I apologize for my previous answer...I didn't read the problem carefully.
Let's focus on the number of ways we can pick 3 skittles such that they are all the different. Let's first look at the number of ways to pick a red, orange, and yellow skittle. There are $12$ red skittles, $16$ orange skittles, and $10$ yellow skittles so there are $12 \cdot 16 \cdot 10$ different ways of picking a red, orange, and yellow skittle.
To get the total number of ways to pick 3 skittles of different colors, just consider all the ways to pick $3$ colors out of $4$ and find the number of ways for each color combination. Doing this, we get a total of
$$12 \cdot 16 \cdot 10 + 12\cdot 16 \cdot 18 + 16 \cdot 10 \cdot 18+12\cdot 10 \cdot 18$$
ways to pick $3$ skittles all of different colors.
Since there are ${12+16+10+18 \choose 3}$ ways to pick $3$ skittles in general, the probability is,
$$\frac{12 \cdot 16 \cdot 10 + 12\cdot 16 \cdot 18 + 16 \cdot 10 \cdot 18+12\cdot 10 \cdot 18}{{12+16+10+18 \choose 3}}$$
We'll take Graham's advice and think about the complements - that's a slightly easier calculation.
We want to know the probability that one of the two blocks is one of red, orange, or yellow. The complement of this event is that neither of the first two blocks is one of these three colors.
There are four blocks in the bag that are not one of the desired colors, and seven total blocks. So, the probability that the first block is not red, orange, or yellow is $\frac{4}{7}$.
If the first block is not one of our colors, then the block that was removed first was one of the four we did not want: green, blue, indigo, or violet. So, three of these remain in the bag. We removed one of the seven, so six total remain in the bag. This means that the probability that the second block is also not red, orange or yellow (given that the first one was not) is $\frac{3}{6}$.
Then, the probability that the first block was not one of the three colors and the second block was not is
$$ \frac{4}{7}*\frac{3}{6} = \frac{2}{7} . $$
We want the probability of the complement, so the probability we're looking for is
$$ 1 - \frac{2}{7} = \frac{5}{7} . $$
Best Answer
Since Paul is presumably interested in eating the candies he selects, it is reasonable to assume that the candies are chosen without replacement. There is also an implicit assumption that the candies are selected without looking in the bag, which ensures that the selection is random.
You made the assumption that the selections are made successively, which is not necessary. We consider two approaches:
The candies are selected randomly in succession without replacement:
There are $6 + 5 + 12 + 9 + 13 = 45$ candies in the bag, of which $9$ are yellow and $45 - 9 = 36$ are not yellow. Thus, on the first draw, the probability that the first candy is not yellow is $36/45$. If the first candy is not yellow, that leaves $44$ candies in the bag, of which $35$ are not yellow. Thus, the probability that the second candy is also not yellow is $35/44$. Hence, the probability that both candies are not yellow is $$\frac{36}{45} \cdot \frac{35}{44} = \frac{4}{5} \cdot \frac{35}{44} = \frac{7}{11}$$
Two candies are selected randomly at once:
There are $\binom{45}{2}$ ways to select a subset of two of the $45$ candies and $\binom{36}{2}$ ways to select a subset of two of the $36$ candies that are not yellow. Hence, the probability that the two candies selected are not yellow is $$\frac{\dbinom{36}{2}}{\dbinom{45}{2}} = \frac{\dfrac{36!}{2!34!}}{\dfrac{45!}{2!43!}} = \frac{\dfrac{36 \cdot 35 \cdot 34!}{2 \cdot 1 \cdot 34!}}{\dfrac{45 \cdot 44 \cdot 43!}{2 \cdot 1 \cdot 43!}} = \frac{\dfrac{36 \cdot 35}{2}}{\dfrac{45 \cdot 44}{2}} = \frac{36 \cdot 35}{2} \cdot \frac{2}{45 \cdot 44} = \frac{36 \cdot 35}{45 \cdot 44} = \frac{7}{11}$$