Is it true that $P(A)\le P(B)\implies P\left( { A }\mid{ C } \right) \le P\left( { B }\mid{ C } \right)$?

Question

$P(A)\le P(B)\implies P\left( { A }\mid{ C } \right) \le P\left( { B }\mid{ C } \right)$

I feel like it's correct but I don't really know how to come up with a proof. So far I've been trying with algebra of sets but doesn't seem clear.

Answer 1

The statement is not generally true, because (for example) $B$ and $C$ may be disjoint.

Counter example

$U =$ the set $\{1, 2, 3, \cdots, 20\}.$

$B$ is the event that a # drawn from $U$ is between $6$ and $15$, inclusive.

$C$ is the event that a # drawn from $U$ is between $1$ and $5$, inclusive.

$A$ is the event that a # drawn from $U$ is between $1$ and $3$, inclusive.

$p(A) = 3/20$.

$p(B) = 10/20$.

$p(A|C) = 3/5$.

$p(B|C) = 0$.