$1.$ 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.
Let's set x = blue balls.
Let's take 2 cases:
Case 1) They are both red. The chances of that is $$\frac{4}{10}\cdot \frac{16}{16+x}$$
Case 2) They are both blue. The chance of that is $$\frac{6}{10}\cdot \frac{x}{16+x}$$
So now we have the total chance: $$\frac{4}{10}\cdot \frac{16}{16+x}+\frac{6}{10}\cdot \frac{x}{16+x}=\frac{44}{100}$$
Simplifying this, we get $$\frac{3x+32}{5x+80}=0.44$$
Solving, we get $$x=\boxed{4}$$
Also, I wouldn't consider this high school math, as I'm in 7th grade :D
Best Answer
Numer of all possible ways of selecting 1 ball from urn 1 and 2 balls from urn 2 is ${\left( \begin{gathered} 10 \\ 1 \\ \end{gathered} \right)\left( \begin{gathered} 20 \\ 2 \\ \end{gathered} \right)}$.
Number of ways to select 1 red ball from urn 1 and 2 red balls from urn 2 is ${\left( \begin{gathered} 4 \\ 1 \\ \end{gathered} \right)\left( \begin{gathered} 16 \\ 2 \\ \end{gathered} \right)}$.
Number of ways to select 1 blue ball from urn 1 and 2 blue balls from urn 2 is ${\left( \begin{gathered} 6 \\ 1 \\ \end{gathered} \right)\left( \begin{gathered} 4 \\ 2 \\ \end{gathered} \right)}$.
Hence, number of ways of selecting 1 ball from urn 1 and 2 balls from urn 2 such that they are of the same color is ${\left( \begin{gathered} 4 \\ 1 \\ \end{gathered} \right)\left( \begin{gathered} 16 \\ 2 \\ \end{gathered} \right) + \left( \begin{gathered} 6 \\ 1 \\ \end{gathered} \right)\left( \begin{gathered} 4 \\ 2 \\ \end{gathered} \right)}$ since they are mutually exclusive.
Since probability of selecting balls of the same color is number of ways that can be done divided by number of all possible ways of selecting 1 ball from urn 1 and 2 balls from urn 2, we have that wanted probability is
$\frac{{\left( \begin{gathered} 4 \\ 1 \\ \end{gathered} \right)\left( \begin{gathered} 16 \\ 2 \\ \end{gathered} \right) + \left( \begin{gathered} 6 \\ 1 \\ \end{gathered} \right)\left( \begin{gathered} 4 \\ 2 \\ \end{gathered} \right)}} {{\left( \begin{gathered} 10 \\ 1 \\ \end{gathered} \right)\left( \begin{gathered} 20 \\ 2 \\ \end{gathered} \right)}} = \frac{{129}} {{475}}$