Consider a permutation sampled uniformly at random from the set of all permutations of \(\{1, 2, 3, \cdots, n\}\) for some \(n \geq 4\). 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)
Explanation
By symmetry, 1 comes before 2 in exactly half of all permutations, so \(P(X) = \frac{1}{2}\); likewise \(P(Y) = \frac{1}{2}\).
The order inside the pair \(\{1, 2\}\) and the order inside the pair \(\{3, 4\}\) are each equally likely either way and do not influence each other, so \(P(X \cap Y) = \frac{1}{4} = P(X)\,P(Y)\). Hence \(X\) and \(Y\) are independent.
- A fails: both events can happen together (probability \(\frac{1}{4}\)).
- C fails: both can fail together, for example in 2, 1, 4, 3, ….
- D fails: the two events are equally likely.
Answer: B