Question 2.26: Let P(C)P(C^c) > 0. Then the inequalities P(A | C) > P......

The inequalities P(A\mid C)\gt P(B\mid C) and P(A\mid C^{c})\gt P(B\mid C^{c}) are equivalent to P(A\cap C)\gt P(B\cap C) and P(A\cap C^{c})\gt P(B\cap C^{c}). Adding up these inequalities, we obtain P(A\cap C)+P(A\cap C^{c})\gt P(B\cap C)+ P(B\cap C^{c})\ {\mathrm{or}}\ P(A)\gt P(B), since A=(A\cap C)\cup(A\cap C^{c}) and B=\left(B\cap C\right)\cup (B\cap C^{c}).

REMARK 4 Once again, that the inequalities of the two conditional probabilities should imply the same inequality for the unconditional probabilities is quite obvious on intuitive grounds. The justification given above simply makes it rigorous.