Problem:
A random triplet is taken from the set of non - negative real numbers satisfying $x+y+z=1$. I am intrigued to find the probability of:
- $x,y,z\leq m $ where $\frac{1}{3}\leq m\leq \frac{1}{2}$ or
- Two of $x,y,z$ are greater than or equal to $m$ where $\frac{1}{3}\leq m \leq\frac{1}{2}$
My attempt:
Please refer to the picture, triangle $ABC$ is equilateral triangle with sides $1$
- $x\vec{A}+y\vec{B}+z\vec{C}$ consists of all points inside triangle $ABC$
- $x\vec{A}+y\vec{B}+z\vec{C}$ with $x,y,z\leq m$ consists of all points inside triangle with sides $3m-1$. Denoted by $S$
- $x\vec{A}+y\vec{B}+z\vec{C}$ with two of $x,y,z$ greater than or equal to $m$ consists of all points inside three triangles with sides $1-2m$. Denoted by $P,Q,R$
Thus the probability are given by ratio of triangles area: $(3m-1)^{2}+3(1-2m)^{2}$
Could You guys please check this solution and drop Your own solutions too. Thanks!