Is $P(B|A) \leq P(B|A\cup B)?$ Is $P(B|A)\leq P(B|A \cup B)$?
I believe the answer is yes and I've simplified to the LHS = $\frac{P(A\cap B)}{P(A)}$ and RHS = $\frac{P(B)}{P(A\cup B)}$ but I'm a little lost from there.
 A: Using Inclusion-Exclusion principle:
$$P(A\cup B) = P(A)+P(B)-P(A\cap B)$$
Note that $P(B) \ge P(A \cap B)$, and observe that:
$$\frac {P(A\cap B) + (P(B) - P(A \cap B))}{P(A) + (P(B) - P(A \cap B))} = \frac {P(B)}{P(A\cup B)}$$
Do you know how to fill in the rest?
A: $Pr(B |A \cup B)$ = $\frac{Pr(B)}{Pr(A \cup B)}$
$Pr(A \cup B)$ = $Pr(A)+Pr(B) - Pr(A,B)$ -(1)
$Pr(B |A)$ = $\frac{Pr(A,B)}{Pr(A)}$, for $Pr(A)>0$ - (2)
$\frac{RHS}{LHS}$ = $\left(\frac{Pr(B)\times Pr(A)}{Pr(A,B)\times \{Pr(A) + Pr(B) - Pr(A,B)\}}\right)$
Where do we go from here? Can we say anything about $\frac{RHS}{LHS}$ in general?
A: Note that $P(B|A)={P [ {B \cap A} ) \over P A}$ and
$P(B | A \cup B) = { P(B) \over P(A \cup B) } = {P ( {B \cap A} ) + P(B \setminus A)\over P A + P(B \setminus A)}$.
Consider the formula $\phi(t) = {a+t \over b+t } $ where $a \le b$, then we see that
$\phi'(t) = { b-a\over (b+t)^2} \ge 0$ and so $\phi$ is non decreasing.
Letting $a=P ( {B \cap A} ),b=pA, t = P(B \setminus A)$ we get the desired result.
