Let A and B be events such that P(A)>0. We denote the probability of B given that A has occurred by
$$P(B|A)\text{ or } P_{A}(B)$$
and \(P(B|A)\) is called the conditional probability of B given A.
Since A is known to have occurred in the conditional probability of B given A, A becomes the new sample space replacing the original sample space S. Then we have
$$P(B|A)=\frac{|A\cap B|}{|A|}=\frac{P(A\cap B)}{P(A)}.$$
Example:
Suppose that a fair dice is to be tossed. Find the probability that a single toss of the die will result in a number greater than 4 when it is noted that the number is even.
Solution:
Let A and B be the events the number is “even” and “greater than 4” respectively. We are looking for \(P_{A}(B)\) or \(P(B|A)\). This is given by
$$P_{A}(B)=\frac{P(A\cap B)}{P(A)}=\frac{2/6}{3/6}=\frac{2}{3}\ \ \ \text{or }\ \ \ P_{A}(B)=\frac{|A\cap B|}{|A|}=\frac{2}{3}$$
Multiplication Formula
Let A and B are events such that \(P(A)\not=0\) and \(P(B)\not=0\). Then
$$P(A\cap{B})=P(A|B)P(B)=P(B|A)P(A)$$
By the definition of the conditional probability, we have
\(P(A|B)\)：the probability of A given that B has occurred \(P(B|A)\)：the probability of B given that A has occurredthen
$$P(A|B)=\frac{P(A\cap{B})}{P(B)},\ P(B|A)=\frac{P(A\cap{B})}{P(A)}$$
holds, and transforming the above equations, we obtain
$$P(A\cap{B})=P(A|B)P(B)=P(B|A)P(A).$$
Independent Events
We say that A and B are independent, if the probability of B occurring is not affected by the occurrence or non-occurrence of A . That is.
$$P(B|A)=P(B)\ \ \ \ \ P(B|A^{c})=P(B)$$
and these are equivalent to
$$P(A\cap B)=P(A)P(B|A)=P(B)P(A)$$
Bayes’ Theorem
Let A and B are events such that \(P(A)\not=0,\ P(B)\not=0\). Then
$$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$$
holds, and this is called Bays’ theorem.
$$P(A\cap{B})=P(A|B)P(B)=P(B|A)P(A)\ \cdots\text{ (1)}$$
holds, and and we obtain Bays’ theorem by transforming (1).
And also, the theorem can be written as
$$P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^{c})P(A^{c})}\ \cdots\text{(2)}$$
We obtain (2) by transforming
$$\begin{eqnarray*}P(B)&=&P(A\cap{B})+P(A^{c}\cap{B})\\&=&P(B|A)P(A)+P(B|A^{c})P(A^{c})\end{eqnarray*}$$
This theorem enable us to find the probability of the event A that can cause B. This is why Bays’ theorem is often called a theorem on the probability of outcomes.