# Probability of forming a triangle from a unit line segment with two random cuts

I know that the answer is $$\dfrac14$$ and also a beautiful way to calculate it.

However, where am I going wrong with the following argument:

Assume that the first cut is at length $$x$$ and the second one is at length $$y$$ and that the line segment is of unit length.

1. For a triangle, $$0 < x < \dfrac12$$. Therefore, $$x$$ can be independently selected with probability $$\dfrac12$$.
2. For each $$x$$ in Step 1, $$\dfrac12 < y < x+\dfrac12$$ for a triangle. That is, $$y$$ will work when it is in the aforesaid range of length $$x$$. On the other hand, the total possible values for $$y$$, for any given $$x$$, are $$x < y < 1$$. The length of this interval is $$1-x$$. Therefor for each $$x$$ in Step 1, $$y$$ can be selected with probability $$\dfrac{x}{1-x}$$.

$$\therefore$$ For $$0 < x < \dfrac12$$, the average value of probability of selecting $$y$$ to form a triangle is $$0.386$$.

Since the events 1 and 2 are independent, the final probability for both the events combined would be $$\dfrac12 \times 0.386 = 0.193$$.

## 1 Answer

You seem to have assumed $$0 \le x \le y \le 1$$.

If so, you would then be wrong to assume that the marginal density for $$X$$ (the lower of the two random numbers) is constant.

It is in fact $$f_X(x)=2-2x$$ and so your calculation should be $$\dfrac{\int\limits_{x=0}^{1/2} (2-2x) \dfrac{x}{1-x} \, dx}{\int\limits_{x=0}^{1} (2-2x) \, dx} = \frac14$$

• Thank you! Could you please also guide on the logical arguments that could help me arrive at that density function? Feb 18 '20 at 8:04
• @RiteshSingh The probability both values are above $x$ is $(1-x)^2$ so $F_X(x)=1-(1-x)^2 = 2x-x^2$ and thus $f_X(x)=2-2x$ Feb 18 '20 at 8:29