P(C)=2P(A) and P(B)=P(A U C). Find probability of A,B ,C?

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P(B)=P(A∪C)=P(A)+P(C)-P(A)P(C)=

P(A)+2P(A)-2(P(A))²=3P(A)-2(P(A))².

If y=P(A) then 0≤y≤1. This implies that 0≤P(B)≤0.5 because 0≤y(3-2y)≤1

If we solve for y: y(3-2y)≤1, 2y²-3y+1≥0, (2y-1)(y-1)≥0 only when y≤0.5. So 0≤y≤0.5.

P(C)=2y, so P(C) is unrestricted (can be any probability).

To find specific values for the probabilities more information is needed. Since the sum of the probabilities is 1 (exhaustive) then P(A)+P(B)+P(C)=1=3y+3y-2y², 2y²-6y+1=0. y=(6±√(36-8))/4=(6±2√7)/4=2.82 or 0.1771 approx. Therefore P(A)=0.1771, P(C)=0.3542, P(B)=0.4686.

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