Geometric Probability problem in 3 unknowns

Question

I've been solving different geometrical probability questions, and there is one, where I'm somewhat stuck.

Suppose we have to choose $3$ numbers, $a,b$ and $c$ such that $a,b,c \in [0,1]$. The numbers are randomply distributed in an uniform distribution between $0$ and $1$. Then I've been asked to find the probability of $a+b\gt2c$.

I'm not being able to represent this in a geometrical way. I've tried fixing the value of $c$ and then figuring out where $a$ and $b$ would lie on a line segment, but that got me nowhere.

How should I approach this particular type of problem?

Answer 1

You could think of $a,b,c$ are the three coordinates of a point in $1 \times 1 \times 1$ cube, then you want to find the volume of portion of the cube where the $z$ coordinate $c$ is less than half the sum of the $x$ and $y$ coordinates.

The plane separating the two disjoint regions is $z = \dfrac{1}{2} (x + y) $

Therefore,

$P = \text{Volume} = \displaystyle \int_{x= 0 }^{1} \int_{y = 0}^{1} \dfrac{1}{2} (x + y) dy dx = \dfrac{1}{2} \int_{x=0}^1 (x + \dfrac{1}{2})dx = \dfrac{1}{2} (\dfrac{1}{2} + \dfrac{1}{2} ) = \dfrac{1}{2} $

Answer 2

\begin{align} \mathsf{P}(a+b>2c)&=\mathsf{E}[\mathsf{P}(a+b>2c\mid c)]=\int_0^1 \mathsf{P}(a+b>2x)\, dx \\ &=\int_0^{1/2} \left(1-2x^2\right)\, dx+\int_{1/2}^1 2\left(1-x\right)^2\, dx =\frac{1}{2}. \end{align}