In a support system in the U.S. space program, a single crucial component works only 85% of the time. In order to enhance the reliability of the system, it is decided that 3 components will be installed in parallel such that the system fails only if they all fail. Assume the components act independently and that they are equivalent in the sense that all 3 of them have an 85% success rate. Consider the random variable X as the number of components out of 3 that fail. (a) Write out a probability function for the random variable X. (b) What is E(X) (i.e., the mean number of components out of 3 that fail)? (c) What is Var(X)? (d) What is the probability that the entire system is successful? (e) What is the 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?

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 distributionThe 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.