Conditional probability derivation with example the formula for conditional probability is 

$P(A|B) = \dfrac{P(A ∩ B)}{P(B)}$

I am giving a simple problem to explain my doubt. (this question is made by me to explain my doubt and may contain errors. if so, please correct me)

2 blue and 3 red balls are in a bag. We take 2 balls without
  replacement.
Suppose A is the event that first ball is blue and B is the event that second ball is red
P(B|A) = P(drawing a red ball when a blue ball is drawn) which is
  $\dfrac{3}{4}$

The above answer is derived using basic conditional probability concepts. But, according to the formula of conditional probability given at the beginning, how do we solve it?
Note:  The derivation given here for derivation of the formula is too difficult for me to understand. My text book does not give a proof for this. So, please explain in simple terms, if possible, using the same example

The events should be defined accurately. $2/4$ is the probability of "drawing a blue ball in the second attempt, given that in the first attempt you have drawn a red ball". So, order is important in this definition of the events.
Events can be defined as below:
A= drawing a blue in the second attempt
B= drawing a red in the first attempt
Using the concept of conditional probability, $P(A|B)=2/4$
Using the formula, $\frac{P(A∩B)}{P(B)}=\frac{3*2/5*4}{3/5}=\frac{2}{4}$
The other way of defining would be:
A= having at least one blue after two attempts
B= having at least one red after two attempts
Using the concept of conditional probability, for this case, is a bit hard.
Using the formula, $\frac{P(A∩B)}{P(B)}=\frac{(3*2*2)/(5*4)}{((3*4*2)-6)/5*4}=\frac{2}{3}$
