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I am studying compound and conditional probability.I am facing some confusion regarding the following formulas . The first formula is about compound probability.

If $E_1,E_2,\ldots,E_n $are mutually exclusive and exhaustive events and $ E $ be any event

$$P(E)=\sum P(E_i)\cdot P(E \mid E_i)$$ If $P(E_i)>0$ and summation is from $i=1$ to $i=n$.

And the second formula is about conditional probability. I know that probability of occurrence of a event $A$ given that $B$ has already occurred is represented by $P(A\mid B)$.but I can't understand why the above expression should be equal to

$$P(A\mid B)=\frac{P(A \cap B )}{P(B)}$$ Please provide me some insight as to why the above formulas holds good.there must be some justification behind the formula. thanks.

Navin
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    The notation $P(\frac{A}{B})$ is not standard for conditional probability -- it looks like you mean $P(A|B)$. – Batman Oct 10 '16 at 03:44
  • What expression is supposed to be equal to what? I don't think $P(E)$ is generally equal to $P(A\mid B)$; there's really no relationship between them without a lot more information given. Possibly you're referring obliquely to some explanation of one of these formulas somewhere, but it's hard to guess what the original explanation was. The conditional probability formula does imply that $P(E_i)P(E\mid E_i)=P(E\cap E_i)$ (replace $A$ and $B$ by $E$ and $E_i$ and multiply both sides by $P(E_i)$) so maybe that's the relevance. – David K Oct 10 '16 at 05:02
  • @David K I seemed to mean why the conditional probability formula imply$\frac{ P(E \cap E_I)}{P(E_I)}=P(E \mid E_I)$ – Navin Oct 10 '16 at 05:31
  • Often that is the formula by which the notation $P(A\mid B)$ is first defined in a theory of probability, so before that formula is given we don't even really have a meaning for $P(A\mid B)$. Do you mean to ask why we say that $P(A\mid B)$, defined equal to $P(A\cap B)/P(B)$, is called "the probability of $A$ given $B$"? – David K Oct 10 '16 at 12:40

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Rewriting it as follows: $$P(A\cap B)=P(B)\cdot P(A\mid B)\tag1$$ makes it more intuitive and you avoid division with zero, so that $(1)$ holds for any $A,B$ even if $P(B)=0$. Now read

The probability of $A$ and $B$ occuring both and simultaneously (LHS), is equal to the probability of $B$ occuring first and then of $A$ occuring given that $B$ has occured (RHS).

Of course the following is also true $$P(A\cap B)=P(A)\cdot P(B\mid A)$$ In other words:

"Two events $A$ and $B$ may both occur, if one occurs and then the other occurs (given that the first has occured). "

Jimmy R.
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