A permutation of the numbers 1, ... 12 is drawn at random, and written on a sheet of paper. For example,
9, 2; 5, 1, 8, 12, 10, 3, 6, 11, 4, 7. Consecutive pairs determine a random interval. In the example, the random
intervals are [2, 9] , [1, 5] , [8, 12] , [3, 10] , [6, 11] , [4, 7]. Let X be the random variable that counts the number of
intervals that intersect all other intervals. In the example, [2, 9] intersects all others, and so does [3, 10]. These
are the only two, hence X = 2. It holds that

Pr (X > = k) = (2 ^ k) / (2k+1 C k)
for k = 0, 1, 2, 3, 4, 5, 6.

(C is the combination choose thing)

1. Find the probability that X = 2

2.) Prove mu (mean)= The summation from k=1 to k=6 of Pr(X > = k)

3. Find mu (mean) for permutations of length n = 12

4. Find the value the limit for mu when n goes to infinity (n--->inf)
mu = The summation from k=1 to k=infinity of (2 ^ k) / (2k+1 C k)
(C is the combination choose thing)

(hint: do some numerical test for 2(mu) and large n)

Thank you very much for looking at this I'm so lost??

Here's an image of the work on mathtype: