Probability without replacement question I have a bag full of 10 marbles: 2 black, 1 blue, 1 yellow, 3 green, 1 brown, and 2 purple. I draw 5 marbles one at a time without replacement. What is the probability of a black marble to be in the five drawn?

At least one black: It is easier to find first the probability of no black.
There are $\binom{10}{5}$ ways to choose $5$ marbles, all equally likely. Note that there are $\binom{8}{5}$ ways to choose $5$ marbles from the $8$ non-black.  So the probability that all the balls are non-black is
$$\frac{\binom{8}{5}}{\binom{10}{5}},$$
and therefore the probability of at least one black is 
$$1-\frac{\binom{8}{5}}{\binom{10}{5}}.$$
B: Exactly one black: There are $\binom{2}{1}$ ways of choosing one black from the two available. For each such way, there are $\binom{8}{4}$ ways to choose the non-blacks to go with it. So the total number of ways to pick exactly one black, and the rest non-black, is $\binom{2}{1}\binom{8}{4}$. Thus our probability is
$$\frac{\binom{2}{1}\binom{8}{4}}{\binom{10}{5}}.$$
Remark: We used general techniques. For A, there is a simpler way. The probability that the first marble chosen is non-black is $\frac{8}{10}$. Given the first was non-black, the probability that the second is non-black is $\frac{7}{9}$, since there are $7$ non-blacks left in a total of $9$. So the probability the first two are non-black is $\frac{8}{10}\cdot\frac{7}{9}$.  Continue in this way. The probability all five are non-black is
$$\frac{8}{10}\cdot\frac{7}{9}\cdot\frac{6}{8}\cdot\frac{5}{7}\cdot\frac{4}{6}.$$
As in the earlier discussion, subtract the above from $1$.
