Why does $P(A \cap B \mid C) = P(A \mid B\cap C) \cdot P(B \mid C)$?

Question

I'm guessing this is some kind of application of the total law of probability and Bayes' theorem, I'm just not exactly sure on the steps of how to show it.

I'm mainly confused by the term $P(A \mid B\cap C)$. More specifically, $B\cap C$. Why is it conditioned on $B$ AND $C$?

Answer 1

$$P(A \cap B \mid C) = \frac{P(A\cap B \cap C)}{P(C)} = \frac{P(A\cap B \cap C)}{P(B\cap C)} \frac{P(B \cap C)}{P(C)} = P(A \mid B\cap C) P(B \mid C)$$

Answer 2

@bruderjakob17 explains why this is true algebraically.

In words, the formula $$P(A \cap B \mid C) = P(A \mid B \cap C) \cdot P(B \mid C)$$ means, that the probability of $A$ and $B$ given $C$ equals:

• the probability of $B$ given $C$ times
• the probability of $A$ given $B$ and $C$

This is best illustrated with an example. For instance, the probability that your 5 year old car won't start and it's because the alternator is rusty equals

1. the probability that your alternator is rusty ($B$) given that the car is 5 years old ($C$), times
2. the probability that the car won't start ($A$) given that the alternator is rusty ($B$) and that the car is 5 years old ($C$).

In statement 2, you need to condition on both $B$ and $C$ because it's possible that a younger car might still start with a rusty alternator. And in general, you need to condition on all you know.

More generally, yes, this is using the chain rule to decompose the joint probability into a product of conditional probabilities (Bayes rule is slightly different and is used to "invert" conditional probabilities).