x = geoinv(y,p) returns
the inverse cumulative distribution function (icdf) of the geometric
distribution at each value in y using the corresponding
probabilities in p.

geoinv returns the smallest positive integer x such
that the geometric cdf evaluated at x is equal
to or exceeds y. You can think of y as
the probability of observing x successes in a row
in independent trials, where p is the probability
of success in each trial.

y and p can be vectors,
matrices, or multidimensional arrays that all have the same size.
A scalar input for p or y is
expanded to a constant array with the same dimensions as the other
input. The values in p and y must
lie on the interval [0,1].

Suppose the probability of a five-year-old car battery not starting in cold weather is 0.03. If we want no more than a ten percent chance that the car does not start, what is the maximum number of days in a row that we should try to start the car?

To solve, compute the inverse cdf of the geometric distribution. In this example, a "success" means the car does not start, while a "failure" means the car does start. The probability of success for each trial p equals 0.03, while the probability of observing x failures in a row before observing a success y equals 0.1.

y = 0.1;
p = 0.03;
x = geoinv(y,p)

x =
3

The returned result indicates that if we start the car three times, there is at least a ten percent chance that it will not start on one of those tries. Therefore, if we want no greater than a ten percent chance that the car will not start, we should only attempt to start it for a maximum of two days in a row.

We can confirm this result by evaluating the cdf at values of x equal to 2 and 3, given the probability of success for each trial p equal to 0.03.

y2 = geocdf(2,p) % cdf for x = 2
y3 = geocdf(3,p) % cdf for x = 3

y2 =
0.0873
y3 =
0.1147

The returned results indicate an 8.7% chance of the car not starting if we try two days in a row, and an 11.5% chance of not starting if we try three days in a row.