A bar of length t is broken into 3 pieces at 2 random points, what is the probability that the shortest piece is shorter than 0.1t?

Question

We have a bar of length $t$. We randomly choose 2 points and break it at these points, chopping the bar into 3 bars. What is the probability that the smallest piece is shorter than $\frac{t}{10}$?

I think by denoting the two points $a$ and $b$ and $a<b$, the three pieces are of length $a$, $b-a$ and $t-b$, we need to first find the distribution of the smallest of the three and then use its c.d.f to calculate the final answer. However I'm 100% sure that's not the best way to do this problem, please help me clear my mind and tell me an elegant way to attack this problem, thank you very much!

Answer 1

Let $A$ be the event in which the smallest piece is longer than $0.1 t$. When we pick the first point, we can choose any value $x$ with $0.1 t < x < 0.5 t$ (values between $0.5 t$ and $0.9 t$ are also possible, but we are dealing with a symmetric problem). We can distinguish two cases:

1. $0.1 t < x < 0.2 t$. In this case, the remaining area of valid choices for the second value equals $0.8 t - x$, since we cannot pick a value to the left of $x + 0.1 t$, or to the right of $0.9 t$.
2. $0.2 t < x < 0.5 t$. In this case, the remaining area of valid choices equals $0.6 t$, since we cannot pick any value to the left of $0.1 t$, to the right of $0.9 t$ or between $x - 0.1 t$ and $x + 0.1 t$.

Taking symmetry into account, we find:

\begin{align} P(A) & = 2 \int_{0.1}^{0.2} (0.8 - x) dx + 2 \int_{0.2}^{0.5} 0.6 dx \\ & = 2 \cdot 0.8 \cdot (0.2 - 0.1) - 2 \cdot \frac{0.2^2 - 0.1^2}{2} + 2 \cdot 0.6 \cdot (0.5 - 0.2) \\ & = 0.16 - 0.03 + 0.36 \\ & = 0.49 \end{align}

Answer 2

We compute the probability of the event $A$ that all three pieces are $\geq0.1$.

The random process produces a uniformly distributed point $(x,y)\in[0,1]^2$. The three pieces then have lengths $$\min\{x,y\}, \quad |y-x|, \quad1-\max\{x,y\}\ .$$ It follows that $$A=\bigl\{(x,y)\in[0,1]^2\bigm| (x,y)\in[0.1,\>0.9]^2, \ |y-x|\geq0.1\bigr\}\ .$$ This set consists of two isosceles right triangles with leg lengths $0.7$, so that $P(A)={0.7^2\over1}=0.49$. It follows that the desired probability of at least one piece having length $<0.1$ is $$P(\neg A)=0.51\ .$$