Prove inequality:
$\mathbb P(A \mid B) \ge 1- \frac{\mathbb P(A')}{\mathbb P(B)}$
I expanded the conditional probability on the left side of the inequality through the formula:
$\mathbb P(A \mid B) = \frac{\mathbb P(A\cdot\ B)}{\mathbb P(B)}$
Then on the right side I brought everything to a common denominator and got:
$1 - \frac{\mathbb P(A')}{\mathbb P(B)} = \frac{\mathbb P(B)- \mathbb P(A')}{\mathbb P(B)}$
After the transformations, I get the following expression and remove the common denominator:
$\mathbb P(A\cdot\ B) \ge \mathbb P(B)- \mathbb P(A')$
But I don’t know what to do next and how to prove the original simplified expression. Tell me please
Easy observe that
$$\mathbb{P}[AB]\ge \mathbb{P}[B]-\mathbb{P}[A']$$
can be rewritten in
$$\mathbb{P}[AB]\ge \mathbb{P}[B]-1+\mathbb{P}[A]$$
that is also
$$\mathbb{P}[A]+\mathbb{P}[B]-\mathbb{P}[AB]\le 1$$
or
$$\mathbb{P}[A\cup B]\le 1$$
which completes the proof