Average of shortest half of two halves and probability of three halves making a triangle

Question

This is a problem that has been bothering me because it seems so easy however the answers don't feel right so...(It showed up in my latest statistics exam and almost everybody got it wrong because we thought it was just too plain easy) "A point is randomly chosen in the [0,1] interval splitting it in two halves."

1.Whats the average length of the shortest half? Isn't the average always 0.5?

2.Now picking two random points in the same interval what's the probability of being able to create a triangle with the 3 segments? Shouldn't this be 1?

Answer 1

1. Let $l$ be the length of the shortest half. Then $$ \mathsf{E}l=\int_0^{1/2}x\,dx+\int_{1/2}^1 (1-x)\, dx = \frac{1}{8}+\frac{1}{8} = \frac{1}{4}. $$

2. Look at this question (your probability is $1-$ the probability that the length of one of the segments is greater than $1/2$).

Answer 2

a) The short piece must be less than or equal to $0.5.$ So, it is not possible for the expected value to be $0.5$

$X$ is a unifrom random variable in $[0,1]$

$Y = \min (x, 1-x)$

Find $E[Y]$

b)

In order for the 3 segments to form a triangle, they must stratify the "triangle inequality." That is the longest of the 3 segments must be shorter, than the sum of the other two segments. (i.e. it must be less than $0.5$)

This vid has a nice explanation and solution.

https://www.youtube.com/watch?v=udxwP26gTwA