Consider a permutation sampled uniformly at random from the set of all…

Answer: The event that 1 occurs before 2 and the event that 3 occurs before 4 are independent.

Key ideas:

• By symmetry, the probability that 1 occurs before 2 is 1/2 (in a random permutation 1 is equally likely to appear before or after 2). Similarly, the probability that 3 occurs before 4 is 1/2.

• To compute the joint probability that 1 occurs before 2 and 3 occurs before 4, look at the relative order of the four elements 1,2,3,4. All 4! = 24 orders of these four elements are equally likely.

• Choose the two positions occupied by {1,2} (there are C(4,2)=6 choices); once those positions are chosen, requiring 1 before 2 fixes their order, and the remaining two positions for {3,4} with 3 before 4 are also fixed. Thus there are 6 orders where both desired relations hold, so P(both) = 6/24 = 1/4.

• Since P(both) = 1/4 = (1/2)(1/2) = P(1 before 2) × P(3 before 4), the events are independent. This argument does not depend on n beyond requiring n ≥ 4, because other elements do not affect the relative order of these four.