Is this a geometric distribution problem? Suppose a student starts with test A, and will proceed to test B, then test C if he passes.
The probability for the student to pass test A is 30%.
The probability for the student to pass test B is 20%.
The probability for the student to pass test C is 10%.
The student has maximum 20 tries to attempt for the tests in TOTAL.
How do I calculate the probability of the student to pass test C, starting from test A, within 20 tries? My hunch is using geometric distribution but I am rather unsure since this problem involves multiple different "stages" with each stage having different probabilities.
Clarification: If the student passes test A, he does not have to take A again, and can use the remaining attempts (20 minus attempts used to pass A) for B and C. Same goes for B.
 A: This can be solved using Markov chains. First define your 4x4 transition matrix (A, B, C, F - final state), probabilities of passing from one state to the next. I am assuming there is only one direction of progression, forward, i.e. there is no rollback to previous tests upon failure. If this is not true then the matrix below can be slightly changed to account for this.
    A   B   C   F
A 0.7 0.3 0.0 0.0
B 0.0 0.8 0.2 0.0
C 0.0 0.0 0.9 0.1
F 0.0 0.0 0.0 1.0

Then you need to raise this matrix to the power of 20 (19), which gives you the probabilities of being in a certain state depending on where you started after 20 turns. The result is
             A          B         C         F
A 0.0007979227 0.03219388 0.2979484 0.6690598
B 0.0000000000 0.01152922 0.2200949 0.7683759
C 0.0000000000 0.00000000 0.1215767 0.8784233
F 0.0000000000 0.00000000 0.0000000 1.0000000

Here you would look at the first row, since you started in state A. The probability of being in state F (the final state) starting in state A after 20 turns is roughly equal to 67 %.
A: No.
Since the probabilities are different at each time point, this data will not follow a geometric distribution.
I also don't quite understand when you say "The student has maximum 20 tries to attempt for the tests in TOTAL." bit, but that's not necessary to answer your question.
A: If we call "an attempt" a single try at tests A, and then B and then C in turn if the preceding test is passed; an attempt fails if any of A, B or C fail.
The probability of passing all three is $0.3 \times 0.2 \times 0.1 = 0.006$. In short P(attempt succeeds) = 0.006.
In that case, the probability of succeeding at least once in 20 attempts could be done from first principles:
P(succeed at least once) = 1-P(don't succeed even once)
= $1 - 0.994^{20} = 0.1134 = 11.34\%$
Or it would typically be calculated using probabilities from a binomial distribution.
The geometric is used when calculating the distribution of the number of trials to the first success; however, you're correct to suppose that it can be used to solve this problem; if the first success occurs after trial $k$ there was no success by trial $k$. So you're just doing a calculation in the other tail. Since tail sums of geometric p.f.s are simple, this is quite doable -- but a little more effort than a more direct calculation in this instance.
