A manned interplanetary space vehicle has four engines each with reliability 0 9

Question

A manned interplanetary space vehicle has four engines each with reliability 0 99. Each engine has a failure detection system which may itself fail. If the engine does fail there is a conditional probability of 0.02 that a success will be signaled. If an engine fails and is not detected the result is catastrophic. However, the mission can be completed with three engines if one engine fails and is detected. What is the probability of mission success? If there is no abort system or escape system, what action would you recommend if two failures are signaled? (Calculate probability of no engine failures plus one detected failure.)

Explanation / Answer

Probability of no engine failore = 0.99*0.99*0.99*0.99 =0.96059

conditional probability of one decated failoure= 0.02

probability of success = .96059 +0.02=.98059

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