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This sort of probability is a bit outside my usual comfort zone, but I'll have a go at it.You have a thin rod of length $\ell$ and you cut it randomly into three pieces. Calculate the probability that the three pieces can form a triangle.
Clearly we can suppose without loss of generality that $\ell=1$. Setting $x_{1}$, $x_{2}$ and $x_{3}$ the length ot the three segments, the requested probability is the probability that none of the lengths of the segments is greater that $\frac{1}{2}$, i.e. ...You have a thin rod of length $\ell$ and you cut it randomly into three pieces. Calculate the probability that the three pieces can form a triangle.
This sort of probability is a bit outside my usual comfort zone, but I'll have a go at it.
The length $\ell$ is immaterial, so I'll assume $\ell=1$.
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Suppose that the cuts occur at $y$ and $x$, with $x,\,y$ independently uniformly distributed in $[0,1]$ and $x$ denoting the larger of them. We must have $x\geqslant1/2$ if the three segments are to form a triangle (because none of the sides can be greater than $1/2$). So one side on the triangle has length $1-x$. The difference between the other two sides must be at most $1-x$, and so $y$ must lie in the interval $\bigl[x-\tfrac12,\tfrac12\bigr]$, of length $1-x$ (see the picture). The probability of that happening (for a given $x$) is $\frac{1-x}x$. However, the probability distribution of $x$ (as the maximum of the two cut points) is not uniform, but has p.d.f. $2x$. So the overall probability of the three segments forming a triangle is $$\int_{1/2}^1 \frac{1-x}x\,2x\,dx = \Bigl[2x-x^2\Bigr]_{1/2}^1 = 1-\frac34 = \frac14.$$
(I hope that's right!)
Because a different solution with different result has been proposed the 'mental construction' I followed has to be explained with the scope to identify my error...Clearly we can suppose without loss of generality that $\ell=1$. Setting $x_{1}$, $x_{2}$ and $x_{3}$ the length ot the three segments, the requested probability is the probability that none of the lengths of the segments is greater that $\frac{1}{2}$, i.e. ...
$$P = \int_{0}^{\frac{1}{2}} \int_{\frac{1}{2}- x_{1}}^{\frac{1}{2}} \frac{1}{1-x_{1}}\ d x_{2}\ d x_{1} = \frac{1}{2}\ \int_{0}^{\frac{1}{2}} \frac{x_{1}}{1 - x_{1}}\ d x_{1} = |- x_{1} - \ln (1-x_{1})|_{0}^{\frac{1}{2}} = - \frac{1}{2} + \ln 2 = .19314718...$$
Kind regards
$\chi$ $\sigma$
My 'error' has been that, setting $\xi_{1}$ and $\xi_{2}$ the coordinates of the first and second selected point, I have considered only the case $\xi_{1}<\xi_{2}$ and the computed probability is...Because a different solution with different result has been proposed the 'mental construction' I followed has to be explained with the scope to identify my error...
If $x_{1}$ is the first segment and we suppose it uniformely distributed in $[0,1]$, then $x_{2}$ is uniformely distributed in $[0,1 - x_{1}]$. The condition that $x_{1}$, $x_{2}$ and $x_{3}$ are the sides of the triangle means that none of them exceeds $\frac{1}{2}$ so that we are searching the probability that $\frac{1}{2} -x < x_{2} < \frac{1}{2}$ conditioned by the probability that $x_{1} = x < \frac{1}{2}$ ...
Kind regards
$\chi$ $\sigma$
You selected the same approach I did.$$P = \int_{0}^{\frac{1}{2}} \int_{\frac{1}{2}- x_{1}}^{\frac{1}{2}} \frac{1}{1-x_{1}}\ d x_{2}\ d x_{1} = ... = - \frac{1}{2} + \ln 2 = .19314718...$$
Kind regards
$\chi$ $\sigma$
Really?I must be missing something, because I thought you could always make a triangle out of any three lengths...
Yes, that's what I was missing. Serves me right for picturing triangles in my mind and manipulating them, rather than picturing different lengths haha.Really?
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Good luck
[JUSTIFY]Specifically the triangle inequality puts a lower bound on the minimum total length required of the two smaller sides in order to form a triangle. If the larger side has length L, then the two other sides must sum to a length strictly more than L in order to form a triangle (less than L and no triangle is possible, equal to L and the best you can do is a degenerate triangle with three colinear vertices).
In Euclidean space, anyway.[/JUSTIFY]
can i ask why the pdf of x is 2x?This sort of probability is a bit outside my usual comfort zone, but I'll have a go at it.
The length $\ell$ is immaterial, so I'll assume $\ell=1$.
![]()
Suppose that the cuts occur at $y$ and $x$, with $x,\,y$ independently uniformly distributed in $[0,1]$ and $x$ denoting the larger of them. We must have $x\geqslant1/2$ if the three segments are to form a triangle (because none of the sides can be greater than $1/2$). So one side on the triangle has length $1-x$. The difference between the other two sides must be at most $1-x$, and so $y$ must lie in the interval $\bigl[x-\tfrac12,\tfrac12\bigr]$, of length $1-x$ (see the picture). The probability of that happening (for a given $x$) is $\frac{1-x}x$. However, the probability distribution of $x$ (as the maximum of the two cut points) is not uniform, but has p.d.f. $2x$. So the overall probability of the three segments forming a triangle is $$\int_{1/2}^1 \frac{1-x}x\,2x\,dx = \Bigl[2x-x^2\Bigr]_{1/2}^1 = 1-\frac34 = \frac14.$$
(I hope that's right!)
The key here is that x is defined to be the maximum of the two cuts. Since we have two (probabilistically) independent cuts, each with a uniform pdf, their maximum will have a pdf of 2x.can i ask why the pdf of x is 2x?
thanks a lotThe key here is that x is defined to be the maximum of the two cuts. Since we have two (probabilistically) independent cuts, each with a uniform pdf, their maximum will have a pdf of 2x.
For my own sake, I will try to put together (and post) a more formal argument for the above statement. It agrees with my intuition, but requires probing.
This comes from a neat little trick I picked up in probstats, which I'm glad I found useful enough to remember.The key here is that x is defined to be the maximum of the two cuts. Since we have two (probabilistically) independent cuts, each with a uniform pdf, their maximum will have a pdf of 2x.
For my own sake, I will try to put together (and post) a more formal argument for the above statement. It agrees with my intuition, but requires probing.
The pdf for a uniformly distributed variable on the unit interval is the constant function $1$. The cumulative distribution function (cdf) is the integral of the pdf, namely the function $x$. The rule (which I don't know how to prove, not being a probabilist) is that for the maximum of two independent random variables, you multiply the cdf's. In this case, that gives you $x^2$ for the cdf of the max. You then differentiate the cdf to get the pdf, so that is where the $2x$ comes from.can i ask why the pdf of x is 2x?
I probably can't pronounce myself a probabilist, but the following is probably true.The rule (which I don't know how to prove, not being a probabilist) is that for the maximum of two independent random variables, you multiply the cdf's.