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.

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.
