Question 130766: I need help with this homework question ASAP. It is not from textbook. it is for the probability and statistics class.
can i solve this question this way:
P(X = 3) = since each component has probability of functioning as 0.9. the probability of not functioning is 1-0.9 = 0.1.
so 3 system out of 5 will have = (0.9)(0.9)(0.9) = 0.729 ? is this correct for part a?
b. atleast 3 out of 5 will function. is this correct?
c. (3)(0.9) = 2.7?
question: A k out of n system is one in which there is a group of n components, and the system will function if at least k of the components function. Assume the components function independently of one another.
1). In 3 out 5 system, each component has probability 0.9 of functioning. What is the probability that the system will function?
2). Based on the information given in 1), what is the expected number of components that function.
3). In a 3 out n system, in which each component has probability 0.9 of functioning, what is the smallest values of n needed so that the probability that the system functions is at least 0.90?
I really appreciate the help with homeworks that you guys provide.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! P(X = 3) = since each component has probability of functioning as 0.9. the probability of not functioning is 1-0.9 = 0.1.
so 3 system out of 5 will have = (0.9)(0.9)(0.9) = 0.729 ? is this correct for part a?
P(3 successes in 5 trials) = 5C3(0.9)^3(0.1)^2 = 0.0729
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b. at least 3 out of 5 will function.
P(at least 3 out of 5) = 1 - binomcdf(5,0.9,2) = 0.99144
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c. (3)(0.9) = 2.7?
I don't understand your "c" question.
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question: A k out of n system is one in which there is a group of n components, and the system will function if at least k of the components function. Assume the components function independently of one another.
1). In 3 out 5 system, each component has probability 0.9 of functioning. What is the probability that the system will function?
Prob = 1-binomcdf(5,0.9,2) =
2). Based on the information given in 1), what is the expected number of components that function.
Expected value is the mean which is np for binomial distributions.
Expected value = 5*0.9 = 4.5
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3). In a 3 out n system, in which each component has probability 0.9 of functioning, what is the smallest values of n needed so that the probability that the system functions is at least 0.90?
We know 3 out of 5 is 0.99
Try 3 out of 4: 1-binomcdf(4,0.9,2)= 0.9477
Try 3 out of 3: 1-binomcdf(3,0.9,2)= 0.729
Looks like n=4 will do the job.
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Cheers,
Stan H.
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