Can someone help me to prove $P(A \cap B|A\cup B)\le P(A\cup B|A).$

Question

What I have done is to say that $(A\cap B)\cap(A\cup B)=A\cap B$ and $(A\cap B)\cap A=(A\cap B).$ And, I cannot see if $P(A\cup B)\ge P(A)$ to use Baye's law $\frac{P(A\cap B)}{P(A\cup B)}\le \frac{P(A\cap B)}{P(A)}.$

Answer 1

You can say

• $A \subset A\cup B$ so $\mathbb P(A\cup B) \ge P(A)$ but more importantly $\mathbb P(A\cup B \mid A) =1$
• $\mathbb P(A\cap B \mid A\cup B)$ is a conditional probability so cannot exceed $1$
• so $\mathbb P(A\cap B \mid A\cup B) \le \mathbb P(A\cup B \mid A)$

You may need $\mathbb P(A) > 0$ for $\mathbb P(A\cup B \mid A)$ to be well-defined; if it is then $\mathbb P(A\cup B) >0 $ so $\mathbb P(A\cap B \mid A\cup B)$ will also be well-defined.