If events A, B, and C are mutually independent (independent overall), then they
[ Select ]
must be
must not be
might be
pairwise independent. If events A, B, and C are pairwise independent, then they
[ Select ]
must be
must not be
might be
mutually independent.

Question

If events A, B, and C are mutually independent (independent overall), then they 
[ Select ]
must be
must not be
might be
 pairwise independent.  If events A, B, and C are pairwise independent, then they 
[ Select ]
must be
must not be
might be
 mutually independent.

BlackTom AI · Accepted Answer

Mutual independence for three events means every subcollection is independent: P(A∩B)=P(A)P(B), P(A∩C)=P(A)P(C), P(B∩C)=P(B)P(C), and P(A∩B∩C)=P(A)P(B)P(C). That last condition implies the three pairwise equalities, so mutual independence always yields pairwise independence.

Step-by-step reasoning:
1. Definitions and key point:
   - Pairwise independent: each pair satisfies P(X∩Y)=P(X)P(Y) for X,Y ∈ {A,B,C}.
   - Mutually independent (overall): every subset (including the triple) satisfies the product rule; in particular it includes the three pairwise relations.
   Therefore if A, B, C are mutually independent, they must be pairwise independent. This justifies the first blank: "must be."

2. Why the converse need not hold:
   Pairwise independence does not guarantee P(A∩B∩C)=P(A)P(B)P(C). A concrete counterexample shows this.
   - Sample space: two fair coin flips with outcomes (first,second) ∈ {(0,0),(0,1),(1,0),(1,1)}, each with probability 1/4.
   - Define A = {first = 1} = {(1,0),(1,1)} so P(A)=1/2.
   - Define B = {second = 1} = {(0,1),(1,1)} so P(B)=1/2.
   - Define C = {bits equal} = {(0,0),(1,1)} so P(C)=1/2.
   Compute pairwise intersections:
   - A∩B = {(1,1)} ⇒ P(A∩B)=1/4 = P(A)P(B).
   - A∩C = {(1,1)} ⇒ P(A∩C)=1/4 = P(A)P(C).
   - B∩C = {(1,1)} ⇒ P(B∩C)=1/4 = P(B)P(C).
   So A,B,C are pairwise independent. But the triple intersection A∩B∩C = {(1,1)} has probability 1/4, while P(A)P(B)P(C) = (1/2)^3 = 1/8. Because 1/4 ≠ 1/8, they are not mutually independent.

3. Why other choices are wrong:
   - For the first blank, "must not be" and "might be" are wrong because mutual independence enforces the pairwise equalities.
   - For the second blank, "must be" is wrong because the counterexample shows pairwise independence can fail to imply mutual independence; "must not be" is wrong because sometimes pairwise independent sets are also mutually independent.

Conclusion: The correct answers are "must be" for the first blank and "might be" for the second blank, matching the provided answer.

If events A, B, and C are mutually independent (independent overall), then they [ Select ] must be must not be might be pairwise independent. If events A, B, and C are pairwise independent, then they [ Select ] must be must not be might be mutually independent.多重下拉选择题

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