Why $P(A \cap B \cap C) = P(A) \times P(B | A) \times P(C | A \cap B)$?

Question

I need help to intuitively understand that $P(A \cap B \cap C) = P(A) \times P(B | A) \times P(C | A \cap B)$? Why it's true?

Answer 1

Intuitive explanation / example

Here's an example that might help with the intuition: If I draw three cards without replacement from a standard deck, what is the probability that all three cards are hearts?

Let $A, B, C$ be the events that the first, second, and third cards are hearts, respectively.

If I want all three cards to be hearts, then I need:

• my first card to be hearts; $\mathbb P(A) = 13/52$
• my second card to be hearts, given that my first was: $\mathbb P(B \mid A) = \frac{12}{51}$
• my third card to be hearts, given that my first two were: $\mathbb P(C \mid A \cap B) = \frac{11}{50}$

The probability of all three happening is the product of all three probabilities, or $\frac{12 \cdot 11 \cdot 10}{52 \cdot 51 \cdot 50}.$ The intuition is that you're "chaining" each probability computation on previous ones, and as you do so you predicate each computation on the previous ones having occurred.

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Algebraic explanation

By definition, $$\mathbb P(D \cap E) = \mathbb P(D) \cdot \mathbb P(E \mid D)$$ so applying this definition with events $D = C$ and $E = A \cap B$ gives $$\mathbb P(C \cap B \cap A) = \mathbb P(A \cap B) \cdot \mathbb P(C \mid A \cap B)$$ and applying the definition on the inner term with $D = A$ and $E = B$ gives the desired result.