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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 $\dfrac{x}{1-x}$.

Now, $\dfrac {\int_0^{1/2} \dfrac{x}{1-x} dx}{\dfrac12} = 0.386$

$\therefore$ 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 $\dfrac12 \times 0.386 = 0.193$.

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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 $$\dfrac{\int\limits_{x=0}^{1/2} (2-2x) \dfrac{x}{1-x} \, dx}{\int\limits_{x=0}^{1} (2-2x) \, dx} = \frac14$$

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  • $\begingroup$ Thank you! Could you please also guide on the logical arguments that could help me arrive at that density function? $\endgroup$ Feb 18 '20 at 8:04
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    $\begingroup$ @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 $f_X(x)=2-2x$ $\endgroup$
    – Henry
    Feb 18 '20 at 8:29

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