Posted by Diego Assencio on 2013.10.12 under Mathematics (Statistics and probability)

Let $A$ and $B$ be two events and let $P(A|B)$ be the conditional
probability of $A$ given that $B$ has occurred. Then
Bayes' theorem
states that:
$$
\boxed{
P(B|A) = \displaystyle\frac{P(A|B)P(B)}{P(A)} =
\displaystyle\frac{P(A|B)P(B)}{P(A|B)P(B)+ P(A|B^c)P(B^c)}
}
$$

In other words, Bayes' theorem gives us the conditional probability of
$B$ given that $A$ has occurred as long as we know $P(A|B)$, $P(A|B^c)$ and
$P(B) = 1 - P(B^c)$.

The proof of this equation is quite simple. First, consider the
following facts:
$$P(A|B) := \displaystyle\frac{P(A \cap B)}{P(B)} \Longrightarrow
P(A \cap B) = P(A|B)P(B)
$$
$$
\begin{eqnarray}
P(A) & = & P((A \cap B) \cup (A \cap B^c)) \nonumber\\[5pt]
& = & P(A \cap B) + P(A \cap B^c) \nonumber\\[5pt]
& = & P(A|B)P(B) + P(A|B^c)P(B^c) \label{post_bcb216aa523d43833789bc008e394275_second_fact}
\end{eqnarray}
$$
where on equation \eqref{post_bcb216aa523d43833789bc008e394275_second_fact} the fact that $A \cap B$
and $A \cap B^c$ are mutually exclusive events was used. Therefore, since
$A\cap B = B\cap A$:
$$
\begin{eqnarray}
P(B|A) &=& \displaystyle\frac{P(B \cap A)}{P(A)} \nonumber\\[5pt]
&=& \displaystyle\frac{P(A|B)P(B)}{P(A)} \nonumber\\[5pt]
&=& \displaystyle\frac{P(A|B)P(B)}{P(A|B)P(B)+ P(A|B^c)P(B^c)}
\label{post_bcb216aa523d43833789bc008e394275_proof}
\end{eqnarray}
$$
as we wanted to prove.
We can also prove a very interesting formula using Bayes' theorem. From
equation \eqref{post_bcb216aa523d43833789bc008e394275_proof} (used with both $B$ and $B^c$) and
equation \eqref{post_bcb216aa523d43833789bc008e394275_second_fact}, we have:
$$
P(B|A) + P(B^c|A) =
\displaystyle\frac{P(A|B)P(B)}{P(A)}
+ \displaystyle\frac{P(A|B^c)P(B^c)}{P(A)}
= \displaystyle\frac{P(A)}{P(A)} = 1
$$
so we obtain:
$$
\boxed{ P(B|A) = 1 - P(B^c|A) }
$$

Comments

MANISH NAIDU on Feb 07, 2017:

thanks a lot for your help...

Chiranjeevi Kanaka on Apr 13, 2018:

given $A \cap B$ and $A \cap B'$ are independent events, but they are actually mutually exclusive events.

That's all perfect, but easier to remember the following.

The definition of conditional probability is
(1) P(B|A)=P(B^A)/P(A)

Multiply both sides of Bayes formula by P(A)
and you get
(2) P(B|A)*P(A)=P(A|B)*P(B)

Use (1) in both sides of (2) and get
(3) P(B^A)/P(A)*P(A)=P(A^B)/P(B)*P(B)

You get an evidence
P(B^A)=P(A^B)

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## Comments

The definition of conditional probability is

(1) P(B|A)=P(B^A)/P(A)

Multiply both sides of Bayes formula by P(A)

and you get

(2) P(B|A)*P(A)=P(A|B)*P(B)

Use (1) in both sides of (2) and get

(3) P(B^A)/P(A)*P(A)=P(A^B)/P(B)*P(B)

You get an evidence

P(B^A)=P(A^B)