Probability of exactly two heads in four coin flips? 
When you flip a coin four times, what is the probability that it will come up heads 
  exactly twice?   

My calculation: 


*

*we have $2$ results for one flip : up or down 

*so flip $4$ times, we have $4\cdot2 = 8$ results total 


Thus the probability is: $2/ 8 = 0.25 $
but the correct answer is $0.375$. Can anyone explain why I'm wrong?

Use binomial probability since there are only two possibilities: success and failure, where success represents getting a heads, and tails being a fail.
Let $X$ = Success (i.e. heads)
Therefore we are trying to find $P(X=2)$, which is $\binom42\cdot(0.5)^2\cdot(0.5)^2=0.375$.
Hope this helped!

The derivation of binomial probability:
Getting two heads out of 4 can be portrayed is, disregarding order:
HHTT (H=heads and T=tails)
Multiplying their probabilities will yield $(0.5)^4$, but as for ordering, we get $4!/(2!\cdot2!)$ due to repetition, which is the same as $4C2$. So our answer is $\binom42\cdot(0.5)^4$ which is $0.375$
