Answer:
Step-by-step explanation:
For each component, there are only two possible outcomes. Either it fails, or it does not. The components are independent. We want to know how many outcomes until r failures. The expected value is given by
[tex]E = \frac{r}{p}[/tex]
In which r is the number of failures we want and p is the probability of a failure.
In this problem, we have that:
r = 1 because we want the first failed unit.
[tex]p = 0.4[\tex]
So
[tex]E = \frac{r}{p} = \frac{1}{0.4} = 2.5[/tex]
The expected number of systems inspected until the first failed unit is 2.5
