Simple probability problems This may be very simple, but I need to know correct answers.
Problem 1.
There are $6$ balls in the box - $2$ black, and $4$ white. We take $3$ balls from the box. What is the probability that we took exactly 1 black ball.
Problem 2.
What is the probability to score $7$ points when you throw two dices.
My answers
Problem 1: ${\binom{2}{1}\binom{4}{2} \over \binom{6}{3}}=0.6$
Problem 2: Denote by $(s,c)$ a pair where s - is a score, c - amount of possible outcomes. Then we have $(2,1) (3,1) (4,2) (5,2) (6,3) (7,3) (8,3) (9,2) (10,2) (11,1) (12,1)$. Amount of possible outcomes is 21. Amount of desired outcomes is 3, hence $p=\frac 3 {21}=\frac 17$
 A: The critical distinction to make in problem 1 is whether you're sampling with or without replacement.  This is, when you pick a ball out of the jar, do you note its color and put in back in, or keep it it.  If it's the former case, you're sampling with replacement.  In that case, the number of black balls in a sample of three is binomially distributed variable (thanks to Dilip for pointing out this & that my original analysis was incorrect) with $n=3$ and $p = \frac{1}{3}$.  Thus, the probability of drawing one black ball is given by: 
$$\binom{3}{1}\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^2 = \frac{4}{9}$$
If you're sampling with out replacement, the solution is a little different.  Consider, you could pick a black ball, then two whites, a white, then a black, then a white, or a two whites and then a black.  The probabilities of those situations will just be the products of the probabilities of each draw.  Thus, they will be: $$\frac{2}{6}\frac{4}{5}\frac{3}{4} ,   \frac{4}{6}\frac{2}{5}\frac{3}{4} , \frac{4}{6}\frac{3}{5}\frac{2}{4}$$ respectively.  The sum, then, will be the total probability of drawing two white balls and one black: $$\frac{2}{6}\frac{4}{5}\frac{3}{4} +   \frac{4}{6}\frac{2}{5}\frac{3}{4} + \frac{4}{6}\frac{3}{5}\frac{2}{4} = \frac{3(4*3*2)}{6*5*4} = \frac{3}{5}$$  This is equivalent to the solution you posted for this problem.
A: For problem 2):
Scoring 7 points means (presumably) the sum of the rolls is 7. Imagine you roll a die twice, in succession. You record the results as an ordered pair giving the result of the first roll, then the result of the second roll. 
A score of 7 occurs  in (and only in) the following ways: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1). 
Out of the 36 equally likely outcomes that result from throwing a die twice in succession, exactly six  will have a score of 7. So the probability of scoring 7 is ${6\over 36}={1\over6}$.

See Dilip' comment below.  With unordered outcomes it's twice as likely to get "a one and a three" as "two threes".


Let's solve the problem
using indistinguishable outcomes:

The outcomes from rolling two dice (dies?) simultaneously are
$$
\matrix{ 
\{1,1\} & \{1,2\}& \{1,3\}&\{1,4\}& \{1,5\} &\{1,6\} \cr
   &  \{2,2\}&\{2,3\}& \{2,4\} &\{2,5\} &\{2,6\}\cr
  &   &\{3,3\}& \{3,4\} &\{3,5\} &\{3,6\}\cr
  &   & & \{4,4\} &\{4,5\} &\{4,6\}\cr
&   & &   &\{5,5\} &\{5,6\}\cr
&   & &   & &\{6,6\}\cr
 }
$$
Now, imagine, please, that the two dice were of different colors, say taupe and ecrue.  
In how many ways could you have obtained the outcome $\{1,2\}$? The answer is two ways: taupe was was 1 and ecrue was 2, or taupe was 2 and ecrue was 1. 
In how many was could you have obtained the outcome $\{2,2\}$? Well, only one way: both of the dies showed "2".
A moment's reflection should convince you that an outcome that has different entries is twice as likely to occur as an outcome that has the same entries.
Further reflection reveals that any two outcomes of the "different entry" variety have the same probability of occurring and any two outcomes of the "same entry" variety have the same probability of occurring. 
From this, we can calculate probabilities of individual outcomes:
let the probability of an outcome with same entries be $a$. Then  the probability of an outcome with different entries is $2a$.
Then, since the sum of the probabilities of all outcomes is 1
$$
1= 6\cdot a+15\cdot (2a).
$$
So $a=1/36$, and we have
$
P(\text{outcome with same entries})={1\over36}
$
and
$
P(\text{outcome with different entries})={2\over36}
$
Now we can find
$$ 
P( \text{score of 7} ) =P(1,6)+P(2,5)+P(3,4) =3\cdot {2\over36}={1\over6}.
$$
That was quite a bit of reflecting to arrive at the answer. As far as "real life" is concerned, you could imagine the two dice (dies?) were distinguishable; and in that case, it is much easier to convince yourself that there are 36 equally likely outcomes.
