Question

probability of a cut lying in second half of a number line is

Answer 1

Firstly note that probability of any two numbers being equal is zero.

Let us assume that x is the minimum of the numbers.

Since x is assumed to be minimum, it is enough to have y and z at a distance smaller than 1/3 from x.

When 0≤x≤2/3, y,z has to lie in the interval [x,x+1/3], which is of length 1/3. So probability is 1/3⋅1/3. Integrate that over the region 0≤x≤2/3.

When2/3≤x≤1, y and z just have to lie in the right side of x. i.e in the interval [x,1], which is of length (1−x).So probability is (1−x)2. Integrate over 2/3≤x≤1.

∫2/3013⋅13dx+∫12/3(1−x)2dx

Which equals 227+181=781

So the required probability is 3⋅781=727.