probability of pick same color ball 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 0.44. Calculate the number of blue balls in the second urn. 
I just found this question on internet ( the problem changes everytime you refresh the browser, maybe you see different problems now).  
I have never learned probability before, actually I'm going to learn it next semester. At first glance, I thought this problem seems like high school math problem, but I'm wrong, I couldn't figure out the right way to solve this problem. if anyone could explain how to approach this question please!
 A: 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
A: Suppose there are $b$ blue balls in the second urn.
The probability that both are red would be
$$\frac4{10} \frac{16}{16+b}$$
The probability that both are blue would be
$$\frac6{10} \frac{b}{16+b}$$
Hence the probability that the balls are of the same color would be 
$$\frac4{10}\frac{16}{16+b}+\frac6{10}\frac{b}{16+b}=0.44$$
You can convert the problem to a linear problem and solve for $b$.
A: Suppose the number of blue balls is $x$. Then $P($Both balls of same colour)$=P($both are red)$+P($both are blue)$=P(${red from 1st urn})*P({red from 2nd urn})$+P(${blue from 1st urn})*P({blue from 2nd urn})$=\frac{4}{10}*\frac{16}{16+x}+\frac{6}{10}*\frac{x}{16+x}=0.44$. Solve this for $x$.
$P(${red from 1st urn},{red from 2nd urn})=$P$(red from 1st urn)*$P$(red from 2nd urn) since the draws are supposd to be independent. Similarly for blue.
A: Let there be $b$ blue balls in the second urn. The only way we can have both of the balls be the same color is if they are both red, or both blue. $\mathbb{P}(R_1)=\frac{2}{5}$, and $\mathbb{P}(R_2)=\frac{16}{16+b}$, meaning $\mathbb{P}(R_1\cap{R_2})=\frac{32}{5(16+b)}$ The probability that they are both blue is $\mathbb{P}(B_1)\mathbb{P}(B_2)=\frac{3}{5}\frac{b}{16+b}=\frac{3b}{5(16+b)}$. Simply add $\mathbb{P}(R_1\cap{R_2})$ and $\mathbb{P}(B_1\cap{B_2})$, and set it equal to 0.44, giving you a linear system that is easily solved. 
