system reliability, probabilities of rocket launch suspended A rocket has 3 computers (a,b,c). If a fails than b will take over and so on. Each computer has 0.01 failure rate. These 3 computers are linked to an expensive sensor. If this sensor fails than all 3 computers will fail, this sensor has a failure rate of 10^-6. What are the chances the rocket launch has to be suspended?
So i tried solving this by using the formulas of system reliability. i put the computers in a parrel and linked them in series to the sensor. 
I came up with the following formula P(s) = 1-(1-10^-6)*(1-0.01^3) = 0.000002
I have a feeling my answer is wrong though. Could anyone help me confirm this?
 A: Your answer looks correct.  My approach (which I think is the same as yours) would be: 


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*(ignoring the sensor) what is the probability a particular computer fails?


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*$0.01$


*(ignoring the sensor) what is the probability all three computers fail? 


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*$0.01^3 = 0.000001$ by independence 


*(ignoring the sensor) what is the probability at least one of the three computers does not fail?


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*$1- 0.000001=0.999999$


*what is the probability the sensor does not fail?


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*$1- 0.000001=0.999999$


*what is the probability the sensor does not fail and (ignoring the sensor) at least one of the three computers does not fail?


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*$0.999999 \times 0.999999 = 0.999998000001$ by independence 


*what is the probability either the sensor fails or all three computers fails?


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*$1- 0.999998000001 = 0.000001999999$ which is essentially your $0.000002$



joriki's questions in the comments are about checking whether ignoring the sensor and independence are correct interpretations of the question
