A bag contains 3 red marbles and 4 blue marbles. Then, marbles are
drawn from the bag, at random, without replacement. If the first
marble drawn is red, what is the probability that the second marble is
blue?

Although this example is too simple and can be solved ad hoc, it is
still important to practice how to model problems such as these in the
correct way, so below is one way you might do it.

                                                     P(2nd blue and 1st red)
We need P(2nd blue|1st red):  P(2nd blue|1st red) =  -----------------------
                                                        P(1st red)
P(1st red) = 3/7


The denominator P(1st red) is straightforward: we have 7 balls in
total, 3 of them are red, so P(1st Red) = 3/7.

The numerator is less obvious, but not by much. We can rewrite it as
P(2nd blue and 1st Red) = (# of pairs where the first is red and the second is
blue) / (total # of possible pairs).

Total num of possible pairs is Perm(7,2) = 7! /(7 - 2)! = 7 * 6 = 42

3 options for the first ball (3 reds) and 4 options for the second
ball (4 blues), so 3 * 4 = 12.

P(2B and 1R) = 12 / 42 = 2/7.

P(2B|1R) = (2/7) / (3/7) = 2/3.

--------------------
FR = first red marble
FB = first blue marble
SR = second red marble
SB = second blue marble

outcomes in experiment:
total number of outcomes: 2P7 = 7! / (7-2)! = 7 * 6 = 42 # Permutation
number of outcomes in event (FR and SB): 3 * 4 = 12

probability:
P(FR) = 3/7
P(FR and SB) = 12 / 42 = 2/7 # Classical definition of probability

P(SB | FR) = P(SB and FR) / P(FR) # Conditional Probability
= P(FR and SB) / P(FR)
= 2/7 / 3/7
= 2/3