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Let P(A)=0.4 , P(B)=0.7, P(AUB)=0.8. Find:1) P(Bc)=2) P(A-B)=3) P(AUBc)=4) P(B/A)=5) P(Bc/A)= |
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Answer» P(A∪B)+P(A∩B)=P(A)+P(B)=0.7+0.2=0.9 P(A∩B)=0.2=P(A)P(B) P(B)=0.9-P(A), P(A)(0.9-P(A))=0.2, P^2(A)-0.9P(A)+0.2=0 P(A)=½(0.9±√[0.81–0.8])={0.5,0.4} P(B)=0.9-P(A)={0.4,0.5}, as P(A)>P(B): P(A)=0.5, P(B)=0.4 Answer : P(A)=0.5, P(B)=0.4 |
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