Question

A computer program has a bug that causes it to fail once in every thousand runs, on average. In an effort to find the bug, independent runs of the program will be made until the program has failed five times.

a. What is the mean number of runs required?
b. What is the standard deviation of the number of runs required?

Answer 1

a) The mean number of runs required is 5000.

b) The standard deviation of the number of runs required is 2235.

Explanation:

For each time the program is run, there are only two posssible outcomes. Either it fails, or it does not. The probability of failing on each run is independent of other runs. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected number of trials for r sucesses is:

`E = (r)/(p)`

The standard deviation is:

`S = \sqrt{(r(1-p))/(p^(2))}`

A computer program has a bug that causes it to fail once in every thousand runs, on average.

This means that
`p = (1)/(1000) = 0.001`

In an effort to find the bug, independent runs of the program will be made until the program has failed five times.

This means that
`r = 5`

a. What is the mean number of runs required?

`E = (r)/(p) = (5)/(0.001) = 5000`

The mean number of runs required is 5000.

b. What is the standard deviation of the number of runs required?

`S = \sqrt{(5*(1-0.001))/((0.001)^(2))} = 2235`

The standard deviation of the number of runs required is 2235.