Beginner in Probability Needs Advice/Guidance So, I work for SP+ and make garage signs for them. Most of the math I ever need is Geometry and maybe some trigonometry. My memory of calculus is faint, and I can only recall basic concepts of probability. 
While was at work, a question about probability and chance occurred to me. 
We had 20 signs made, but 3 had a missing punctuation. I know that means that 15% of the whole stack has to be fixed and that there is a 3 in 20 chance that whichever random sign I picked would need to be altered. 
My coworker took half the stack. This is when I thought to myself that I didn't know the chances of getting any of the three signs if half of the randomized stack was taken. 
My first thought was that half of 15% would be 7.5%. In other words, I would then have a 7.5% chance of getting a sign with a missing piece of vinyl (let's just refer to it as M for simplicity's sake). 
That seems too simple though. I don't think that would work, and even if it did how would I be able to figure out the chances of getting 1, 2, or all 3 signs in my stack? Adding 7.5 with itself doesn't feel correct. 
I tried asking my wife (she was a math major a few years back), and she didn't know nor really like probability. I tried googling it, but I'm not even sure what this concept would be called. I stumbled upon something called combining multiple probabilities, but I'm unfamiliar with some of the signs and concepts that are used in the equation. I'm basically leaping from basic probability to this complex problem.
I need help in trying to figure this out because the question is continuously nagging at me. Where would I start in trying to figure this out? Is there a special name for this particular type of thing in math? And how would I be able to come to an answer for a similar question in the future?
 A: Surely you remember this familiar old concept:
$$P(E)=\frac{\operatorname{number of events in E}}{\operatorname{total number of events}}$$
The total number of ways to partition the $20$ signs into groups of $2$ is
$$\frac{20!}{10!10!}$$
which can be obtained by the partitions formula. This formula is that the number of ways to partition $n$ objects into groups of $g_1,...,g_n$ objects so that $g_1+...+g_n=n$ is given by
$$\frac{n!}{g_1!...g_n!}$$
Now we need to find the number of arrangements in which you have at least one bad sign in your pile. This is, of course, given by
$$\frac{20!}{10!10!}-N^*$$
Where $N^*$ is the number of arrangements in which you do not have any bad signs. If you have no bad signs, that is the same as just giving all three bad signs to your friend and then divvying up the good signs into a group of $10$ signs for you and $7$ more for your friend. By the partitions formula again, this is
$$N^*=\frac{17!}{10!7!}$$
So the number of arrangements in which you get no bad signs is
$$\frac{20!}{10!10!}-\frac{17!}{10!7!}$$
and the final probability is
$$\frac{\frac{20!}{10!10!}-\frac{17!}{10!7!}}{\frac{20!}{10!10!}}$$
$$\frac{20!-8\cdot9\cdot10\cdot17!}{20!}$$
Which is $\approx 0.89$ that at least one bad sign will show up in your pile. So you will probably get at least one bad sign.
A: Usually a good way of first testing a hypothesis is to try and take the extreme cases. If $100\%$ of the stack needed to be fixed then you would have a $100\%$ chance of needing to fix a sign after the stack was split rather than $50\%$.
On a side note, if you wanted to know the expected number of pieces you needed to fix, then the answer would indeed be $7.5\% \times 20$ due to linearity of expectation.

For the problem in hand, you want to know the probability, $p$, of getting at least one sign you need to fix. This is $1-P(\text{getting no bad signs})$
Calculating this is quite simple and is just $1-P(\text{getting a good sign $10$ times in a row})$. So therefore
$$p=1-\frac{17}{20}\frac{16}{19}...\frac{8}{11}=1-\frac{(10)(9)(8)}{(20)(19)(18)}=1-\frac{8}{2^2(19)}=\frac{17}{19}$$
(Note that in the calculation we are treating it as if we were drawing the signs one at a time from the pile. We can do this because that actually is no different from drawing $10$ signs 'all at once')
A: Instead of thinking of choosing a randomized half, you can equivalently think of it as shuffling the 20 signs, and then taking the first 10.
Each ordering of the 20 signs is equally likely in a fair shuffle, so the probability you seek is
$$P(\text{first 10 signs are not defective}) = \frac{\text{number of orderings of the 20 signs, such that the first 10 signs are not defective}}{\text{number of orderings of the 20 signs}}.$$
[If you want the complement (probability that there is at least one defective sign in your selection), then subtract the above probability from $1$.]
The denominator is $20! = 20 \cdot 19 \cdot 18 \cdots 2 \cdot 1$
For the numerator, we want the three defective signs to be on the last 10 signs (not the 10 that you pick). In the ordering, there are $10 \cdot 9 \cdot 8$ positions in the last 10 signs for the three defective signs (e.g., #12, #15, #18 is one possibility, #18, #15, #12 is another separate possibility, as is #11, #13, #12). Then there are $17! = 17 \cdot 16 \cdots 1$ ways to order the 17 good signs in the remaining positions.
Thus, the probability is
$$\frac{10 \cdot 9 \cdot 8 \cdot 17!}{20!} = \frac{10 \cdot 9 \cdot 8}{20 \cdot 19 \cdot 18}.$$
