super naively, I would say that the problem is just asking for the conditional probability of second marble being blue, given that the first is red, this is P(B = blue given A = red).

When we have extracted the first red marble, the number of possible desired outcomes is 4 and the total possible outcomes is 6, so we arrive to the same solution.

Does anybody knows if this result is the same just 'by chance'?

I arrived at answer in a simple way:

Say initial state of the bag is: R1 R2 R3 B1 B2 B3 B4

Imagine first marble removed is Red(R1).

The remaining is: R2 R3 B1 B2 B3 B4 = 6 marbles

Now I have to pick one more marble.

Total number outcomes = Number of permutations when I pick 1 marble from a set of 6 = P(6,1) = 6

We just want the probability of a blue marble coming in that experiment.

There are 4 possible favorable outcomes: B1,B2,B3,B4

So The probability of that outcome is = 4/6

P(R) = 3/7

P(B) = 4/7

P(R|B) = 3/6 = 1/2

P(R|Bc) = 2/6 = 1/3

P(Bc) = 1-4/7 = 3/7

P(B|R) = P(R|B) * P(B) / P(R|B) * P(B) + P(R|Bc) * P(Bc)

P(B|R) = { (1/2) * (4/7) } / { (1/2) * (4/7) + (1/3) * (3/7) }

P(B|R) = { (4/14) } / { (4/14) + (1/7) }

P(B|R) = { (4/14) } / { (4/14) + (2/14) }

P(B|R) = (4/14) / (6/14)

P(B|R) = (4/6)

P(B|R) = (2/3)

2 / 3