I am new to this so don't know if I am asking the right question as I just read about its usage but didn't know what exactly a cumulative Binomial probability is.

So my question is: What are cumulative Binomial probabilities? Any example will be of great help.

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    $\begingroup$ What was the context? There's only one right answer here, but how best to explain it depends on how much you already know. If you can say what the context was, I think I can write a more helpful answer. $\endgroup$ – shadowtalker Jul 15 '14 at 11:35

TL;DR - Skip down to the ===== section to read just the example:

I'll start off with a random variable that is not binomial, but will provide an easy to understand example: a uniform random variable. An example of an event that has a uniform distribution is rolling a dice; each of the outcomes is equally likely to occur (1/6 probability for a 6 sided die). So if I asked, "what is the probability that a 1 is rolled?" You would say 1/6. In notation, if $X$ is a (discrete) random variable with a uniform distribution over the integers {1,2,3,4,5,6} then $$\text{Pr}\left\{X = 1\right\} =\cdots=\text{Pr}\left\{X = 6\right\}= 1/6$$

One often writes this as a function in terms of $x$, i.e. $p(x) = \text{Pr}\left\{X = x\right\}$. This is often called the probability mass function or pmf (for continuous random variable there is an analog called the probability density function or pdf).

The cumulative mass/density function (or sometimes called distribution function) is $$F(x) = \text{Pr}\left\{X \leq x\right\}$$ So in our uniform example it is what is the probability of rolling an $x$ or lower (e.g. What is probability of rolling a 3 or less?).

A binomial random variable, $X\sim \text{Binomial}(n,p)$, is characterized by two parameters, $n$ and $p$, then number of trials and the success probability at each trial. See wikipedia for more information on definition of a binomial. The binomial has pmf $$\text{Pr}\left\{X = x\right\} = {n \choose x}p^x(1-p)^{n-x}$$ and distribution function $$F(x) = \text{Pr}\left\{X \leq x\right\}=\sum_{k=1}^x {n\choose k}p^k(1-p)^{n-k}$$ So that was a lot of math how about some examples:

A binomial is characterized by a random phenomenon in which there are (1) $n$ independent trials, each (2) dichotomous (i.e. success/failure, yes/no, 1/0), and (3) probability of success at each trial is constant, $p$.


A classic example is flipping a coin $n$ times. Lets say we flip a fair coin 10 times and we are interested in the number of heads. Then the random variable $X = \text{Number of heads out of 10 flips}$ is $X\sim\text{Binomial}(n=10,p=1/2)$. We could ask questions such as, what is the probability we get exactly 5 heads? We could answer this using the pmf: $$\text{Pr}\left\{X = 5\right\} = {10\choose 5}(1/2)^5(1-1/2)^{10-5}=252*(0.03125)*(0.03125) = 0.2460938$$ or we can ask a question like: "What is the probability of getting 5 or less heads?" This is a cumulative binomial probability. We use the distribution function to get an answer:

$\begin{align*} \text{Pr}\left\{X \leq 5\right\} &= \sum_{k=1}^5 {10\choose k}(1/2)^k(1-1/2)^{10-k}\\ &= (0.5)(0.0009765625) + 10*(0.5)(0.001953125) + 45(0.25)(0.00390625) + 120(0.125)(0.0078125) + 210(0.0625)(0.015625) + 252(0.03125)(0.03125)\\ &= 0.6230469 \end{align*}$

it is cumulative in the sense that it 'accumulates' probability, i.e. probability of getting 5 or less heads is equal to probability of getting 0 heads PLUS probability of getting 1 head PLUS probability of getting 2 heads ...

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    $\begingroup$ It should start with k=0 for the last comment. $\endgroup$ – Matemóvil May 25 at 15:17

I thought k=0 then B(x;n,p) If k=1 then it should be B(x-1;n,p) where B( ) cumulative dists. There is an identity b(x;n,p)=B(x;n,p)-B(x-1:n,p)

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