Independent solution

How to solve this Event Independence question

Setup

Setup

Check pairwise independence and mutual independence separately. The three marginal probabilities are all one half.

Pr(A)=Pr(B)=Pr(C)=12\Pr(A)=\Pr(B)=\Pr(C)=\frac12

Model

Model

Each pairwise intersection has probability one quarter, equal to the product of the corresponding marginal probabilities. Thus every pair is independent.

Pr(AB)=Pr(AC)=Pr(BC)=14\Pr(A\cap B)=\Pr(A\cap C)=\Pr(B\cap C)=\frac14

Compute

Compute

Mutual independence would require a triple-intersection probability of one eighth, but the actual triple intersection is empty. Pairwise independence therefore holds without mutual independence.

Pr(ABC)=018\Pr(A\cap B\cap C)=0\ne\frac18

Answer

Answer

Every pair is independent, while the three events together are not.

pairwise independent, not mutually independent(A)\boxed{\text{pairwise independent, not mutually independent}\quad\text{(A)}}