For each device, there are only two possible outcomes. Either it fails, or it does not fail. The probability of a device failling is independent of other devices. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

A total of 15 devices will be used.

This means that $n = 15$

Assume that each device has a probability of 0.05 of failure during the course of the monitoring period.

This means that $p = 0.05$

What is the probability that one of the devices fail?

This is $P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{15,1}.(0.05)^{1}.(0.95)^{14} = 0.3658$

36.58% probability that one of the devices fail

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3 years ago