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Question: Divide a given line segment into two other line segments. Then, cut each of these new line segments into two more line segments. What is the probability that the resulting line segments are the sides of a quadrilateral?

I am stuck on this problem. I think I am close, but I am not sure if it is correct. Any help or conformation on this would be helpful.

My thoughts on the problem:

Let us say that the line segment is of length 1. The only restriction for these for line segments to form a quadrilateral is that no one side > .5 (correct me if I am wrong).

With our first cut we have two smaller line segments, one larger than the other. We only need to look at the longer on of these two line segments. Let us call $y$ the smaller line segment and $x$ the larger one. $x$ will be between 0.5 and 1.

When we cut each of these new line segments we need to find where it does not work for it to be a quadrilateral. We only have to look at $x$. Let us call $a$ the length that we cut.

If we use the example $x=0.6$ we can see that $a$ cannot be less than 0.1 or greater than 0.5.

We can generalize this for any $x$. $a$ cannot be less than $(x-1/2)$ or greater than $1/2$.

This is where I get stuck.

I believe that the probability for any $x$ value that these four line segments will not be a quadrilateral is $$\frac{2(x-1/2)}{x}$$

If this is correct is the total probability that it cannot be a quadrilateral $$\int\limits_{1/2}^1\frac{2(x-1/2)}{x}\mathrm{d}x?$$

Any help is much appreciated. Thank you

Yousuf Soliman
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    +1 for showing your work. In the very last integral, $\mathrm dx$ should be $2\mathrm dx$. Apart from that, your reasoning looks fine. To compute the last integral, use $\int(4-2/x)\mathrm dx=4x-2\log x$, which yields the numerical value $2-2\log2$. – Did Jan 09 '13 at 06:06
  • Why should it be $2dx$? – Yousuf Soliman Jan 09 '13 at 06:09
  • Because the total mass of $\mathrm dx$ on $[1/2,1]$ is $1/2$ and you want a total mass of $1$. – Did Jan 09 '13 at 06:15
  • I am sorry, I'm not sure I entirely understand this. I am fairly new to Calculus in general and have never come across a situation like that. Could you please explain a little more in depth? Thank you! – Yousuf Soliman Jan 09 '13 at 06:19
  • The probability density of $X$ the longest of the two first lengthes is $2$ on the interval $[.5,1]$. For example, the probability that $X\leqslant.6$ is $2(.6-.5)=.2$, not $.1$. And the probability that $X\leqslant1$ is $2(1-.5)=1$, naturally, not $.5$. – Did Jan 09 '13 at 06:23
  • @did, I believe you have provided an answer to this question, so I suggest you post it as such. – MvG Jan 09 '13 at 06:44
  • Isn't that what the coefficient of 2 is in front of the $(x-1/2)$? Or am I supposed to have another 2 before the $dx$? – Yousuf Soliman Jan 09 '13 at 06:44
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    @MuadDib42 Another reason why there is a 2, is because you did a symmetric argument and only considered the case where the break occurred in $[0.5, 1]$, instead of when the break occurred in the entire domain $[0,1]$. – Calvin Lin Jan 09 '13 at 07:07
  • @CalvinLin Thank you! That also makes sense! – Yousuf Soliman Jan 09 '13 at 07:09
  • The problem is not well formulated. You need to give a probability distribution for each of the choices made, and the answer will depend on that. Also "are the sides of a quadrilateral" needs some relaxation, or else the probability will be $0$: since these are all segments of the same line, they never are the sides of any one quadrilateral. – Marc van Leeuwen Jan 09 '13 at 07:33
  • @MarcvanLeeuwen The question should really be something along the lines of "What is the probability that the resulting line segments can form a quadrilateral?"

    What do you mean that I need to give a probability distribution for each of the choices made?

     – Yousuf Soliman Jan 09 '13 at 07:53
  • In probability theory you need to give some measure on the space of possbile outcomes before you can talk about probabilty of an event (subset of outcomes), like here "forming a quadrilateral". If the number of outcomes is finite there is the uniform distribution that gives every outcome the same probability, but with infinitely many outcomes (as is the case here) requiring each individual outcome equally probable just doesn't define the probability of events. In most cases every individual outcome will have probability $0$ anyway. See http://en.wikipedia.org/wiki/Probability_distribution – Marc van Leeuwen Jan 09 '13 at 08:14

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In your very last integral, $\mathrm dx$ should be $2\mathrm dx$ $(*)$. Apart from that, your reasoning looks fine. To compute the last integral, use $\displaystyle\int(2−1/x)\mathrm dx=2x−\log x$, which yields the numerical value $2−2\log2\approx61\%$ for the probability of no quadrilateral.

$(*)$ To understand why, note that the total mass of $\mathrm dx$ on $[1/2,1]$ is $1/2$ and that one wants a total mass of $1$. This indicates that the probability density function of $X$ the longest of the two first lengthes is $2$ on the interval $[1/2,1]$. Once $X=x$ is known, the probability of no quadrilateral is $p(x)=2(x-1/2)/x$, as you aptly showed, hence the total probability of no quadrilateral is $$ \int_{1/2}^1p(x)\,(2\mathrm dx)=2\int_{1/2}^1(2-1/x)\mathrm dx. $$

Did
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