SOLUTION: Suppose that 63% of the women who gave birth at a certain hospital last year were over 30 years old, and that 55% were unmarried. If 68% of the women were over 30 or unmarried (or

Algebra.Com
Question 388443: Suppose that 63% of the women who gave birth at a certain hospital last year were over 30 years old, and that 55% were unmarried. If 68% of the women were over 30 or unmarried (or both), what is the probability that a woman who gave birth at the hospital was both unmarried and over 30? I'm just so stumped on how probability works, its been such a long time!
Answer by robertb(5830)   (Show Source): You can put this solution on YOUR website!
Let AB denote the intersection of the two events A and B. Then by the general addition law, P(AUB) = P(A) + P(B) - P(AB), or P(AB) = P(A) + P(B) - P(AUB). Then P(unmarried AND over 30) = P(unmarried) + P(over 30) - P(unmarried OR over 30) = 0.55 + 0.63 - 0.68 = 0.50.
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