In conditional probability, we find the occurrence of an event given that another event has already occurred. So the formula of P(A|B) = P(intersection of A and B) over P(B).
My doubt is, if event B has already occurred , it would mean that our reduced sample space is the entire set of B. So shouldn't P(B) = 1 just like how we say P(S) = 1 where S is the sample space?
Best Answer
What you are thinking of is that it is certain that $B$ occurs when given that $B$ occurs.$$\mathsf P(B\mid B)=1$$
That says nothing about the unconditional probability that $B$ occurs (ie: when it is not given).
What is the probability that event $A$ occurs when given that event $B$ occurs? It is the weighted measure of outcomes in event $A$ that occur within the event of $B$. Thus, it is the measure of the intersection, $A\cap B$, divided by the measure of $B$.$$\mathsf P(A\mid B)~=~\dfrac{\mathsf P(A\cap B)}{\mathsf P(B)}$$
For example: The roll of standard six-sided dice consists of six equally probable events $\{1,2,3,4,5,6\}$.
Let $B$ be the event of rolling an even number on a standard six sided dice. This event consists of the three equally probable outcomes, $B=\{2,4,6\}$, so we claim the probability of this event is $1/2$.$$\mathsf P(B)=1/2$$
Let $A$ be the event of rolling a prime number. There is only one outcome among the three equally probable outcomes of event $B$ that is also an outcome of event $A$; ie $A\cap B=\{2\}$.
Hence the conditional probability for $A$ when given $B$ is $1/3$.$$\begin{align}\mathsf P(A\mid B)&=\dfrac{\mathsf P(A\cap B)}{\mathsf P(B)}\\[2ex]&=\dfrac{1/6}{1/2}\\[2ex]&=\dfrac 13\end{align}$$