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 $5^{\text{th}}$ 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.
