Question:

If \( g = 2 \), what is the minimum possible number of times for which the weighing machine is to be used?

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Very low sampling ability forces more decision layers—plan for confirmatory steps in your strategy.
Updated On: Jul 28, 2025
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The Correct Option is C

Solution and Explanation

With only 2 coins per box, we have very limited scope in each weighing. We again rely on binary logic.
To find 1 faulty box among 10 using binary decisions: \[ 2^n \geq 10 \Rightarrow n = 4 \text{ (ideal)} \] However, since we can use at most 2 coins per box, combinations for sampling reduce. Additional step is required for confirmation, making 5 weighings the minimum required.
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