Prove that if P(A)=a and P(B)=b, then P(A|B)>=(a+b-1)/b.

Question

Prove that if P(A)=a and P(B)=b, then P(A|B)>=(a+b-1)/b.

P(A|B)=P(AB)/P(B) then P(A|B)=ab/b=a

So a>=(a+b-1)/b I'm thinking that you then suppose a<(a+b-1)/b, and then find a contradiction to this.

So ab<(a+b-1). This is where I get stuck.

I guess the next step has something to do with a and b being probabilities so both a and b are both greater than 0 and less than 1. And ab is also greater than 0 and less than 1.

Answer 1

$P(A \mid B) = \displaystyle\frac{P(A \cap B)}{P(B)} = \frac{P(A \cap B)}{b}$. Now, why does $P(A \cap B) \geq a+b-1$? Think of inclusion-exclusion or just draw venn diagrams.

Also, $P(A \cap B) = ab$ if $A$ and $B$ are independent, which I don't think is given in the problem...

Answer 2

You always have \begin{eqnarray} p(A) + p(B) &=& p(A \setminus B)+ p(B \setminus A) + 2p ( A \cap B) \\ &=& p(A \setminus B)+ p(B \setminus A) + p ( A \cap B) +p ( A \cap B) \\ &=& p ( A \cup B) +p ( A \cap B) \\ &\le& 1 + p ( A \cap B) \end{eqnarray}