The 90-second cap is real and they mean it.
First, clarify that 'at least 2 reds' means exactly 2 or exactly 3 reds, and compute the probability using combinations: P = [C(r,2)C(b+g,1) + C(r,3)] / C(r+b+g,3). Then, for the mental-math shortcut, use the binomial approximation with p = r/(r+b+g) and note that when the total number of balls is large relative to 3, the without-replacement draws are nearly independent, so P ≈ 3p²(1-p) + p³ = 3p² - 2p³.
Pro tip: Optiver values speed and intuition: state the exact formula but emphasize the approximation and its error bound (O(1/N)), showing you can balance rigor with practicality.
Restate that 'at least 2 reds' includes exactly 2 reds and exactly 3 reds, and that draws are without replacement.
Write the probability as [C(r,2)C(b+g,1) + C(r,3)] / C(r+b+g,3), explaining each term.
Let p = r/(r+b+g). For large N, approximate with replacement: P ≈ 3p²(1-p) + p³ = 3p² - 2p³.
Mention that the error is small when N is large (e.g., O(1/N)), and give a quick sanity check with numbers.
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