Arguments. x. number of successes, or a vector of length 2 giving the numbers of successes and failures, respectively. n. number of trials; ignored if x has length 2. p. hypothesized probability of success. In R, it's easy enough: sum (1/6, 1/6, 1/6) This gives 1/2 which is the correct answer. However, I have in the back of my mind (where it possibly should remain) that I should be able to use the binomial distribution for this. I've tried various combinations of arguments for pbinom and dbinom, but I can't get the right answer.

10. Hypothesis Testing: p-values, Exact Binomial Test, Simple one-sided claims about proportions. See Questions 4, 5, and 6, towards the bottom of this page for R scripts to calculate the p-value and a barplot for simple claims about proportions.

The usage and help pages are modeled on the d-p-q-r families of functions for the commonly-used distributions (e.g., dbinom) in the stats package. Heuristically speaking, this distribution spreads the standard probability mass at integer x to the interval [x, x + 1] in a continuous manner. As a result, the distribution looks like a smoothed But then by the linearity of expectation, we have E(X) = E(B1 + B2 + ⋯ + Bn) = E(B1) + E(B2) + ⋯ + E(Bn). It is easy to verify that E(Bi) = p, so E(X) = np. You wrote down another expression for the mean. So the above argument shows that the combinatorial identity of your problem is correct. You can think of it as a mean proof of a
Note that binomial coefficients can be computed by choose in R. If an element of x is not integer, the result of dbinom is zero, with a warning. \ (p (x)\) is computed using Loader's algorithm, see the reference below. The quantile is defined as the smallest value \ (x\) such that \ (F (x) \ge p\), where \ (F\) is the distribution function.
. 395 338 894 549 919 369 364 570

how to use dbinom in r