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We have two rods of length $a$ and $b$ ($b> a$) we randomly cut each rod to $2$ pieces. Then we choose $3$ pieces of $4$ pieces. What is the probability that these pieces form a triangle?

I think if we consider first piece if a as $x$ then $x$ has uniform distribution $(0,a)$ and the other piece has length of $a-x$. For the other rod we consider one piece as y and the other as $b-y$. We have $4$ different combinations for choosing pieces . I think I have to consider each combination, then find the probability of making triangle using triangle inequality but I don't know how to apply it. Also I think I don't have to calculate for each of $4$ states (because of symmetry).

Roger
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Little Stack
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1 Answers1

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WLOG, assume $a=1$. There are two cases: the left out piece corresponds to the long rod (length $b$), or it corresponds to the short rod (length $1$). Both cases are equiprobable.

Denote by $x \in [0,1]$ the cutting position for the short rod, $y \in [0,b]$ the cutting position for the long rod.

Lets compute the probability of success (a triangle can be formed) assuming first case.

The lengths of the pieces are then $a=x$, $b=1-x$, $c=y$. The triangle inequalities ($a<b+c$, $b<a+c$, $c<a+b$) give the conditions

$$[y<1], [y>2x-1], [y>1-2x] $$

Graphing this on the $x,y$ plane we get a triangle of area $1/2$, over the total zone (of uniform probability) of size $b$. Hence the probability is $$p=\frac{1}{2b}$$

As similiar calculation for the other case gives the same result.

In conclusion, in general:

$$p=\frac{\min(a,b)}{2 \max(a,b)}$$

leonbloy
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  • I didn't understand the last part. How did you conclude that p= 1/(2b)? when I draw graph for " a "(not 1) the area is a^2/2 for the first case and b^2/2 for the next. How did you assume a= 1? – Little Stack Jan 12 '21 at 09:41
  • so I think the answer should be a/(2b) for the first case. But b/(2a) for second case – Little Stack Jan 12 '21 at 10:50
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    The scale should be irrelevant $(a=3,b=5)$ should give the same result that $(a=30,b=50)$ . Hence we can assume $a=1$ – leonbloy Jan 12 '21 at 11:09