Question

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?

Answer 1

(a) The probability function for the random variable X is P(X = x) = $^{3}C_{x}(0.15)^x(0.85)^{3-x}$ at X=0,1,2 and 3 otherwise it is 0.

(b) The mean number of components out of 3 that fail, E[X] is 0.45.

(c) The var(X) is 0.3825.

(d) The probability that the entire system is successful, is 0.9966.

(e) The probability that the system fails is 0.0034.

(a) In the given question we have to write out a probability function for the random variable X.

Let X denote number of component out of 3 that fail. SO the random variable X can take 0, 1, 2 and 3.

Here the components act independently and that they equivalent in the sense that all 3 of them have an 85% success rate.

So the failure rate, 1-0.85 = 0.15

Thus X follows binomial distribution with parameters n=3 and p=0.15.

The probability function for the random variable X is

P(X = x) = $^{3}C_{x}(0.15)^x(0.85)^{3-x}$ at X=0,1,2 and 3 otherwise it is 0.

(b) Now we have to find E[X], the mean number of components out of 3 that fail.

E[X] = np

E[X] = 3*0.15

E[X] = 0.45

(c) Now we have to find the var(X).

var(X) = np(1-p)

var(X) = 3*0.15p(1-0.15)

var(X) = 0.3825

(d) Now we have to find the probability that the entire system is successful.

P(at least one component will work) = 1-P(All the component fail)

P(at least one component will work) = 1-P(X = 3)

P(at least one component will work) = 1-$^3C_{3}(0.15)^3(0.85)^{3-3}$

P(at least one component will work) = 1-0.0034

P(at least one component will work) = 0.9966

(e) Now we have to find the probability that the system fails.

P(All the component fail) = P(X = 3)

P(All the component fail) = $^3C_{3}(0.15)^3(0.85)^{3-3}$

P(All the component fail) = 0.0034

To learn more about probability link is here

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