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,
$$\frac{12 \cdot 16 \cdot 10 + 12\cdot 16 \cdot 18 + 16 \cdot 10 \cdot 18+12\cdot 10 \cdot 18}{{12+16+10+18 \choose 3}}$$
With regard to how much money you can win by drawing the blue ball, assuming you earn $10$ dollars every time you draw the blue ball, the scenario without replacement has only two possible outcomes:
$0$ blue balls, probability $P(X=0) = 0.7,$ gain $0.$
$1$ blue ball, probability $P(X=1) = 0.3,$ gain $10.$
The expected value of your winnings is
$$ E(10X) = 0(0.7) + 10(0.3) = 0 + 3 = 3.$$
(Note that $\frac3{10}\times 10 = 0.3 \times 10 \neq 0.3.$ Be careful what you write!)
The scenario with replacement has these possible outcomes:
$0$ blue balls, probability $P(Y=0) = \left(\frac9{10}\right)^3 = 0.729,$ gain $0.$
$1$ blue ball, probability
$P(Y=1) = \binom31 \left(\frac1{10}\right) \left(\frac9{10}\right)^2 = 0.243,$
gain $10.$
$2$ blue balls, probability
$P(Y=2) = \binom32 \left(\frac1{10}\right)^2 \left(\frac9{10}\right) = 0.027,$
gain $20.$
$3$ blue balls, probability $P(Y=3) = \left(\frac1{10}\right)^3 = 0.001,$
gain $30.$
Note that if we just care about whether we draw blue ball at least once, we can find the probability whether that happens either by adding up all the cases where it does happen,
$$P(Y=1) + P(Y=2) + P(Y=3) = 0.243+0.027+0.001 = 0.271,$$
or by taking $1$ and subtracting the probability that it does not happen,
$$1 - P(Y=0) = 1 - 0.729 = 0.271.$$
But when you gain $10$ for each time you draw the blue ball, the expected value of your winnings is
\begin{align}
E(10Y) &= 0(P(0)) + 10(P(1)) + 20(P(2)) + 30(P(3)) \\
&= 0 + 10(0.243) + 20(0.027) + 30(0.001) \\
&= 0 + 2.43 + 0.54 + 0.03 \\
&= 3.
\end{align}
In fact the expected number of blue balls drawn is exactly the same with or without replacement,
and the expected payment (receiving $10$ dollars each time blue is drawn) also is exactly the same in each case.
Most people would interpret "the probability of getting blue" in each scenario as the probability of getting blue at least once.
This probability is less in the scenario without replacement
(or to put it another way, the probability of not even once drawing blue is greater),
but once you start counting the number of times blue is drawn (and paying $10$ dollars each time)
the chance of getting a double or triple payout makes up for the increased chance of getting no payout.
When you ask about a probability, you are asking about something that either happens or does not happen, two possible outcomes.
It is possible to set up an expected value that also deals with only two possible outcomes, but expected value very often deals with more than two possible outcomes (in your example with replacement, four possible outcomes), and in those cases it tends to give answers different than you would get by looking only at two possibilities.
Best Answer
The probability of getting at least $3$ blue skittles in a bowl is $$ 1-\sum_{k=0}^2\frac{\binom{30}{k}\binom{60}{9-k}}{\binom{90}{9}}\approx.632509$$ There are $\binom{30}{k}$ ways to choose $k$ blue skittles, $\binom{60}{9-k}$ ways to choose $9-k$ non-blue skittles, and $\binom{90}{9}$ ways to choose $9$ skittles, so the fraction gives the probability of getting exactly $k$ blue skittles. The probability of getting at least $3$ blue skittles is the same as the probability of not getting $0,\ 1,$ or $2$.
The number of ways to put $3$ blue skittles in each bowl is $$\frac{30!}{(3!)^{10}}\approx$$ because we can line the skittles up in $30!$ ways, put the first $3$ in the first bowl, the next $3$ in the second bowl, and so on. However, the order in which the skittles go into the bowl doesn't matter, so we have to divide by $3!$ in each case. Similarly, there are $$\frac{60!}{(6!)^{10}}$$ ways to put $6$ non-blue skittles in each bowl, and $$\frac{90!}{(9!)^{10}}$$ ways to put $9$ skittles in each bowl, so the probability that there are $3$ blue skittles in each bowl is $$\frac{\binom{9}{3}^{10}}{\binom{90}{30}}\approx.00002598$$
I don't understand the last part. If one of the bowls has $5$ skittles, then the probability of having $3$ skittles in each bowl is $0$.