Number the red and yellow balls from $1$ to $n$, where $n=r+y$. Don't bother to number the blue balls.
Let $X_i=1$ if the $i$-th numbered ball is drawn before any blue ball, and let $X_i=0$ otherwise. Then the total number of draws is $1+(X_1+\cdots+X_n)$. It follows that the expected number of draws is
$$1+E(X_1+\cdots+X_n).$$
By the linearity of expectation, this is
$$1+E(X_1)+\cdots +E(X_n).$$
All the $X_i$ have the same expectation. so let us find $E(X_1)$.
The probability that ball with label $1$ comes before any blue ball is $\frac{1}{b+1}$. This is because all orderings of ball $1$ and the blue balls are equally likely. So $E(X_1)=\frac{1}{b+1}$.
It follows that the number of draws has mean
$$1+\frac{n}{b+1}.$$
Remark: We used the powerful tool of indicator random variables. One can alternately find the probability distribution of the number of draws, and then find the mean by using the standard formula, and a manipulation of binomial coefficients. That is substantially more messy. It is useful to know that in many cases the expectation of a random variable $Y$ can be computed without explicit knowledge of the probability distribution function of $Y$.
$1.$ With replacement: There are three ways this can happen, red, red, red; blue, blue, blue; and green, green, green.
The probability of red, red, red is $\left(\dfrac{5}{19}\right)^3$. Find similar expressions for the other two colurs, and add up.
$2.$ Without Replacement: The probability of red, red, red is $\left(\dfrac{5}{19}\right)\left(\dfrac{4}{18}\right)\left(\dfrac{3}{17}\right)$. Find similar expressions for the other two colours, and add up.
Remark: We could also do the problem by a counting argument. Let's look at the first problem. Put labels on the balls to make them distinct. Then for with replacement, there are $19^3$ strings of length $3$ of our objects. (The three objects in the string are not necessarily distinct.) All of these $19^3$ strings are equally likely.
There are $5^3$ all red strings, $11^3$ all blue, and $8^3$ all green. This gives probability $\dfrac{5^3+6^3+8^3}{19^3}$.
One can do a similar calculation for the without replacement case.
For without replacement, there is a third approach. We can choose $3$ objects from $19$ in $\dbinom{19}{3}$ ways. And we can choose $3$ of the same colour in $\dbinom{5}{3}+\dbinom{6}{3}+\dbinom{8}{3}$ ways.
Best Answer
Let us assume the above question appears on an exam and the professor has a rule: no questions answered during the exam. How to best interpret the problem as given?
Question : A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
We are told that two balls are drawn at random. Since we are not told that the balls were returned to the bag, the simplest interpretation would be that the balls were drawn without replacement and should thus be solved accordingly.
If the instructor then came back and said, "i meant with replacement," then i as a professor of statistics for almost 30 years would say that the question was poorly worded and the student must get the benefit of the doubt. If the instructor did not intentionally set out to trick the student, then the instructor would agree. If the instructor did intend to trick the student, then i would find another instructor.
i hope this reply helps, best.