Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 5 down vote accepted

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.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.