Solving the problem of conditional probability without using P($A | B$). I was trying to differentiate $A|B$ and $A \cap B$. I think understanding the following problem will clear my doubt.
There are two urns namely $U_1$ and $U_2$. $U_1$ contains three white balls and  five red balls while $U_2$ contains four white balls and five red balls. One Urn is chosen randomly and a white ball is drawn. What is the probability of drawing the white ball from the $U_1$?
If I name the event of choosing the urn is $A$ and drawing a white ball is $B$ , Then what will be the event $A \cap B$ and how the sample space will look like.
I want to determine P($A \cap B$)  by determining  $ \frac{n(A\cap B)}{n(S)} $. $S$ denotes the sample space. I know how to solve it by using conditional probability. I can not see the sample space. Can anyone elaborately explain me what will be the sample space $S$.  I want to solve this problem by the way Bram28 has solved here
Can anyone please help me to understand?
 A: You call $A$ "the event of choosing an urn," but events need to be specific things. As you've stated it, $A=S$, meaning every event in your sample space involves picking an urn. That's just part of the problem setup.
Since you are looking for $P(A|B)$ and the problem is to find the "Probability of picking $U_1$ given that a white ball was drawn," you probably meant that $A$ is the event of picking $U_1$.
You say you want to find $P(A\cap B)$ (the probability of picking an urn and then picking a white ball) by finding $\frac{n(A\cap B)}{n(S)}$ (the number of ways you can draw a white ball divided by the size of the sample space). You appear to believe that these quantities are equal, but as a general rule,
$$P(A\cap B)\neq\frac{n(A\cap B)}{n(S)}.$$
You can only count on these being equal if each of the ways in which you could draw a white ball is equally likely to occur (for the urn problem, they're not). Let's change the numbers to make this obvious: if $U_1$ had one red, one white and $U_2$ had one red, 100 white, then $n(A\cap B)=1$ because there is 1 way to draw the white ball form $U_1$. The total number of possible outcomes is the number of balls, $n(S)=103$. So 
$$\frac{n(A\cap B)}{n(S)}=\frac1{103}.$$
But the probability of drawing the white ball from $U_1$ is $\frac12\times\frac12=\frac14$
You say that you know how to do the problem with conditional probability so you don't want any conditional probability solutions. I am quite certain that there is no correct solution method that does not involve conditional probability. You can get rid of conditional probabilities quickly with Bayes' rule, of course:
$$P(A|B)=\frac{P(A\cap B)}{P(B)}.$$
You might want to post your solution as a comment and ask for help checking its accuracy.
edit:
joriki is correct that $(A|B)$ is not an event. In fact, without talking about probability, $(A|B)$ doesn't really have a meaning. $P(A|B)$ is the probability of the event $A$, but measured in a different way. By putting "$|B$" in there, what you mean is that your experiment has fundamentally changed. $P(A|B)$ means you want the chances of $A$ (picking $U_1$) in an experiment where, if event $B$ does not occur (a red ball is drawn), you ignore the result and start over.
$A\cap B$ is an event: the event that both event $A$ and event $B$ occur.
edit again:
You can't solve this problem in the same way as that example. In Bram28's example, there is a die being thrown, and there are 6 events that each have equal chances of occurring. These are $S=\{1,2,3,4,5,6\}$. Bram28 constructs two more events of interest: $A=\{4,5,6\}$ and $B=\{2,4,6\}$. He then calculates
$$
P(A\cap B)
=\frac{n(A\cap B)}{n(S)}
=\frac{n(\{4,6\})}{n(\{1,2,3,4,5,6\})}
=\frac26
$$
This shortcut is only correct because each event has the same probability of occurring:
$$
P(A\cap B)
=
P(\{4,6\})
=
P(\{4\})+P(\{6\})
=
\frac16+\frac16
=
\frac26.
$$
In my 103 ball example, the events in the sample space do not have equal chances. There is more than one way to write the sample space. I choose to write members of $S$ as pairs with the number of the chosen urn first and a letter for the color of the drawn ball second. I'm going to pretend that every white ball in $U_2$ has a number which will be a subscript, like this:
$$
S=\{(1,W),(1,R),(2,W_1),(2,W_2),\dots,(2,W_{100}),(2,R)\}.
$$
I said that Bram28's calculation method won't work for this example because
$$
\frac{n(A\cap B)}{n(S)}=\frac{n(\{(1,W)\})}{103}=\frac1{103}.
$$
A: Based on the events of interest you can consider the following sample space to describe your experiment:
$ S = \{ E_1, E_2, E_3, E_4\}$
where, 
$A$: Pick Urn-1
$A^c$ :Pick Urn-2
$B$ : Pick a white ball 
$B^c$: Pick Black ball
$E_1 = A\cap B, E_2 = A^c \cap B, E_3 =  A \cap B^c, E_4 =  A^c \cap B^c$
($A\cap B$ means "Pick a white ball from urn-1")
In particular, what you are trying to find is $P(E_1)$.
However, note that the elements of S have unequal weights (i.e, they are not equally likely). So, you can not use the naive definition of probability to find $P(E_1)$ (i.e., something like $\frac{n(A\cap B)}{n(S)}$ is not valid).
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