Respuesta :
Answer:
36.58% probability that one of the devices fail
Step-by-step explanation:
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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
A total of 15 devices will be used.
This means that [tex]n = 15[/tex]
Assume that each device has a probability of 0.05 of failure during the course of the monitoring period.
This means that [tex]p = 0.05[/tex]
What is the probability that one of the devices fail?
This is [tex]P(X = 1)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{15,1}.(0.05)^{1}.(0.95)^{14} = 0.3658[/tex]
36.58% probability that one of the devices fail
