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below are two fundamental formulas in probability theory:

Conditional Probability: $$P(A | B) = \frac{P(A \cap B)}{P(B)}$$ Independent Events: $$P(A \cap B) = P(A)P(B)$$ The derivation/explanation of the formula for conditional probability is given in Grinstead and Snell's "Introduction to Probability" book as follows:

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For me, the connection between the intuitive definition of conditional probability and the formula for calculating the conditional probability is not obvious and the above explanation nicely fills this gap.

My question is whether there is such an explanation/derivation for the formula used to define independent events. Similar to conditional probability, the formula used to define independent events is not immediately obvious to me. I cannot see the connection between the intuitive definition of independent events (that is, A is independent of B if the occurrence of A does not change the probability of occurrence of B) and the formula $P(A \cap B) = P(A)P(B)$. Almost all of the textbooks give a derivation, if they give any, of this formula from the formula for conditional probability. However, what I look for is an explanation without using conditional probability, i.e., an explanation using more fundamental concepts such as sample spaces, distribution functions, and of course logic. I'm looking for this since I think the notion of independence is a fundamental notion in probability theory and does not need the notion of conditional probability.

PS: This question can be seen as a continuation of this and this question. Even though there are very good answers I'm still not satisfied, so that is why I asked this question.

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    $\begingroup$ "Independence" essentially is the product in the category of probability spaces. Simply put, there is a standard way to create new mathematical objects out of given ones called a "product" and when this method is applied to probability spaces, the associated probability is the one given by independence. That strikes me as being about as "fundamental" as it could possibly be--but whether you find it satisfying would seem to be a matter for psychology, not statistics or mathematics. $\endgroup$ – whuber Oct 5 '17 at 15:58
  • $\begingroup$ By not satisfied I mean the independence formula still does not fully make sense for me. And I don't want to continue to my learning journey in probability theory by memorizing a formula without understanding it. May be I need to solve more exercises on independence and eventually it will settle down. $\endgroup$ – Sanyo Mn Oct 5 '17 at 21:13

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