Probability of picking an item r times out of n attempts Trying to remember my high school formulas, and coming up dry.
Say I have two choices: $A$ and $B$. 
$P(A) = 0.25$ ; $P(B) = 0.75$
There are no conditional probabilities or anything. Each choice is independent of the last.
How do I go about calculating the probability that I will choose A exactly twice, having chosen ten items?
 A: You want to pick A twice and B eight times, so the probability of doing this in a certain order is $P(A)^2P(B)^8$.  But for you, the order doesn't matter, so you have to divide by the number of ways to choose two things from 10.  This is $\binom{10}{2}=\frac{10!}{8!2!}=45$.  So all in all you get $\frac{(45)3^8}{4^{10}}$, which you can calculate yourself.
In general the formula for $\binom{n}{r}=\frac{n!}{(n-r)!r!}$, where the exclamation point means "factorial," $n!=n\times(n-1)\times\dotsb\times 2\times 1$.  Since $n!$ represents the number of arrangements of $n$ things, you can interpret this formula as giving you the number of arrangements of your $n$ things, and then cancelling out the ways you can arrange the $r$ things you want, and the $n-r$ things you don't want.
A: You want the binomial distribution with $p = 0.25$ and $n = 10$ and $k = 2$.
See Wikipedia's article on the binomial distribution, which says,
"The probability of getting exactly $k$ successes in $n$ trials [when each trial has success probability $p$] is given by ...
$$ \binom{n}{k} p^k (1-p)^{n-k}."$$
Paul VanKoughnett's answer gives you an explanation of why this formula is correct.
