MHB What is the number of blue balls in the second urn?

[Houdini1]

Question:
An urn contains 10 balls: 4 red and 6 blue. A second urn contains 16 red balls and an unknown number of blue balls. A single ball is drawn from each urn. The probability that both balls are the same color is .44. Calculate the number of blue balls in the second urn.

My attempt
I don't know the best notation to use for this situation. I am going to try [math]P(R_1)[/math] as the probability of drawing a red ball from the first urn. So we have [math]P(R_1)=.4 \text{ and } P(B_1)=.6[/math]. To express the probability of both balls being a single color it seems there are two cases to consider which we should add: [math]P(R_1 \cap R_2)+P(B_1 \cap B_2)[/math]. Am I correct in thinking that for mutually exclusive events that's the same as [math]P(R_1 \cdot R_2)+P(B_1 \cdot B_2)[/math]?

I know the basic ways to manipulate these using DeMorgan's Laws but I'm missing the first step or have set up the problem entirely incorrectly. I have the solution key but I don't want the full solution yet.

 

[Plato2]

Question:
An urn contains 10 balls: 4 red and 6 blue. A second urn contains 16 red balls and an unknown number of blue balls. A single ball is drawn from each urn. The probability that both balls are the same color is .44. Calculate the number of blue balls in the second urn.

I would solve the following:
\frac{4}{10}\frac{16}{16+x}+\frac{6}{10}\frac{x}{16+x}=\frac{44}{100}.

 

[Houdini1]

Makes perfect sense. I've been trying to use all of these set rules that I missed it. So x=4 and plugging that in confirms.