What is the probability that a fair coin comes up tail three out of four flips? As the question asks, what is the probability that a fair coin comes up tail three out of four flips? I know the probability of getting tails on one flip is 1/2, but I'm not sure how to solve this for three/four flips. I was thinking about using a permutation of P(4,3). I don't have to do out all the arithmetic.  
 A: A $5^{\text{th}}$ graders attempt at your question :
$$ 
\mathrm {
\color{red}{HHHH},\\
\color{red}{HHH}\color{blue}{T}, \color{red}{HH}\color{blue}{T}\color{red}{H}, \color{red}{H}\color{blue}{T}\color{red}{HH}, \color{blue}{T}\color{red}{HHH},\\
\color{red}{HH}\color{blue}{TT}, \color{blue}{T}\color{red}{HH}\color{blue}{T}, \color{blue}{TT}\color{red}{HH}, \color{red}{H}\color{blue}{TT}\color{red}{H}, \color{red}{H}\color{blue}{T}\color{red}{H}\color{blue}{T}, \color{blue}{T}\color{red}{H}\color{blue}{T}\color{red}{H},\\
\boxed{\color{red}{H}\color{blue}{TTT}, \color{blue}{T}\color{red}{H}\color{blue}{TT}, \color{blue}{TT}\color{red}{H}\color{blue}{T}, \color{blue}{TTT}\color{red}{H}}\ ,\\
\color{blue}{TTTT}}
$$ 
Of the $2^4$ paths in the sample space, there are $4$ paths containing $\color{blue}{\mathrm T}$ thrice.
Hence, you have a $\dfrac{4}{2^4} = \dfrac{1}{4} = 25\%$ chance of landing a tail $3$ times out of $4$ coin flips.
A: The number of different outcomes is $2^4 = 16$. There are 4 outcomes that give three heads (easily seen that the first, second, third or fourth flip must be tails with all the rest heads), so the probability for three heads is 4/16 = 1/4.
More generally, in n flips there are $2^n$ outcomes, and for $m \le n$ there are $^nC_m = n!/m!(n-m)! $  ways to get m heads, giving a probability of $n!/m!(n-m)!2^n$
A: You note that for independent events $A_{i}$ we have the formula:
$$P\left(\bigcap_{i}A_{i}\right)=\prod_{i}P(A_{i})\tag{1}$$
Which can loosely be read as the probability of all of the $A_{i}$ independent events happening is equivalent to the product of their individual probabilities of happening.
Applying this to your problem, we have 4 independent coin tosses, and so the probability of only one heads appearing is:
$$P(\text{only one H}) = P(\text{TTTH})+P(\text{TTHT})+P(\text{THTT})+P(\text{HTTT})$$
But we note that $P(TTTH)=P(TTHT)=P(THTT)=P(HTTT)$ as they all consist of probabilities of a set of independent events all occuring, so we have:
$$P(\text{only one H})=4P(\text{TTTH})=4P(T)P(T)P(T)P(H)=4\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)=\frac{1}{4}$$

We can also note that in general this leads us to a probability distribution called the Geometric distribution, which is the probability of $n-1$ failures followed by a success where the probability of success is a constant $p$.
We have:
$$P(n)=P(n-1 \text{ failures followed by a success})=P\left(\bigcap_{i=1}^{n-1}F_{i}\cap S\right)$$
Where $F_{i}$ are the $i$th failure event and $S$ is the success event. We can use $(1)$ to thus write:
$$P(n)=\prod_{i=1}^{n-1}(1-p) \times p=p(1-p)^{n-1}$$
