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Question in the title. It seems like a silly question, but I literally cannot find an answer to this anywhere.

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2 Answers

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In general the answer in no. Besides, this is exactly the definition of independence of events: two sets $A,~B$ in a (probablity) space $\Omega$, equipped with a probalbility measure, are called independent if $P(A\cap B)=P(A)\cdot P(B)$. And in general, two events $A,~B$ are not independent. For an easy example, take an event $A$ whose probability $p$ satisfies $0<p=P(A)<1$. Take $B=A$ (so $A,~B$ are not independent) and $p=P(A)=P(A\cap B)\neq P(A)P(B)=p^2$.

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$\mathsf P(E~F)=\mathsf P(E)\cdot\mathsf P(F)$ only for independent events; otherwise there would be no point to Bayes' Theorem.

Conditional Probability exists as a distinct concept because not all events are independent.

$$\mathsf P(E\mid F) ~=~ \frac{\mathsf P(E~F)}{\mathsf P(F)}\quad\mathop{\Large =}\limits^{\textsf{iff independent}}\quad \frac{\mathsf P(E)\cdot\mathsf P(F)}{\mathsf P(F)}~=~\mathsf P(E)$$

Instead, Bayes' Theorem is derived by: $$\mathsf P(E\mid F) ~=~ \frac{\mathsf P(E~F)}{\mathsf P(F)}~=~ \frac{\mathsf P(E)\cdot\mathsf P(F\mid E)}{\mathsf P(F)}$$

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