a manned interplanetary space vehicle has four engines each with reliability .99

Question

a manned interplanetary space vehicle has four engines each with reliability .99. Each engine has a failure detection system which may itself fail. If the engine does fail there is a conditional probability of .02 that a success will be signalled. 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 signalled? (Hint: Calculate probability of no engine failures plus one detective failure.) answer=(0.9986; ignore second signal and pray)

Explanation / Answer

P( success signalled | engine failed ) = 0.02

P( failure ) = 0.01

Mission can succeed with 4 engines working and also 3 engine working provided the 4th one which failed is detected.

Now we know that , P( success signalled | engine failed ) = 0.02

Therefore, P( Failure signalled | engine failed ) = 1- P( success signalled | engine failed ) = 0.98

therefore probability of success of mission =

0.994 + 4* 0.993 * 0.01 * 0.98 = 0.9606 + 0.038 = 0.9986

In the above equation the first term represents probability of no engine failure and

second term represents probability of 1 engine failing and is being detected to have failed.

Now for second part if there is no abort system or escape system and 2 failures are signalled, we cant really do anything although there would be a higher probability that there must be 2 failures because the probability of detecting system failure in case of a failure is very high. Therefore we can just hope that one of the 2 signals was wrong and thus could be ignored.

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