<s>[INST] 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 $p=\frac 3 {21}=\frac 17$
 [/INST] 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.
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