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, $$\begin{align}P(\text{yellow chosen}) &= P(\text{yellow first})+P(\text{yellow second})+P(\text{yellow third})+P(\text{yellow fourth})\\ &= \frac15+\frac15+\frac15+\frac15\\ &= \frac45.\end{align}$$
Without replacement means one-by one. The colour of the first influences the probability of the colour of the second. So consider it as two discrete events, picking of the first ball, and then picking the second ball
$$P(2R) = \frac{3}{5}.\frac{2}{4} = \frac{3}{10}$$
$$P(1R,1W) = \frac{3}{5}.\frac{2}{4} + \frac{2}{5}.\frac{3}{4} = \frac{6}{10}$$
Best Answer
Your approach is correct, since you are covering all possible events that satisfy the condition, and you are not over or undercounting.
An alternate perspective, which can confirm your answer, is that we have $ 6 \times 5=30 $ possible ways of conducting this draw. Of this, to satisfy our condition, we need the first ball to be white and the second one to be green, or vice versa. There are $ 2 \times 1$ ways to do both of the above, so 4 ways total.
We just enumerated all the possible cases, and all the possible cases that satisfy your probability, and we obtain 4/30. Note that I am summing 2+2 and then dividing by 30 in the same way that your method considers two equal probabilities. It is pure coincidence that they are equal.