On conditional probability – let's say there's two events $A$ and $B$ with probabilities $\Pr(A)$ and $\Pr(B)$. Is the conditional probability
$$\Pr(A\cap B | A\cup B) = \frac{\Pr\big((A\cap B)\cap(A\cup B)\big)}{\Pr(A\cup B)} = \frac{\Pr(A\cap B)}{\Pr(A\cup B)}\text?$$
I.e., is the probability of both events happening "and" at least one event happening equivalent to the probability of both events happening?
Best Answer
$(A\cap B) \subset \begin{cases}A\\B \end{cases} \subset (A\cup B)$ so $(A\cap B)\subset (A\cup B)$.
And if $M \subset N$ then $M\cap N= M$. I.e. the interection of a subset with its superset is the subset as it is completely contained in it.
So $(A\cap B) \cap (A\cup B) = (A\cap B)$
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And if $M = N$ the $P(M) = P(N)$ as they are the exact same thing.
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So yes we can presume $P([A\cap B]\cap [A\cup B]) = P(A\cap B)$.
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So if we do the following
$P(A\cap B|A\cup B) \underbrace{=}_{\text{conditional probability law}}$
$\frac {P\color{blue}{([A\cap B]\cap [A\cup B])}}{P(A\cup B)} \underbrace{=}_{\color{blue}{([A\cap B]\cap [A\cup B])} = \color{green}{A\cap B}}$
$\frac {P\color{green}{(A\cap B)}}{P(A\cup B)}$
that is valid conclusion.