1

Five intervals are each selected according to the following procedure: two points are sampled from U[0,1], the larger becoming the right endpoint and the smaller becoming the left endpoint. What is the probability that there exists a point of intersection between all five intervals? The answer can be written as a simplified fraction of the form p/q. Find p+q

My attempt was write all the five Intervals then Max{Li}<=Min{Ri}. Total ways to arrange are n! But what to do next or I am going the wrong way

user1657506
  • 63
  • 4
  • 1
    Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Jun 30 '25 at 01:46

1 Answers1

0

We ignore the case the sampled points contains repeated points, as it happens with a probability of $0$. By arranging the $10$ sampled points in increasing order, there exists a point of intersection between all $5$ intervals if and only if the first $5$ points each corresponds to different intervals. Now, all $10!$ permutations of sampled points happen with a probability of $\frac{1}{10!}$, and within the $10!$ permutations, exactly $2^5(5!)^2$ permutations satisfies the conditions. (For the first $5$ points, choose which interval it corresponds to. Same for the last $5$ points. Then for interval decide which of the $2$ points will be the one sampled first.) Therefore the probability is $\frac{2^5 (5!)^2}{10!} = \frac{8}{63}$．

Sugai Ryomei
  • 324
  • 6