Find the probability of $a>b+c$, where $a$, $b$, $c$ are $U(0,1)$ What is the probability that $a > b + c$?
$a, b, c$ are picked independently and uniformly at random from bounded interval [0,1] of $\mathbb{R}$.
 A: 
Probability as the volume of a pyramid $V = \frac{1}{3}Sh = \frac{1}{3}\cdot\frac{1}{2}\cdot1 = \frac{1}{6}.$
A: Let $d = b + c$.
$d$ has the Irwin-Hall distribution with $n=2$ independent variables.
Its CDF is equal to $F_d(x)=\frac{1}{n!}\sum_{k=0}^{\lfloor x\rfloor}(-1)^k\binom{n}{k}(x-k)^n$.
Now,
$$P(a<d) = \int_{-\infty}^{\infty} F_a(x)\, F'_d(x)\,dx.$$
Explanation of this equality. $a$ and $d$ are independent and continuous random variables with cumulative distributions $F_a(x)$ and $F_d(x)$.
So, we can write:
$$P(a > b + c) = P(a > d) = 1 - P(a < d) = 1 - \int_{-\infty}^{\infty} F_a(x)\, F'_d(x)\,dx.$$
$F'_d(x)=0$ for $x\in(-\infty,0)\cup[2, \infty)$. Thus, integral is non-zero only between two intervals:

*

*$x \in [0,1). \lfloor x\rfloor = 0.$
$$F_a(x) = x$$
$$F_d(x) = \frac{1}{2}\sum_{k=0}^{0}(-1)^k\binom{2}{k}(x-k)^2 = \frac{1}{2}x^2$$
$$F_d'(x) = x$$


*$x \in [1,2). \lfloor x\rfloor = 1.$
$$F_a(x) = 1$$
$$F_d(x) = \frac{1}{2}\sum_{k=0}^{1}(-1)^k\binom{2}{k}(x-k)^2 = \frac{1}{2}x^2 - \frac{1}{2}\cdot2(x-1)^2 =-\frac{1}{2}x^2+2x-1$$
$$F_d'(x) = -x+2$$
Therefore:
$$P(a > b + c) = 1 - \int_{0}^{1}F_a(x)\, F'_d(x)\,dx - \int_{1}^{2}F_a(x)\, F'_d(x)\,dx =$$
$$1 - \int_{0}^{1}x\, x\,dx - \int_{1}^{2}1\cdot(-x+2)\,dx = 1 - \frac{1}{3} - \frac{1}{2} = \frac{1}{6}.$$
A: Let's pose that the bounded interval of $\mathbb{R}$ is $I = [0, 1]$.
Fixed $b$ and $c$, we have to find out the conditional probability that $a > b+c$. Namely:
$$\int_{\max(b+c, 1)}^1 f_a(a) da = 1 - \max(b+c, 1)$$
Then, we have to integrate with respect to $b$ and $c$:
$$\int_{0}^1\int_{0}^1 \left( 1 - \max(b+c, 1) \right) f_b(b)f_c(c) db ~dc = 1 - \int_{0}^1\int_{0}^1\max(b+c, 1)db~dc$$
Let's consider $\max(b+c, 1)$. It is equal to $b+c$ when $b < 1 - c$, and $1$ elsewhere. Then:
$$\int_{0}^1\int_{0}^1\max(b+c, 1)db~dc = \int_{0}^1\left[\int_{0}^{1-c}(b+c)db + \int_{1-c}^1 1 \cdot db\right]dc = $$
$$=\int_{0}^1\left[\frac{(1-c)^2}{2} + c(1-c) + c\right]dc = \frac{5}{6}$$
Finally, the probabilty you are looking for is
$$1 - \frac{5}{6} = \frac{1}{6}$$
A: Others have provided graphical, manual integration and distribution theory approaches ---> so here's an automated computer algebra system approach, as a one-liner:

Easy and simple ... calculated here using the mathStatica add-on to Mathematica (I am one of the authors of the former); presumably Maple or other computer algebra packages could handle it too (I am not an author of those packages).
