Question
An exhibition display can draw power through either of two routes. The main route has one component that works throughout the day with probability 90%. A spare route has a switch and a transformer in series: both must work, with probabilities 80% and 75% respectively.
All three components fail or work independently, and the routes share no hardware. The display stays powered if at least one entire route works. What is the probability it stays powered all day?
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Hint 1
The spare route works only if its switch and transformer both work.
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Solution
Let \(M\) mean the main component works, and \(S\) and \(T\) mean the spare switch and transformer work. The main route succeeds with probability 0.90. Because the spare components are independent, the spare route succeeds with probability \(0.80\times0.75=0.60\), so it fails with probability 0.40.
The main component is independent of both spare components, so the two route-success events are also independent. The display loses power only if both routes fail:
\[\Pr(\text{power lost})=(1-0.90)(1-0.60)=0.10\times0.40=0.04.\]
Therefore \(\Pr(\text{powered})=1-0.04=0.96\).
As a component-state check, both routes fail either when the main and spare switch fail (transformer arbitrary), probability \(0.10\times0.20=0.02\), or when the main fails, switch works and transformer fails, probability \(0.10\times0.80\times0.25=0.02\). The disjoint cases total 0.04. It would be wrong to treat the spare switch working on its own as sufficient for an alternative power route.
Assumptions
The three component working events are mutually independent.
Each probability refers to the component working for the entire specified day.
Either complete route suffices to power the display, with no shared failure point.
Final Answer
96%.