Probability taking out two balls from an urn 
In an urn there is five red balls and three blue balls. Two balls are taken randomly from the urn with return. (The colors of the two balls are seen and taken back to the urn).
What is the probability of taking out two red balls?


I know how to calculate probabilities like taking one ball after the other (multiplying probabilities)... Not when there is two balls at the same time. How would I resolve it? Do I need to draw to myself a big probability space $\Omega$ and just count how many possible options there is? It doesn't work at least with tree diagram.
 A: This is a lot easier than you think. The probability of taking out a red ball is just $\frac{5}{8}$. Since you return the ball, it doesn't change for the second turn, so you get $\frac{5}{8}\frac{5}{8}=\frac{25}{64}$.
The emphasis on with return suggests that he doesn't take them at the same time. If he would, it would just be the model without return. So drawing two balls at the same time can be treated as drawing them after each other without return. But again, I think that is not the case for this question.
EDIT:
If the balls are meant to be drawn at the same time, we can apply the model of drawing balls without return. In that case we get $\frac{5}{8}\cdot\frac{4}{7}=\frac{5}{14}$. I want to mention however that it is somewhat problematic to state it with "with return" when the balls are meant to be drawn at the same time which is equivalent to drawing without return. This isn't intented as criticism just advice on how to distinguish between the models with/without return clearly.
A: If you draw two balls only once, why does it matter whether you return them or not?
There are $\binom{8}{2} = 28$ ways to choose two balls out of the eight.  Only $\binom{5}{2} = 10$ ways produce two red balls.  Hence the probability of drawing two red balls is $\frac{10}{28} = \frac{5}{14}$.
If you're not familiar with binomial coefficients, you can proceed as though the two balls are taken sequentially (even if they're not).  You draw the first ball, and then don't replace it.  Then the second ball is taken from the seven remaining balls.  The probability of drawing two red balls is then the probability that you draw a red ball ($\frac{5}{8}$) times the probability that you draw a second red ball given that you drew a first red ball ($\frac{4}{7}$), or again, $\frac{5}{14}$.
A: If neither ball is returned before the other is drawn then the answer would be $\frac58 \times \frac47 = \frac5{14}$
If the first ball is returned before the second is drawn then the answer would be $\frac58 \times \frac58 = \frac{25}{64}$
