Applying Bayes' rule to conditional probabilities with more than 2 variables

Question

Bayes' rule gives us the following equation:

$$P(X|Y) = \frac{P(X,Y)}{P(Y)} = \frac{1}{Z}P(X)P(Y|X)$$

where Z is a normalizer.

Now I saw that the following two equations holds according to Bayes' rule:

$$P(A|B,C,D) = \frac{1}{Z} P(B,C|A,D)P(A,D)$$

$$P(A|B,C) = \frac{P(B|A,C)P(A|C)}{P(B|C)}$$

Why is this true according to Bayes' rule? In opposite to Bayes' rule we condition on 2 (B,C,D) and 3 (B,C) variables and it is not clear to me how in this case Bayes' rule works.

Answer 1

The two equations can obtained by applying several times Bayes' rule and grouping the event : \begin{align*} \mathbb P[A|B,C,D] &= \frac{\mathbb P[A,(B,C,D)]}{\mathbb P[B,C,D]}\\ &=\frac{\mathbb P[A,B,C,D]}{\mathbb P[B,C,D]}\\ &=\frac{\mathbb P[(B,C),(A,D)]}{\mathbb P[B,C,D]}\\ &=\frac{\mathbb P[B,C|A,D]\mathbb P[A,D]}{\mathbb P[B,C,D]} \end{align*} and \begin{align*} \mathbb P[A|B,C] &= \frac{\mathbb P[A,(B,C)]}{\mathbb P[B,C]}\\ &=\frac{\mathbb P[B,(A,C)]}{\mathbb P[B,C]}\\ &=\frac{\mathbb P[B|A,C]\mathbb P[A,C]}{\mathbb P[B|C]\mathbb P[C]}\\ &=\frac{\mathbb P[B|A,C]\mathbb P[A|C]\mathbb P[C]}{\mathbb P[B|C]\mathbb P[C]}\\ &=\frac{\mathbb P[B|A,C]\mathbb P[A|C]}{\mathbb P[B|C]}\\ \end{align*}