# 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?

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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.

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Why would you divide by the number of ways it could be done? Shouldn't you multiply by it? – Smashery Oct 20 '10 at 5:18
Wow, thanks for catching that. You're absolutely right. – Paul VanKoughnett Oct 20 '10 at 5:36
I was thinking to myself, "Surely the probability of it not being in a particular order would be higher than being in an order." ;-) Thanks for your help! – Smashery Oct 20 '10 at 5:47

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.

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