Picking Objects Simultaneously A sack contains $10$ balls: $7$ are white and $3$ are black. You choose $2$ simultaneously; what is the probability of getting $2$ white balls?
Does the fact that the balls are chosen simultaneously change the approach? Is it just $\frac{7}{10} \cdot \frac{6}{9}$?
 A: 
Does the fact that the balls are chosen simultaneously change the approach? Is it just (7/10) * (6/9)?

It won't change your approach to the problem. If picking simultaneously really bothers you, one can use the binomial coefficient. 
There are a total of $\binom{10}{2}=45$ times two balls can be chosen. Meanwhile, one can get two white balls a total of $\binom{7}{2}=21$. Therefore, the answer will still be $\frac{21}{45} = \frac{7}{15}.$

The binomial coefficient $\binom{n}{k}$ is the number of ways of picking $k$ unordered outcomes from $n$ possibilities (source). 

A: The only difference between "selecting simultaneously" and "selecting one-at-a-time-without-replacement", is the loss of information about order of extraction (which ball is "first" et cetera).   This does not affect the probability for counts of ball types among the selection, although new students often seem to feel that it somehow should.
So, you may measure the probability for the event as though you had extracted the balls in sequence then shuffled them.
$\tfrac{7}{10}\cdotp\tfrac{6}{9}$ is the probability for extracting two from seven white balls when drawing two balls from ten.   This is also ${}^7\mathrm C_2\div{}^{10}\mathrm C_2$
$\tfrac{7}{10}\cdotp\tfrac{3}{9}{+}\tfrac{3}{10}\cdotp\tfrac{7}{9}$ is the probability for extracting one from seven white and one from three red balls when drawing two balls from ten.  This is also ${}^7\mathsf C_1\cdotp{}^3\mathrm C_1\div{}^{10}\mathrm C_2$
A: (Not commenting because I do not have enough karma)
It would be a hypergeometric experiment (sampling without replacement). In your case: 
P(2 white) = $7 \choose 2$$3 \choose 0$ / $10 \choose 2$
