As a follow-up, I will answer my own questions...
MarkFL said:
Out of all the possible outcomes from tossing a coin $n$ times, how many of those will be $n$ heads?
There is only one outcome in which all tosses will be heads, since there is only 1 way to choose $n$ from $n$:
$${n \choose n}=\frac{n!}{n!((n-n)!}=1$$
MarkFL said:
For each individual coin toss, how many outcomes are possible?
There are two possible outcomes, either heads or tails, and for a fair coin, both are equally likely.
MarkFL said:
How many of those are heads?
Only 1 outcome is heads, therefore we can compute the probability of getting heads on one toss as follows:
We are certain that we will either get heads (H) or tails (T):
$$P(\text{H})+P(\text{T})=1$$
Since we assume the coin is fair, we know:
$$P(\text{H})=P(\text{T})$$
And so we have:
$$P(\text{H})+P(\text{H})=1$$
$$2P(\text{H})=1$$
$$P(\text{H})=\frac{1}{2}$$
This corresponds with the fact that out of the two possible outcome, only one is favorable:
$$P(\text{H})=\frac{\text{number of ways to get heads}}{\text{total number of outcomes}}=\frac{1}{2}$$
MarkFL said:
Using the fundamental counting principle, how many outcomes will there be for $n$ coin tosses?
Since for each toss, there are two outcomes, the total number $N$ of outcomes for $n$ tosses is:
$$N=\prod_{k=1}^{n}(2)=2^n$$
And so we may conclude that the probability of getting $n$ heads for $n$ tosses is:
$$P(\text{n heads})=\frac{1}{2^n}$$
So, if $n=3$, the probability of getting all 3 heads is:
$$P(\text{3 heads})=\frac{1}{2^3}=\frac{1}{8}$$