The number of ways to select four marbles, one of which is yellow, would in this case be $${}_1C_1\cdot{}_4C_3=1\cdot 4=4,$$ so the probability of selecting the yellow marble is $$\frac{4}{{}_5C_4}=\frac45.$$
Alternately, we can proceed stepwise as follows: There's a $\frac45$ chance that the first marble isn't yellow. If the first marble isn't yellow, then there's a $\frac34$ chance that the second marble isn't yellow. If the first two marbles aren't yellow, then there's a $\frac23$ chance that the third marble isn't yellow. If the first three marbles aren't yellow, then there's a $\frac12$ chance that the fourth marble isn't yellow. Therefore, there's a $$\frac45\cdot\frac34\cdot\frac23\cdot\frac12=\frac15$$ chance that none of the four marbles drawn is yellow, so there's a $$1-\frac15=\frac45$$ chance that one of the four marbles is yellow.
As a third approach (which I'll discuss in more detail), since there's only one yellow marble, then to get the probability that the yellow marble was chosen, we need only add the probability of the following distinct events: (i) the yellow marble was chosen first, (ii) the yellow marble was chosen second, (iii) the yellow marble was chosen third, (iv) the yellow marble was chosen fourth. Hopefully, you see why these events have no overlap (mutually exclusive), and why together they comprise all the possible ways that the yellow marble could be chosen in this circumstance.
We already know that $$P(\text{yellow first})=\frac15.\tag{i}$$ If yellow is chosen second, then some other marble was chosen first--there are ${}_4C_1=4$ ways this can happen out of ${}_5C_1=5$ possibilities--leaving $1$ yellow marble out of a total of $4$ remaining, so $$P(\text{yellow second})=\frac45\cdot\frac14=\frac15.\tag{ii}$$ If yellow is chosen third, then two non-yellow marbles were chosen first--there are ${}_4C_2=6$ ways this can happen out of ${}_5C_2=10$ possibilities--leaving $1$ yellow marble out of a total of $3$ remaining, so $$P(\text{yellow third})=\frac{6}{10}\cdot\frac13=\frac15.\tag{iii}$$ If the yellow marble is chosen fourth, then three non-yellow marbles were chosen first--there are ${}_4C_3=4$ ways this can happen out of ${}_5C_3=10$ possibilities--leaving $1$ yellow marble out of a total of $2$ remaining, so $$P(\text{yellow fourth})=\frac{4}{10}\cdot\frac12=\frac15.\tag{iv}$$ Thus, $$\begin{align}P(\text{yellow chosen}) &= P(\text{yellow first})+P(\text{yellow second})+P(\text{yellow third})+P(\text{yellow fourth})\\ &= \frac15+\frac15+\frac15+\frac15\\ &= \frac45.\end{align}$$
You are decreasing the denominators of your fractions, but you shouldn't do that because you are putting the marbles back after each pick, so there are always 21 marbles in the jar.
Best Answer
Presumably we are sampling without replacement.
A: There are $13$ balls, so $\dbinom{13}{3}$ ways to choose $3$ balls. All these ways are equally likely. There are $\dbinom{4}{3}$ ways to choose $3$ green. Divide.
B: Same denominator. There are $\dbinom{6}{2}$ ways to choose $2$ blue. For each of these ways, there are $\dbinom{3}{1}$ ways to choose $1$ red, for a total of $\dbinom{6}{2}\dbinom{3}{1}$. Now divide.
C: The probability that the first is blue is $\dfrac{6}{13}$. Given this has happened, the probability the next is green is $\dfrac{4}{12}$. Given these two things have happened, the probability the last is red is $\dfrac{3}{11}$. Multiply.