Consider a permutation sampled uniformly at random from the set of all permutations of $\{1, 2, 3, \cdots, n\}$ for some $n \ge 4$. Let $X$ be the event that 1 occurs before 2 in the permutation, and $Y$ the event that 3 occurs before 4. Which one of the following statements is TRUE?

By symmetry, in a random permutation of $\{1, 2, \dots, n\}$:

• 1 is equally likely to appear before or after 2, so $P(X) = \frac{1}{2}$.

• Similarly, 3 is equally likely to appear before or after 4, so $P(Y) = \frac{1}{2}$.

Since the set of elements $\{1, 2\}$ is disjoint from $\{3, 4\}$, any relative ordering of $\{1, 2\}$ is completely independent of the relative ordering of $\{3, 4\}$. Specifically, for any chosen 4 positions out of $n$, all $4! = 24$ orderings of the distinct elements $1, 2, 3, 4$ are equally likely. In 6 of these 24 orderings, 1 appears before 2 and 3 appears before 4 simultaneously:

Since $P(X \cap Y) = P(X) \cdot P(Y) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$, the events $X$ and $Y$ are independent.

Thus, option (B) is TRUE.