Conditional Probability theorem

Question

Conditional probability is denoted as P(A|B)=Probability of occurrence of A when B occurs OR Probability of Event A when B becomes a sample space.

Let us take an example:Let there be 5 white and 4 red balls.Two balls are drawn from the bag one after another without replacement. If we consider these events:

A= Drawing a white ball in first draw

B= Drawing a red ball in second draw.

Now,

P(B|A)= Probability of drawing a red ball from a bag containing 4 white and 4 red balls.

Here I fail to understand how event A can be considered the sample space for the given conditional probability.The sample space for Event A is just white balls while that of this probability is : total number of balls-one white ball from event B.Also what is the intuition behind considering Event A as a sample space if it is?

Answer 1

Adapting to your example let's give the balls numbers $1,2,\dots,9$ where the balls $1,2,3,4,5$ are white and the balls $6,7,8,9$ are red. Then by the experiment of drawing $2$ balls without replacement we can take $\Omega=\{\langle i,j\rangle\mid i,j\in\{1,2,\dots,9\}\wedge i\neq j\}$ to be the sample space.

The distinct outcomes are equiprobable.

For any event $E$ we have $P(E)=\frac{|E|}{|\Omega|}$.

Especially note that the denominator equals the cardinality of $|\Omega|$.

Event $A$ can be recognized as $\{\langle i,j\rangle\in\Omega\mid i\in\{1,2,3,4,5\}\}$ and event $B$ as $\{\langle i,j\rangle\in\Omega\mid j\in\{6,7,8,9\}\}$

Conditional probability $P(B\mid A)$ equals $\frac{P(A\cap B)}{P(B)}=\frac{|B\cap A|/|\Omega|}{|A|/|\Omega|}=\frac{|B\cap A|}{|A|}$.

Now note that here the denominator of the last expression equals the cardinality of $|A|$.

In some sense it has taken over the role of $\Omega$.

This because by computation of $P(B\mid A)$ we only look at outcomes that are in event $A$.

Answer 2

Interpreting the condition of a contional probability as the restricted sample space may not always be straightforward. I would say the condition is simply what has happened. In your problem, the bag contains $5$ white and $4$ red balls. Given that a white ball is drawn first (event $A$ has happened), the bag now contains $4$ white balls and $4$ red balls. So the probability to pick a red ball in the second draw (for event $B$ to happen) is $P(B|A)=1/2$.

Answer 3

I will simplify the above problem to 3 white and 2 red balls; let's define event A and B to be exactly the same:

A - Drawing white ball in the first draw

B - Drawing red ball in the second draw.

Now, let's write our overall event space:

{(RRWWW), (RWRWW), (RWWRW), (RWWWR), (WRRWW), (WRWRW), (WRWWR), (WWRRW), (WWRWR), (WWWRR)}

Now, let's condition on drawing a white ball on the first draw, i.e. only choose the events that have W as the first draw:

{WRRWW), (WRWRW), (WRWWR), (WWRRW), (WWRWR), (WWWRR)}.

Now, if you look at the probability that the second draw is red, you can see that three of these events have an R in the second draw out of 6.

Of course, while you are usually doing the math, you do not have to write it all out, but since you asked for the explanation, here it is. You are restricting your event space; i.e. you are only considering the cases where A has happened.