $\endgroup$
    – Matti P.
    Commented Dec 18, 2020 at 10:11

1 Answer 1

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Let us use some geometrical intuition (this is possible because we have uniform probability and no number has a particular privilege with respect to another in the extraction). Simplify the problem to the case of three numbers $x_1,x_2,x_3$: picking them with uniform probability distribution you are asking what is the probability that $x_1^2+x_2^2+x_3^2\leq 1$ with the condition $x_1,x_2,x_3\in[0,1]$. This is basically computing the volume of $1/8$ of a sphere of radius 1 (you are excluding all negative coordinates) over the volume of a cube of edge 1 (which is 1 in any dimension). In four dimensions, analogously, you can compute the volume of the 4D sphere and then exclude the negative coordinates combinations: namely you should divide the volume of the aforementioned sphere by 16. This leads to the final result which should be $\pi^2/32$.

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