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- Thread starter TomJerry
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Show some work. There are a few formulas about independence you can use to get you started.

- #3

mathman

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Not true!

Counter example: Probability space is unit square, with probability = area.

A: 0≤x≤1/2, 0≤y≤1

B: 0≤x≤1, 0≤y≤1/2

C: two pieces C1 + C2

C1: 0≤x≤1/2, 0≤y≤1/2

C2: 1/2<x≤1, 1/2≤y≤1

P(A)=P(B)=P(C)= 1/2

P(AandB)=1/4, P(AandC)=1/4, A independent of B and C

P(BUC)=3/4, P(AandBUC)=1/4, while P(A)P(BUC)=3/8, not independent!

- #4

mathman

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Consider a probability space with 4 elementary events k,l,m,n where each event has probability 1/4.

Let A={k,l}, B={k,m}, C={k,n}. Then:

P(A)=P(B)=P(C)= 1/2

A∩B={k}, A∩C={k}, A∩(BUC)={k}, while BUC={k,m,n}.

P(A∩B)=1/4, P(A∩C)=1/4, A independent of B and C

P(BUC)=3/4, P(A∩(BUC))=1/4 ≠ P(A)P(BUC)=3/8, not independent!

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