Probability of getting exactly one head (and one tail) from two coin flips Suppose I flip a fair coin twice and ask the question,
"What is the probability of getting exactly one head (and tail) ?"
I was confused on whether I would treat this as a combination or permutation. My original thought was that it is a combination as we don't care about the order and just want the case of one head (or tail) giving the probability of 1/3.
 A: The possible outcomes are
$$HH, HT, TH, TT$$
Two out of four qualify, so the probability is $1/2$.
A: I think it is worth stressing that you can think of the outcomes as ordered or not, as you wish.  That does not change the problem, nor does it change the final answer, but it does alter the computation.
Ordered:  There are $4$ possible outcomes $$HH, HT, TH, TT$$  each of which has probability $\frac 14$.  By inspection, the winning outcomes are $HT, TH$ so the answer is $\frac 12$.
Unordered:  Now there are $3$ possible outcomes $$\{H,H\}, \{H,T\}, \{T,T\}$$
but these three are not equi-probable.  Indeed, there is only one way to get $\{H,H\}$, say, which makes the probability of that outcome $\frac 14$.  Same for $\{T,T\}$.  There are two ways to get $\{H, T\}$ so that has twice the probability, again yielding $\frac 12$.
This comes up for dice throws, for example.  If you consider ordered outcomes, then each outcome has probability $\frac 1{36}$. If you consider them as unordered, then the probability of getting distinct values, like $\{1,2\}$ is $\frac 2{36}=\frac 1{18}$ while the probability of getting a double like $\{1,1\}$ is still $\frac 1{36}$.  Personally, I try to stick to equi-probable cases wherever possible, as that simplifies the computations. But it's up to you, and in some situations you really can't avoid considering outcomes with differing probabilities.
A: Probability of getting exactly one head, this can happen in two ways : $HT + TH$
Favorable cases = 2
Total Cases = 4
So, required probability is $\frac{1}{2}$
