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Part CCSIR NET December 2025swapping-the-two-iid-dice-negates-x-minus-y-over-x-plus-y-that-symmetry-alone-forces-its-mean-to-0

Swapping the two iid dice negates x minus y over x plus y that symmetry alone forces its mean to 0

Suppose two fair dice are thrown independently at random. Let X and Y be the numbers on the upper face of the first die and that of the second die, respectively. Then which of the following statements are true?

  1. A.P(X−Y=0 ∣ X+Y=2) = P(X−Y=0 ∣ X+Y=12)✓
  2. B.E((X−Y)/(X+Y)) = 0✓
  3. C.Cov(X+Y, X−Y) = 0✓
  4. D.(X+Y) and (X−Y) are independent

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

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50 are analysed free — try those first.

Related counterexample: Pairwise independent events are mutually independent

More on this topic

The chapter behind this: Probability axioms, conditioning and independence — free to read

From Probability › Axioms, conditional probability, independence, Bayes

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