3
$\begingroup$

What is the probability that $a > b + c$?

$a, b, c$ are picked independently and uniformly at random from bounded interval [0,1] of $\mathbb{R}$.

$\endgroup$
  • 2
    $\begingroup$ Welcome to Math.SE! Please, consider updating your question to include what you have tried and where you are getting stuck. That way, people on this site will know exactly what help you need. $\endgroup$ – Did Oct 14 '13 at 9:18
  • 1
    $\begingroup$ what interval is involved here? It is important to know that. For instance if it is something like $\left(5,6\right)$ then automatically $a<b+c$. So the probability you mention is $0$ in that case. $\endgroup$ – drhab Oct 14 '13 at 9:23
9
$\begingroup$

abc probability

Probability as the volume of a pyramid $V = \frac{1}{3}Sh = \frac{1}{3}\cdot\frac{1}{2}\cdot1 = \frac{1}{6}.$

$\endgroup$
4
$\begingroup$

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:

  1. $x \in [0,1). \lfloor x\rfloor = 0.$

    $$F_a(x) = x$$ $$F_d(x) = \frac{1}{2}\sum_{k=0}^{0}(-1)^k\binom{2}{k}(x-k)^2 = \frac{1}{2}x^2$$ $$F_d'(x) = x$$

  2. $x \in [1,2). \lfloor x\rfloor = 1.$

    $$F_a(x) = 1$$ $$F_d(x) = \frac{1}{2}\sum_{k=0}^{1}(-1)^k\binom{2}{k}(x-k)^2 = \frac{1}{2}x^2 - \frac{1}{2}\cdot2(x-1)^2 =-\frac{1}{2}x^2+2x-1$$ $$F_d'(x) = -x+2$$

Therefore:

$$P(a > b + c) = 1 - \int_{0}^{1}F_a(x)\, F'_d(x)\,dx - \int_{1}^{2}F_a(x)\, F'_d(x)\,dx =$$ $$1 - \int_{0}^{1}x\, x\,dx - \int_{1}^{2}1\cdot(-x+2)\,dx = 1 - \frac{1}{3} - \frac{1}{2} = \frac{1}{6}.$$

$\endgroup$
3
$\begingroup$

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 $$1 - \frac{5}{6} = \frac{1}{6}$$

$\endgroup$
1
$\begingroup$

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).

$\endgroup$
  • 5
    $\begingroup$ You might want to continue to mention I should add that I am one of the authors of mathStatica (or a similar formulation) when posting answers based on this software (in three recent answers, you forgot to do so). $\endgroup$ – Did Oct 14 '13 at 12:24
  • $\begingroup$ Wouldn't that be boasting? $\endgroup$ – wolfies Oct 14 '13 at 13:05
  • $\begingroup$ Are you going to add the mention to this answer and to the others, or not? $\endgroup$ – Did Oct 14 '13 at 17:03
  • 2
    $\begingroup$ @wolfies: it doesn't seem like boasting. There is a difference between an impartial endorsement and a paid endorsement. Granted, this is somewhere in-between, but if you don't say something, it looks like an impartial endorsement, which it really is not. $\endgroup$ – robjohn Oct 14 '13 at 21:16
  • $\begingroup$ @robjohn No problem. $\endgroup$ – wolfies Oct 15 '13 at 3:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.