Show that if event A is completely independent of Event B

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Event A being completely independent of events B and C does not imply that A is independent of the union of B and C (BUC). Counterexamples demonstrate this, using a probability space defined as a unit square. In one example, A, B, and C each have a probability of 1/2, yet the intersection probabilities show A is not independent of BUC. Another simpler example with four elementary events illustrates the same principle, confirming that independence does not extend to unions. Therefore, the assertion that independence of A from B and C guarantees independence from BUC is false.
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Problem : Show that if event A is completely independent of Event B and C then A is independent of BUC?
 
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Show some work. There are a few formulas about independence you can use to get you started.
 
TomJerry said:
Problem : Show that if event A is completely independent of Event B and C then A is independent of BUC?
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!
 
Simpler counter example (actually a simplification of the above).
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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