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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?

  1. The events $X$ and $Y$ are mutually exclusive
  2. The events $X$ and $Y$ are independent
  3. Either event $X$ or $Y$ must occur
  4. Event $X$ is more likely than event $Y$
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Just by reading the question, it is evident that -

A. Events X and Y are not mutually exclusive, both events can occur simultaneously.

C. It is also possible that both even can not occur.

D. Both events are symmetric, their probabilities are exactly same.

B. Occurring of either event doesn't affect occurring of another event. Thus, independent.

P(X) = P(Y) = 0.5, P(X,Y) = 0.25 = P(x) * P(Y). This shows that indeed events X and Y are independent.

Total number of permutations = n!

Number of permutations where 1 comes before 2 = ${n \choose 2} * (n-2)!$

Number of permutations where 3 comes before 4 = ${n \choose 2} * (n-2)!$

Number of permutations where 1 comes before 2 and 3 comes before 4 = ${n \choose 2} * {n-2 \choose 2} * (n-4)!$

Answer - B.
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