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The full question is: A rope of length L is cut at a random point. Find the probability that the ratio of the length of the shorter piece of rope to the length of the longer piece of rope is less than 1/3.

So I started by saying that X is uniformly distributed over (0,L) and we're trying to find $P\left \{ X <\frac{L}{3} \right \}$.

So I said the probability density function is $$f(x) = \left\{\begin{matrix}\frac{1}{L} &0 < x < L \\ 0 & otherwise\end{matrix}\right.$$

and so $$P\left \{ X <\frac{L}{3} \right \} = \int_0^\frac{L}{3} \frac{1}{L} dx$$

However this doesn't seem right to me. Am I on the right track?

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    $\begingroup$ First, the interval $(1,L)$ does not have length $L$. Second, you want the probability that the shorter piece is $<1/3$ the length of the larger piece, not $<1/3$ the length of the entire rope. $\endgroup$ – angryavian Oct 25 '17 at 23:37
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    $\begingroup$ The cut needs to occur in which part of the rope for the ratio to be that small? $\endgroup$ – Joffan Oct 25 '17 at 23:37
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    $\begingroup$ Also, when you correct $1/3$ to $1/4$, what about if $X > (3/4)L$? By symmetry, you csn just multiply by $2$, but don't forget to do that. $\endgroup$ – quasi Oct 25 '17 at 23:38
  • $\begingroup$ HHH, hopefully helpful hint: the length of the rope is irrelevant. $\endgroup$ – Joffan Oct 25 '17 at 23:40
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    $\begingroup$ Also, using an integral is overkill. Just use length over length (i.e., length of the success region over total length). $\endgroup$ – quasi Oct 25 '17 at 23:41
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Imagine your rope as one unit long. Putting it on the number line, the endpoints will be at 0 and 1. Cutting at $1/4$, the ratio is $1:3$, and any cuts left of that also give smaller ratios. By symmetry, the cuts at greater or equal to $3/4$ also give $3:1$ ratios. Thus, we have two quarter length segments that work, in a segment of unit length, giving the probability $1/2$.

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