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Home » Expected number = 3.5% of 100 = 0.035 × 100 = 3.5 - 500apps

Expected number = 3.5% of 100 = 0.035 × 100 = <<0.035*100=3.5>>3.5 - 500apps

📅 March 13, 2026 👤 scraface

Mar 13, 2026

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📌 5Question: A volcanologist monitors 4 active volcanoes, each of which can erupt in one of 3 distinct intensity levels: low, medium, or high. If the eruptive behavior of each volcano is independent and the order of eruption does not matter, how many distinct combinations of eruption profiles can be observed?

📌 Solution: Each of the 4 volcanoes independently exhibits one of 3 eruption intensities: low, medium, or high. Since the volcanoes are distinguishable (due to different locations), but the eruption *profile* (i.e., the multiset of intensities) only considers counts of each type, and the volcanoes are distinguishable, we are counting the number of 4-tuples where each element is from a 3-element set (low, medium, high), and the order does **not** matter in terms of labeling—wait, correction: since each volcano is a distinct entity (e.g., monitored individually), the classification is based on assigning an intensity to each volcano, and even though eruptive profiles are unordered in reporting, the underlying assignment to specific volcanoes **is** tracked. Therefore, we are counting **functional mappings** from 4 distinguishable volcanoes to 3 intensity categories, **with repetition allowed**, and **order of assignment does not affect group counts**—but since volcanoes are distinguishable, each different assignment is unique unless specified otherwise.

📌 However, the key phrase is: "the eruptive behavior... can erupt in one of 3 distinct intensities" and "combinations of eruption profiles", with *order not matters*—this suggests we are counting **multisets** of eruption types assigned to volcanoes, but since volcanoes are distinct, it's better interpreted as: we assign to each volcano one intensity level, and although the profile is unordered in presentation, the underlying assignment is specific. Thus, the total number of assignments is simply $3^4 = 81$, since each volcano independently chooses one of 3 levels.

📌 But "distinct combinations of eruption profiles" where profile means the multiset of intensities (regardless of volcano identity) would be different—yet the context implies monitoring individual volcanoes, so a profile includes which volcano has which level. But since the question says "combinations... observed" with vertices monitored (distinguishable), and no specification of symmetry-breaking, standard interpretation in such combinatorics problems is that labeled objects are distinguished.

📌 Thus, each volcano independently chooses one of 3 levels, so total distinct ordered assignments: $3^4 = 81$. But if profiles are considered unordered across volcanoes (e.g., just the count distribution), then we must count the number of integer solutions to $x_L + x_M + x_H = 4$, $x_i \geq 0$, which is $\binom{4 + 3 - 1}{3 - 1} = \binom{6}{2} = 15$.

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