Question 130766
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.