For each component, there are only two possible outcomes. Either it fails, or it does not fail. Components perform independently. 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.

Assume that a component passes a test is 0.85

So they fail with probability of $p = 1 - 0.85 = 0.15$

What is the probability that the third failure will occur on the tenth component tested

First 9 components: Two failures, that is, P(X = 2) when n = 9.

10th component: Failure with probability 0.15.

$P = 0.15P(X = 2)$

In which

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

$P(X = 2) = C_{9,2}.(0.15)^{2}.(0.85)^{7} = 0.2597$

$P = 0.15P(X = 2) = 0.15*0.2597 = 0.0390$

3.90% probability that the third failure will occur on the tenth component tested

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3 years ago