Show that $P(A\cap B) \geq 1-P(A^{c})-P(B^{c})$.

Question

So far I have:

$$P(A\cap B) = 1-P((A\cap B)^{c}) $$ (complement axiom) $$= \[1-P(A^{c}\cup B^{c})\]$$ (DeMorgan's) =... I'm not sure what to do after this ;(

Answer 1

Hint: Use $1-P(A^c)-P(B^c) = P(A) + P(B) - 1$

Answer 2

Rearrange the inequality in the following way

$$ P(A \cap B) \geq P(A) + P(B) - 1 $$

And thus

$$ 1 \geq P(A) + P(B) - P(A \cap B) $$

which implies

$$ 1 \geq P(A \cup B) $$

which is true always

Answer 3

Note that for any event $C$, $P[C]\le 1$. So $P[A\cup B]\le 1\iff -P[A\cup B]\ge -1$. Call this inequality $(a)$. Note that $P[A\cup B]=P[A]+P[B]-P[A\cap B]\implies P[A\cap B]=P[A]+P[B]-P[A\cup B]$. Using $(a)$ in here, what do you get?