# Thread: one probability prove question

1. ## one probability prove question

Suppose that P(A)=a and P(B)=b. show that P(A|B)>=(a+b-1)/b.

2. http://www.mathhelpforum.com/math-he...ial-19060.html
Please, please learn to post in symbols? You can use LaTeX tags.
$$P(A|B)\ge \dfrac{a+b-1}{b}$$ gives $\displaystyle P(A|B)\ge \dfrac{a+b-1}{b}$.
Having misread this posting, I wasted a good bit of time.

All it takes is to notice $\displaystyle P(A\cap B)\ge P(A)+P(B)-1$.

3. Originally Posted by Plato
http://www.mathhelpforum.com/math-he...ial-19060.html
Please, please learn to post in symbols? You can use LaTeX tags.
$$P(A|B)\ge \dfrac{a+b-1}{b}$$ gives $\displaystyle P(A|B)\ge \dfrac{a+b-1}{b}$.
Having misread this posting, I wasted a good bit of time.

All it takes is to notice $\displaystyle P(A\cap B)\ge P(A)+P(B)-1$.
sry, how do you notice this?

4. Originally Posted by wopashui
sry, how do you notice this?
Have you studied the axioms of probability?

5. this is not one of the axioms of probability, is it, I'm afraid that we need to prove this is true before we use it

6. Originally Posted by wopashui
this is not one of the axioms of probability, is it, I'm afraid that we need to prove this is true before we use it
Well you surely know that $\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)$
AND $\displaystyle \left( {\forall X} \right)\left( {0 \leqslant P(X) \leqslant 1} \right)$.

Put it together for yourself.