How should I understand conditional probability?

Question

I think now everyone know the formula for calculating conditional probability $$P(B|A) = \frac{P(A \bigcap B)}{P(A)}$$ But I'm having a hard time understanding it via a problem. Let say we have 3 marble, 1 red, 1 blue, 1 yellow. Now intuitively I know that the probability of taking a red one randomly after taking a blue one (also randomly) is 50% * 33% = 16.5% because the probability of taking a blue one is about 33% at first. After that the probability of taking a red one is 50% because there is only 2 marbles left to choose from. But if I try to use the formula, things become hard to understand, especially because the $P(A \bigcap B)$ part. What is $A \bigcap B$ in this case ? By definition, intersection of 2 sets will result in a new set being constructed, contains what A and B have "in common". So is it {O} in this case ? As far as I know, there is nothing "in common" between the probability of taking a blue one and a red one. Please kindly explain that part to me.

Answer 1

You seem to be confused by the meaning of $P\left(A \bigcap B\right)$ and $P\left(B|A\right)$.

$P\left(A \bigcap B\right)$ means the probability of both $A$ and $B$ happening, whereas $P\left(B|A\right)$ means the probability of $B$ happening given that $A$ has already happened.

In your case $P\left(A \bigcap B\right)$ would mean the probability of getting a blue ball first and then a red ball, whereas $P\left(B|A\right)$ would mean the probability of getting a red ball after you have already got a blue ball.

$P\left(B|A\right) = \frac{P\left(A \bigcap B\right)}{P\left(A\right)}$ is very logical if you write it as $P\left(A \bigcap B\right)=P\left(A\right)\cdot P\left(B|A\right)$. Which means, the probability of both $A$ and $B$ happening is the multiplication of $A$ happening and probability of $B$ happening given that $A$ has already happened.

In your case,

$$ \begin{align} P\left(A\right)&=\frac{1}{3}\\ P\left(B|A\right)&=\frac{1}{2}\\ P\left(A \bigcap B\right)&=\frac{1}{3}\cdot\frac{1}{2}=\frac{1}{6} \end{align} $$

Answer 2

What is A⋂B in this case?

I you have 3 marbles, R,B and Y and you pick two you can have the following sample space

$$\{RB,RY,BR,BY,YR,YB\}$$

thus in your case, $A\cap B$ is the event $BR$ with probability $\frac{1}{3}\cdot\frac{1}{2}$

as you can argue, any events among the 6 possible are equiprobable, with probability $\frac{1}{6}$