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 trialpequals 0.03, while the probability of observingxfailures in a row before observing a successyequals 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 ofxequal to 2 and 3, given the probability of success for each trialpequal to 0.03.

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

y2 = 0.0873

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

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.