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Suppose $k$ events form a partition of sample space $\Omega$. Assume that $P(B) \gt 0$. Prove that if $P(A_1|B) \lt P(A_1)$ then $P(A_i|B) \gt P(A_i)$ for some $i = 2, \ldots, k$

My problem is that I just can't see how to use either Bayes' Theorem or the Law of Total Probability to prove this.

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$\displaystyle \sum_{i=1}^k P(A_i|B) = 1 = \sum_{i=1}^k P(A_i) $ so $\displaystyle \sum_{i=1}^k (P(A_i|B) - P(A_i))=0 $ and thus $$\displaystyle \sum_{i=2}^k (P(A_i|B) - P(A_i))=P(A_1) - P(A_1|B) .$$

If the right hand side is positive, then at least one term in the left hand sum must also be positive.

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Hint:

If: $$\sum_{i=1}^kx_i=\sum_{i=1}^ky_i$$ then $x_1<y_1$ implies the existence of some index $i\in\{2,\dots,k\}$ with $x_i>y_i$.

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