5
$\begingroup$

A stick of length $a$ is broken in three parts. Find the probability that the length of each part is less than $b$, where $b>a/3$.

A sample space, $\Omega$, is defined as:

$$\Omega=\{(x,y): x>0,y>0,a-x-y>0\}$$ $$=\{(x,y): x>0,y>0,x+y<a\}$$

where $x,y,a-x-y$ are lenghts of broken parts.

Event $A$: "The length of every part is less than $b,b>a/3$."

$$A=\{(x,y)\in\Omega\::0<x<b,0<y<b,0<a-x-y<b\}$$ $$=\{(x,y)\in\Omega\::0<x<b,0<y<b,a-b<x+y<a\}$$

Now, what I don't understand is the following:

In my book's solution it says that we consider two cases:

First case $$0<\frac{a}{3}<b\le \frac{a}{2}$$

In this case, $\Omega$ is a right angled triangle with sides $a$.

The problem in this case is how to determine event $A$ (I am given the geometric approach).

In my book's solution it says that event $A$ is also a right angled triangle. How?

Shouldn't it be quadrilateral surface?

We have that $$m(A)=\int_{a-2b}^b(b-(-x+a-b))dx$$

How?

Second case $$b > \frac{a}{2}$$

Here, $\Omega$ is defined as a square with side $b$, and event $A$ as a hexagonal surface.

How?

We have that $$m(A)=b^2-\frac{1}{2}(a-b)^2-\frac{1}{2}(2b-c)^2$$

Why do we choose $\frac{a}{2}$ as a bound in both cases?

$\endgroup$
  • 3
    $\begingroup$ You are in effect asking about the cumulative distribution function for the shortest of the three parts. However we need to be clear about the way the "breaks" of the stick are chosen. One might ask that two points along the stick be chosen independently as the places to break it. Another version might break the stick once at a uniformly random place, and then break one or the other pieces in a randomly chosen point in that piece. The exact setup can affect the answer, as may be seen in the discussions of similar previous Questions. $\endgroup$ – hardmath Mar 17 '17 at 13:13
  • $\begingroup$ Closely related: expected length of broken stick $\endgroup$ – hardmath Mar 17 '17 at 13:16
  • 3
    $\begingroup$ @hardmath, I think that points are chosen independently. $\endgroup$ – user300045 Mar 17 '17 at 13:19
6
+250
$\begingroup$

The most natural sample space is the square $\Omega:=[0,1]^2$ with area as probability measure. A point $(x,y)\in\Omega$ signifies that the stick $[0,1]$ is partitioned into three pieces by making two cuts independently at $x\in[0,1]$ and at $y\in[0,1]$. Assume that a number $\beta\in\bigl[{1\over3},1\bigr]$ is given. We want to know the probability $p_\beta$ that all three pieces have a length $\leq\beta$.

In the following figures only the case $x\leq y$ is represented in detail. In this case the lengths of the pieces are $x$, $y-x$, and $1-y$. It follows that a point $(x,y)$ is admissible iff $$0\leq x\leq \beta\quad\wedge\quad x\leq y\leq \beta+x\quad\wedge\quad y\geq1-\beta\ .$$It turns out that we have to distinguish the cases ${1\over3}\leq\beta\leq{1\over2}$ and ${1\over2}\leq\beta\leq 1$. In the figures the admissible area is shaded, and $p_\beta$ is twice this area.

enter image description here

From the figures we can read off $$\eqalign{p_\beta&=2\cdot{1\over2}\bigl(\beta-(1-2\beta)\bigr)^2=(3\beta-1)^2\qquad\left({1\over3}\leq\beta\leq{1\over2}\right)\>,\cr p_\beta&=2\left(\beta^2-{1\over2}(1-\beta)^2-{1\over2}(2\beta-1)^2\right)=-3\beta^2+6\beta-2\qquad\left({1\over2}\leq\beta\leq1\right)\ .\cr}$$

$\endgroup$

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.