Calculate CDF of Binomial Distribution

Question

I need to solve the following problem:

Given $n$ yes/no experiments, and a success probability $p$, at least how many successes $k$ can I expect (say, with a 'confidence' of $c=95\%$ or more)?

Another way to put it is: I want to be 95% sure to have at least $k$ successes. Everything is given, and I am interested in $k$.

As the Cumulative Distribution Function (CDF) for the Binomial Distribution is $$ F(k; n, p) = P(X \leq k) = I(1 - p; n - k, k + 1), $$ where $I(\cdot; \cdot, \cdot)$ denotes the regularized incomplete beta function, and I'm mostly interested in calculating the numerical outcome, the way I wanted to do it is to solve $$ P(X \geq k) = 1 - P(X \leq k - 1) = I(p; k, n - k + 1) = c $$ for $k$ using Maxima (although any other approach/tool would be good, too).

Doing some sanity checks with the CDF, though, I get unexpected results. For example, I would expect to get a probability value for

F(1; 100, .63)
= beta_incomplete_regularized(.37, 99, 2)

However, Maxima outputs:

1.898744430721408*10^-16*%i+1.014897496320844

Any comments/hints to this output, or to the approach in general? What other tools would be simple to use here?

Answer 1

I used Mathematica to solve this:

The general rule to find k is

InverseCDF[BinomialDistribution[n, p], confidence interval] = k

Suppose you have $n=10$, $p=0.4$ and you want the value of $k$ at 95% confidence.

InverseCDF[BinomialDistribution[10, 0.4], 0.95]

You get answer 7.

Answer 2

Found the solution:

As the CDF is $F(k, n, p) = 1 - F(n - k - 1, n, 1 - p)$ (wikipedia), solving for at-least-$k$-with-95%-confidence means to substract the 95%-quantile of $F(n - k, n, 1 - p) = 1 - F(k - 1, n, p)$ from $n$.

This way it also works in Maxima:

100 - quantile_binomial(.95,100,.37);
                                55

For example, with $n=100$ yes/no experiments, and a success probability $p=0.63$, I can be 95% sure to have at least $k=55$ successes.