Given $\mathbb{P}(A\cap B)\le\mathbb{P}(A)\cdot\mathbb{P}(B)$. Prove: $\mathbb{P}(A^c\cap B^c)\le\mathbb{P}(A^c)\cdot\mathbb{P}(B ^c)$

Question

Let $\mathbb{P}$ be a probability measure and $A$ , $B\in \Omega$ events such that: $\mathbb{P}(A\cap B)\le\mathbb{P}(A)\cdot\mathbb{P}(B) $

Prove: $\mathbb{P}(A^c\cap B^c)\le\mathbb{P}(A^c)\cdot\mathbb{P}(B ^c)$

Any help would be appreciated.

Answer 1

$$ P(A\cap B) \leq P(A)P(B) $$

$$1-P((A\cap B)^c) \leq (1-P(A^c))(1-P(B^c)) $$

$$ P(A^c)+P(B^c)- P(A^c\cup B^c)\leq P(A^c)P(B^c)$$

$$P(A^c\cap B^c) \leq P(A^c)P(B^c)$$