Discussion
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?
Answer: B
The events \(X\) and \(Y\) are independent.
By symmetry, 1 precedes 2 in half of all permutations, so \(P(X) = 1/2\), and likewise \(P(Y) = 1/2\). The relative order of 1 and 2 and the relative order of 3 and 4 involve disjoint pairs of elements, and all four combinations of orders are equally likely, so
\[ P(X \cap Y) = \frac{1}{4} = P(X)\,P(Y) \]
They are not mutually exclusive, since both can happen (for example 1, 2, 3, 4, ...). Neither must occur: both fail together with probability 1/4. And X is not more likely than Y, since both have probability 1/2.
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