# Conditional probability and intersections of events

Conditional probability is : $$P(A\mid B)= \frac{P\left(A \bigcap B\right)}{P(B)}$$ Why did I find this as a solution of an exercise : $$P\left(\left(E_1 \bigcap E_2 \bigcap E_3\right)\mid H\right) = P\left(E_1\mid H\right) \cdot P\left(E_2\mid \left(E_1 \bigcap H\right)\right) \cdot P\left(E_3\mid \left(E_1 \bigcap E_2 \bigcap H\right)\right)$$ I have two questions :

1. where's the P(B) of the definition in the denominator?
2. how do I go ahead when I have an intersection of more than 1 event given an event in the conditional probability?
• That formula should make intuitive sense if you think through what it means. David already gave the formal derivation. – BallBoy Jan 12 '18 at 0:19

\eqalign{{\rm RHS} &=\frac{P(E_1\cap H)}{P(H)}\,\frac{P(E_2\cap E_1\cap H)}{P(E_1\cap H)}\, \frac{P(E_3\cap E_1\cap E_2\cap H)}{P(E_1\cap E_2\cap H)}\cr &=\frac{P(E_1\cap E_2\cap E_3\cap H)}{P(H)}\cr &={\rm LHS}\ .\cr}