# sum of four squares is less then one [closed]

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.

• The sum of the numbers or the sum of the squares? Commented Dec 18, 2020 at 10:00
• What does the "volume", that this condition creates, look like? Commented Dec 18, 2020 at 10:11

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 $$\pi^2/32$$.