Slicing a bar in three pieces - probability I have the following problem: If you split a bar of length 1 in three pieces by choosing two cut points randomly. What is the probability that the lenght of at least one of them is less than 1/5?.

I think that $P(a<1/5)=P(1-b<1/5)=P(b-a<1/5)=1/5$ But now I am confused how to use this in the porpouse of compute the total probability. I would bet that 3/5 is the answer (because you either have that a<1/5 or 1-b<1/5 or a-b<1/5) but it is not crystal clear to me.
 A: In your calculations you have supposed that the pieces have lengths $a$, $b-a$, and $1-b$ and that $a$ and $b$ are independent uniformly distributed between $0$ and $1$. However, this is not the case. If $a$ and $b$ are independent and uniformly distributed between $0$ and $1$, then we do not necessarily have $a<b$. Said another way, if we suppose $a<b$ so the pieces have lengths $a$, $b-a$, and $1-b$, then $a$ and $b$ are not independent and uniformly distributed - we have to condition on $a<b$.
Next, when calculating the probability that at least one event happens, it is often easier to calculate the probability that none of the events happens, and then take the complement. I.e.,
$$P(\text{at least one segment is $<1/5$})=1-P(\text{all segments are $\geq 1/5$}).$$
Let $X$ and $Y$ be the two points at which the bar is cut, with $X$ and $Y$ independent and uniformly distributed on $[0,1]$. Then in order to have all three segments $\geq 1/5$, we require:

*

*$1/5 \leq X \leq 4/5$

*$1/5 \leq Y \leq 4/5$

*$|X-Y| \geq 1/5$.

Graphically, the set of $(X,Y)$ for which the above holds lies in the intersection of the red, green, and blue shaded regions in this graph. These are two right triangles with leg length $2/5$. Since $(X,Y)$ is uniformly distributed over $[0,1]^2$, the probability of this event is the area of the region where the event occurs. Hence,
$$P(\text{all segments are $\geq 1/5$})= \frac{1}{2}\left(\frac{2}{5}\right)^2+\frac{1}{2}\left(\frac{2}{5}\right)^2=\frac{4}{25}.$$
Thus, $P(\text{at least one segment is $<1/5$})=1-4/25=21/25$.
A: A geometrical approach

which should be quite clear
