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Back to 3.2 Problems on Conditional Probability

3.2 Exercise 16

Avg difficulty 2.5

Show that if P(A|C)>P(B|C)P \left(\right. A \left|\right. C \left.\right) > P \left(\right. B \left|\right. C \left.\right) and P(A|Cc)>P(B|Cc)P \left(\right. A \left|\right. C^{c} \left.\right) > P \left(\right. B \left|\right. C^{c} \left.\right), then P(A)>P(B)P \left(A\right) > P \left(B\right). Is the converse true? Prove or give a counterexample.

Pfeiffer, Applied Probability, 3.2 Problems on Conditional Probability, exercise 16

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Adapted from Applied Probability © 2009 Paul E. Pfeiffer. Licensed under CC BY 3.0. Source: Applied Probability. Imported at revision d7a0420.