A stick has a length of 5 units. The stick is then broken at two points, chosen at random. What is the probability that all three resulting pieces are SHORTER THAN THREE UNITS.
Many people have posted 4/25, 16/25 but those are wrong, PLS HELP.
+66 use the logic from previous answers to answer this, it shouldn't be that hard
+2668 I'm not sure if my logic is right, but anyways...
Think about it this way... You break it into 2 points initially. There are 2 cases. 1 stick is longer than 3 units (1) or both sticks are shorter than 3 units (2).
The probability of (2) happening is \(1 \over 5\)
Once you split it into 2 sticks, each less than 3, you are guaranteed to get a 3rd stick that is also less than 3.
It is the region that is on the line \(x+y=5\), but bounded by the inequalities \(x \leq 3\) and \(y \leq 3\).
Here is a graph if you would like a more "visual" proof: https://www.desmos.com/calculator/bqrv2fkhhz
The probability of (1) happening is very complicated because the probability depends on where the "breaking point" was.
The best-case scenario is when you have 2 sticks that are split into a stick with length 3 and a stick with length 2.
Here, you have a \(3 \over 5\) chance of success, because if you break it anywhere in the segment with a length of 3, you split it into 2 sticks each with a length less than 3
However, the worst-case scenario is when you split it into lengths 5 and 0 (or smth very close to it).
Here, you only have a \(1 \over 5\) chance of splitting the stick into 2 lengths of 3, as shown in (2).
Because the probability depends, you take the average, which is \(2 \over 5\) (I think...)
Thus, the probability is \({1 \over 5 }+ {2 \over 5} = \color{brown}\boxed{3 \over 5}\)
