Independent Events happening at the same time I'm trying to wrap my head around a pretty basic concept. But I'm not 100% sure my reasoning is right.
I have an event that happens 10 times, at the same time, for thousands of occurrences. All of these 10 events are independent.
For each of these events, there is a 5% chance I will have a bad outcome, and 95% chance I will have a good outcome. 
So first occurrence:


*

*95% Good Outcome

*95% Good Outcome


Etc... until 10 times. 
Would it be fair to say that over a few thousand occurrences, that the bad event will happen at least once 50% of the time in each occurrence, twice 25%, thrice 12.5% etc... since there is a 5% chance of it happening for every event, and there are 10 events each occurrence?
 A: One issue is you don't really explain why you'd expect this to hold. You just suggest "maybe it works like this", and that's not the way to convince someone else something is true. You aren't even convinced yourself.
Start by checking small cases if possible (which it is in this case) to see if it could possibly be true. So for example, what if there was a 25% chance the outcome is B, and just 2 events happening at the same time. Then try to generalize based on that.
In this case, AA has a 9/16 chance of happening, AB and BA together have a 6/16 chance of happening, and BB has a 1/16 chance of happening. Thus the pattern you were hoping for doesn't seem to hold.
A: Just to be clear about the subdivisions within the experiment, let us talk about individual trials, groups of 10 trials, and a run of 1000 groups.  
We are looking for: the expected number of groups in the run that have a certain number of trials with bad outcomes.
The probability of a bad outcome on a trial is: $p=0.05$ as given.
The number of bad outcomes in a group will have a binomial distribution. $$X \sim \mathcal{Bin}(10,p)$$  That means the probability of exactly $x$ bad outcomes in a group is: 
$$\begin{align}
\mathsf P(X=x) & =\binom{10}{x} p^x(1-p)^{10-x} & ,\text{ for }x\in\{0, 1, .., 10\}
\\ & = \binom{10}{x}\frac{19^{10-x}}{20^{10}}
\end{align}$$
Let $N_x$ be the number of groups with exactly $x$ bad outcomes in a run.  This will also have a binomial distribution, with parameters $1000$ and $\mathsf P(X=x)$.
$$N_x \sim\mathcal{Bin}(1000, \mathsf P(X=x))$$
The expected value of a random variable with a binomial distribution $\mathcal{Bin}(n,p)$ is $np$.   So the expected number of groups with exactly $x$ bad outcomes in a run will be:
$$\begin{align}
\mathsf E[N_x] & = 1000\mathsf P(X=x) 
\\ & =\binom{10}{x}\frac{10^3 \cdot 19^{10-x}}{20^{10}}
\end{align}$$
For instance $\mathsf E[N_0] \approx 598.7, \mathsf E[N_1] \approx 315.1, \mathsf E[N_2] = 74.6,$ et cetera. 

Well, that's the expected number of groups with exactly $x$ bad outcomes.
Can you now do the expected number of groups with at least $x$ bad outcomes?
