Discussion
Which of the following would not help identify which of several units has failed?
Answer: B
Correct answer: Fitting a second unit
A second unit does not help identify which has failed. Two units allow a comparator to show that the outputs disagree, but the data contain nothing to say which of the two is wrong, so both must be treated as suspect.
The other three all help. A third unit makes the outlier identifiable, since it is the one disagreeing with both others; median voting among three performs that exclusion automatically; and comparison against the standby instruments provides an independent source to arbitrate between two disagreeing units. Detection needs two sources, isolation needs three.
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