The probability of drawing a type of card from a deck with 2 draws. Let's propose that I have a deck with 16 cards, with the following cards: 4 A cards, 4 B cards, 4 C cards and 4 D cards.
I know how to calculate the chance of drawing an A card from this deck with a single draw: good draws / number of cards which cames up to 25%.
How could I calculate the same chance percentage, if I am able to draw two cards from this deck? Could you give me a formula to use, for these kind of calculations, in which I can change the deck size, the deck composition or even the number of possible draws also?
 A: There are several ways. 
(1) You could create the sample space of all two-drawings. Numbering the cards $A1, A2, A3, A4, B1, ...$ etc.  You'll have $(A1,A2), (A1,A3),$ $..., (B1, A1)$, etc. A total of $16\cdot 15$ possibilities (without replacement), of which you can form the ratio the same way as you did for the simpler problem.
(2) You do not need to write out the sample space, however. You could simply count. Consider the mutually exclusive events: 
$(E) \,\, \text{Exactly one A is drawn, and it is drawn 1st... (} \, 4\cdot 12 \text { ways.)}$
$(F) \,\, \text{Exactly one A is drawn, and it is drawn 2nd.. (} \,12\cdot 4 \text { ways.)}$
$(G) \,\, \text{Exactly two A's are drawn... (} \, 4\cdot 3 \text { ways.)}$
The event At least one $A$ then is given by the union of these events, and so:
$$P(\text{At least one A}) = P(E\cup F \cup G) = \dfrac{4\cdot 12}{16\cdot 15}+\dfrac{12\cdot 4}{16\cdot 15}+\dfrac{4\cdot 3}{16\cdot 15}$$
(3) A (perhaps) quicker approach would be to use a formula:
$$P(\text{At least one A}) = 1- P(\text{no A is Drawn})$$
The second term above, the probability no A is drawn, can be calculated as
$$P(1^{st} \text{ not an A})\cdot P(2^{nd} \text{ not an A, assuming the first is not an A})$$
$$ = \dfrac{12}{16}\cdot \dfrac{11}{15}$$
A: If, like in your case, you have $16$ cards, four of each color, red, green, blue and black, then you're right that the first card being black has a probability of $.25$. 
In order to find out the probability of at least one black card, for example, it's easier to find out the probably of no black cards, and then subtract that from $1$, the total probability space. This is somewhat useful in this case, but much more applicable in larger spaces. 
In order to find the probability of missing a black card both times, you just figure out the probability of drawing any other type of card, which is just $\frac{12}{16}$. If we then draw another card, we know that we've lost one non-black card from our total, which is now $15$, so the probability that we draw a second non-black card is $\frac{11}{15}$. This continues on for each non-black card you don't draw. Once you pick as many non-black cards as you want ($n$ of them to be general), you end up with:
$$\frac{12}{16}\times\frac{11}{15}\times...\times\frac{12-n+1}{16-n+1}$$
Once you get this number, which is the probability of missing a black card $n$ times, you can subtract it from $1$, the total probability space, to get you the probability of at least one black card. 
