Respuesta :
Answer:
a) [tex]P(X = x) = C_{3,x}.(0.15)^{x}.(0.85)^{3-x}[/tex]
b) [tex]E(X) = 0.45[/tex]
c) [tex]Var(x)=0.3825[/tex]
d) There is a 61.41% probability that the entire system is succesful.
e) There is a 0.34% probability that the system fails.
f) Three components are suficient.
Step-by-step explanation:
For each component, there are only two possible outcomes. Either it works, or it does not work. This means that we can solve this problem using concepts of the binomial probability distribution.
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 combinatios 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.
In this problem, we have that:
In a support system in the U.S. space program, a single crucial component works only 85% of the time. Consider the random variable X as the number of components out of 3 that fail. This means that p is the probability of a failure of a single component. So [tex]p = 0.15[/tex]. There are 3 components, so [tex]n = 3[/tex].
(a) Write out a probability function for the random variable X.
[tex]P(X = x) = C_{3,x}.(0.15)^{x}.(0.85)^{3-x}[/tex]
(b) What is E(X) (i.e., the mean number of components out of 3 that fail)?
[tex]E(X) = np = 3*0.15 = 0.45[/tex]
(c) What is Var(X)?
[tex]Var(X) = np(1-p) = 3*0.15*0.85 = 0.3825[/tex]
(d) What is the probability that the entire system is successful?
That is P(X = 0), that is, no component fails.
[tex]P(X = 0) = C_{3,0}.(0.15)^{0}.(0.85)^{3} = 0.6141[/tex]
There is a 61.41% probability that the entire system is succesful.
(e) What is the probability that the system fails?
The system only fails if all three component fails. So this is [tex]P(X = 3)[/tex].
[tex]P(X = 3) = C_{3,3}.(0.15)^{3}.(0.85)^{0} = 0.0034[/tex]
There is a 0.34% probability that the system fails.
(f) If the desire is to have the system be successful with probability 0.99, are three components sufficient? If not, how many are required?
With three components, there is only a 0.34% probability that the system fails. This probability is lower than 1%, so three components are suficient.
a) [tex]P(X=x) C3, x, (0.15)^x, (0.85)^3^-^x[/tex]
b) [tex]E(X) = 0.45[/tex]
c) [tex]Var(x) = 0.3825[/tex]
d) There is a 61.41% probability that the entire system is successful.
e) There is a 0.34% probability that the system fails.
f) Three components are sufficient.
Calculations and Paramters:
We can note that only two outcomes are possible, so we use the binomial distribution.
Hence, the probability that all would work successfully is
[tex]P(X=0) = C3,0.(0.15)^0.((0.85)^3[/tex]
= 0.6141.
Hence, there is a 61.41% probability that the entire system is successful.
Read more about probability here:
https://brainly.com/question/25870256
