Question

Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area

Answer 1

Let
`a,b,c` be the randomly selected lengths. Without loss of generality, suppose
`a</p><p>[tex]P(A + B \ge C) = P(A + B - C \ge 0)`

where
`A,B,C` are independent random variables with the same uniform distribution on [0, 1].

By their mutual independence, we have

`P(A=a,B=b,C=c) = P(A=a) * P(B=b) * P(C=c)`

so that the joint density function is

`P(A=a,B=b,C=c) = \begin{cases}1 & \text{if }(a,b,c)\in[0,1]^3 \\ 0 & \text{otherwise}\end{cases}`

where
`[0,1]^3=[0,1]*[0,1]*[0,1]` is the cube with vertices at (0, 0, 0) and (1, 1, 1).

Consider the plane

`a + b - c = 0`

with
`(a,b,c)\in\Bbb R^3`. This plane passes through (0, 0, 0), (1, 0, 1), and (0, 1, 1), and thus splits up the cube into one tetrahedral region above the plane and the rest of the cube under it. (see attached plot)

The point (0, 0, 1) (the vertex of the cube above the plane) does not belong the region
`a+b-c\ge0`, since
`0+0-1=-1<0`. So the probability we want is the volume of the bottom "half" of the cube. We could integrate the joint density over this set, but integrating over the complement is simpler since it's a tetrahedron.

Then we have

`\displaystyle P(A+B-C\ge0) = 1 - P(A+B-C < 0) \\\\ ~~~~~~~~ = 1 - \int_0^1\int_0^(1-a)\int_(a+b)^1 P(A=a,B=b,C=c) \, dc\,db\,da \\\\ ~~~~~~~~ = 1 - \int_0^1 \int_0^(1-a) (1 - a - b) \, db \, da \\\\ ~~~~~~~~ = 1 - \int_0^1 \frac{(1-a)^2}2\,da \\\\ ~~~~~~~~ = 1 - \frac16 = \boxed{\frac56}`