# Can anyone solve $P(A|B)\leq P(A) \implies P(A|B^c)\geq P(A)$?

Does anyone know how to prove $P(A|B)\leq P(A) \implies P(A|B^c)\geq P(A)$ ? It is from the book "introduction to probability"

By the law of total probability and our hypothesis we have $$P(A) = P(A \mid B)P(B) + P(A \mid B^c)P(B^c) \leq P(A)P(B) + P(A \cap B^c).$$

Hence, $$P(A)(1 - P(B)) \leq P(A \cap B^c)$$

Since $1 - P(B) = P(B^c)$,

$$P(A) \leq \frac{P(A \cap B^c)}{P(B^c)} = P(A \mid B^c).$$

To get you started:

$P(A\mid B)\leq P(A) \\ \Updownarrow \text{ (if$P(B)\neq 0$)} \\ P(A\cap B)\leq P(A)P(B) \\ \Updownarrow \\ P(A)-P(A\cap B^\complement)\leq P(A)\,(1-P(B^\complement)) \\ \Updownarrow \\ \\ \ddots \\$

• it starts with $P(A|B)≤P(A)$. I guess it doesn't matter since we can just flip the sign to another direction. – Newbornalive Oct 6 '16 at 5:48
• @Newbornalive Yes, indeed. – Graham Kemp Oct 6 '16 at 23:48