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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?
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  • $\begingroup$ That formula should make intuitive sense if you think through what it means. David already gave the formal derivation. $\endgroup$ – BallBoy Jan 12 '18 at 0:19
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$$\eqalign{{\rm RHS} &=\frac{P(E_1\cap H)}{P(H)}\,\frac{P(E_2\cap E_1\cap H)}{P(E_1\cap H)}\, \frac{P(E_3\cap E_1\cap E_2\cap H)}{P(E_1\cap E_2\cap H)}\cr &=\frac{P(E_1\cap E_2\cap E_3\cap H)}{P(H)}\cr &={\rm LHS}\ .\cr}$$

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  • $\begingroup$ What do you mean with "RHS" and "LHS"? $\endgroup$ – FBIT Jan 11 '18 at 23:18
  • $\begingroup$ Right/left hand side. Of the equation. $\endgroup$ – David Jan 11 '18 at 23:19
  • $\begingroup$ If I have an intersection of two or more events P(E1∩H)/P(H) in the numerator or in the denominator can I solve it with a moltiplication EX: P(E1) x P(E2) x P(H) or should I go ahead with inverse formula of conditional probability ? $\endgroup$ – FBIT Jan 11 '18 at 23:24
  • $\begingroup$ You can multiply if the events are independent. $\endgroup$ – David Jan 11 '18 at 23:58
  • $\begingroup$ Is there a way to know if two events are independent without using conditional probability formula? $\endgroup$ – FBIT Jan 12 '18 at 14:16

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