Question on basic probability conditioning

Let $$\Omega, F, P$$ be a probability space and $$A, B, D \in F$$. If $$P( A \mid D ) \ge P(A)$$ and $$P( B \mid D) \ge P(B)$$. Show that if $$A \cap B = \emptyset$$ then $$P(A^c \cap B^c \mid D) \le P(A^c \cap B^c)$$.

I know that given that $$A \cap B = \emptyset$$ then they are mutually exclusives, this means that $$P(A \cap B) = 0$$

So, $$P( A \cup B) = P(A) + P(B)$$, and since $$P( A \cup B) = 1 - P(A^c \cap B^c)$$, this gives:

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

I got an expression for $$P(A^c \cap B^c)$$ but I don't know how to develop the proof from here, $$P(A^c \cap B^c \mid D)$$ can be expanded using the Bayes theorem, but this lead's me to nowhere, Im stuck.

• Can you check if the condition is $P(D|A) \geq P(A)$ or $P(A|D) \geq P(A)$? Dec 3, 2023 at 5:53
• Sorry @Zhanxiong, the condition is $P(A \mid D) \ge P(A)$, im editing the original question. Dec 3, 2023 at 14:20

Using the observation that you have made (i.e., the De Morgan law and the complementary law) shows that it is equivalent to prove \begin{align*} P(A \cup B) \leq P(A \cup B|D). \tag{1}\label{1} \end{align*} The condition $$A \cap B = \varnothing$$ implies that $$(A \cap D) \cap (B \cap D) = \varnothing$$, whence \begin{align*} P(A \cup B|D) &= \frac{P((A \cup B)\cap D)}{P(D)} \\ &= \frac{P(A \cap D) + P(B \cap D)}{P(D)} \\ &= P(A|D) + P(B|D) \\ & \geq P(A) + P(B) = P(A \cup B). \end{align*} This completes the proof.