Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

How would I find the cumulative distribution function of a binomial? I know I'd have to integrate it with its given parameters but how would someone go about doing that?

share|cite|improve this question
You don't have to integrate since this is a discrete distribution. Check this:… – jay-sun Nov 6 '12 at 23:20
up vote 2 down vote accepted

Binomial distribution is discrete, so you can't integrate it, but rather sum. This is what you should look into. If $X \sim Binomial(n,p)$, then CDF of $X$ is $$ P(X\leq m)=\sum_{k=0}^{m}\binom{n}{k}p^{k}(1-p)^{n-k} $$ This expression does not exist in closed form, since partial sum of rows of Pascal triangle do not exist in closed form. Depending on what exactly you need you can find various approximations to this sum, e.g. using Central Limit Theorem for large $n$, since Binomial RV is a sum of Bernoulli RVs.

EDIT: OK, as the OP specified the parameters are $n=3,p=\frac{1}{2}$, so the CDF is just $$ P(X \leq m)=\sum_{k=0}^{m}\binom{3}{k}\frac{1}{2^{k}}\bigg(1-\frac{1}{2}\bigg)^{3-k}=\frac{1}{8}\sum_{k=0}^{m}\binom{3}{k} $$

share|cite|improve this answer
Well I'm looking for the CDF with parameters (3,$\frac12$), so I guess I'm still a little confused on how you would do this. – TheHopefulActuary Nov 6 '12 at 23:30
Sorry that last comment was a fail but that was what I was going to put. Thanks! – TheHopefulActuary Nov 6 '12 at 23:38
The question is, do you know what exactly you need, since closed-form solution does not exist in this unless you know $m$, then you can easily sum up over $k$ for such small values – Alex Nov 6 '12 at 23:40

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.