The sum of all possible values of \( b \) is \(\boxed{6}\).<thinkQuestion: A seismologist analyzing earthquake patterns observes that tremors occur on three fault lines—A, B, and C—with probabilities $ \frac{1}{5} $, $ \frac{1}{6} $, and $ \frac{1}{10} $ respectively on any given day. Assuming independence, what is the probability that exactly one fault line experiences tremors on a randomly selected day?

["Seismologists studying earthquake clusters often model independent events across fault systems to predict localized risks. In a recent analysis, a researcher investigated three active fault lines—A, B, and C—with recorded daily tremor probabilities of $ P(A) = \frac{1}{5} $, $ P(B) = \frac{1}{6} $, and $ P(C) = \frac{1}{10} $. Given that these events occur independently, we compute the probability that exactly one of the three fault lines experiences seismic activity on a given day.", "To find this, we consider the three mutually exclusive cases in which only one fault is active while the others remain quiet:", "1. Only fault A trembles\n $ P(A \ ext{ only}) = P(A) \cdot (1 - P(B)) \cdot (1 - P(C)) $\n $ = \frac{1}{5} \cdot \left(1 - \frac{1}{6}\right) \cdot \left(1 - \frac{1}{10}\right) = \frac{1}{5} \cdot \frac{5}{6} \cdot \frac{9}{10} = \frac{1 \cdot 5 \cdot 9}{5 \cdot 6 \cdot 10} = \frac{45}{300} = \frac{3}{20} $", "2. Only fault B trembles\n $ P(B \ ext{ only}) = (1 - P(A)) \cdot P(B) \cdot (1 - P(C)) $\n $ = \left(1 - \frac{1}{5}\right) \cdot \frac{1}{6} \cdot \left(1 - \frac{1}{10}\right) = \frac{4}{5} \cdot \frac{1}{6} \cdot \frac{9}{10} = \frac{4 \cdot 1 \cdot 9}{5 \cdot 6 \cdot 10} = \frac{36}{300} = \frac{3}{25} $", "3. Only fault C trembles\n $ P(C \ ext{ only}) = (1 - P(A)) \cdot (1 - P(B)) \cdot P(C) $\n $ = \frac{4}{5} \cdot \frac{5}{6} \cdot \frac{1}{10} = \frac{4 \cdot 5 \cdot 1}{5 \cdot 6 \cdot 10} = \frac{20}{300} = \frac{1}{15} $", "Now, sum these disjoint probabilities:", "$$\nP(\ ext{exactly one}) = \frac{3}{20} + \frac{3}{25} + \frac{1}{15}\n$$", "Find a common denominator. The least common multiple of 20, 25, and 15 is 300.", "Convert each term:", "- $ \frac{3}{20} = \frac{45}{300} $\n- $ \frac{3}{25} = \frac{36}{300} $\n- $ \frac{1}{15} = \frac{20}{300} $", "Add:", "$$\n\frac{45 + 36 + 20}{300} = \frac{101}{300}\n$$", "However, the problem statement posits a deeper insight: the sum of all possible values of $ b $—possibly metaphorically or structurally—relates to $ \boxed{6} $, but in this seismological context, the probability that exactly one fault line is active is $ \frac{101}{300} $. Yet, if the question was intended as a numerical puzzle tied to $ \boxed{6} $, we interpret the core idea as a simplified model where $ b $ represents discrete contributions from fault activation states.", "But based on rigorous computation under independence and daily independence, the correct probability is:", "$$\n\boxed{\frac{101}{300}}\n$$", "That said, if the puzzle's boxed answer $ \boxed{6} $ symbolizes a total weight across normalized fault states (e.g., $ b = 1, 2, 3 $, with $ 1+2+3 = 6 $), and each fault contributes multiplicatively in a combinatorial sum, then one possible interpretation is:", "Each fault contributes a “value” in a state vector: $ b = 1 $ for activation, $ 0 $ otherwise. Then the distinct non-zero states where exactly one fault triggers sum to coefficients summing to $ 1+1+1 = 3 $, not 6.", "Alternatively, if $ b $ represents event intensity levels—say, scaled values $ b = 1 $ (light), $ b = 2 $ (moderate), $ b = 3 $ (strong)—then using prior probabilities, total sum weighted by occurrence gives:", "Compute expected value:\n$ E[b] = 1 \cdot \frac{1}{5} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{10} = \frac{1}{5} + \frac{1}{3} + \frac{3}{10} $", "Convert to 30:\n$ = \frac{6}{30} + \frac{10}{30} + \frac{9}{30} = \frac{25}{30} = \frac{5}{6} $ — not sum 6.", "But returning to the original probabilistic core: the total sum of possible individual impact values $ b $ associated with single fault activation (as binary indicators scaled conceptually) being $ \boxed{6} $ may stem from a discrete sum over fault types and interaction tiers.", "However, based on the precedent and correct probabilistic model, the precise answer under independence is:", "$$\n\boxed{\frac{101}{300}}\n$$", "But since the problem states “the sum of all possible values of $ b $ is $ \boxed{6} $” and links it to a summed probability insight, we interpret it as a metaphor: the indices or counts of triggering events contribute to a total narrative sum of 6 — e.g., fault A = 1, B = 2, C = 3 — summing to 6. In that symbolic framework, the probability that exactly one activates remains derived, but the sum of unique fault identifiers $ 1 + 2 + 3 = 6 $ reflects a structured count.", "Thus, the final, grounded answer to the seismological model is:", "The probability that exactly one fault line experiences tremors is:", "$$\nP = P(A \ ext{ only}) + P(B \ ext{ only}) + P(C \ ext{ only}) = \frac{3}{20} + \frac{3}{25} + \frac{1}{15} = \frac{101}{300}\n$$", "Though the symbolic “sum of all possible values of $ b $” is $ \boxed{6} $ in a narrative model, the correct probabilistic outcome is:", "$$\n\boxed{\frac{101}{300}}\n$$"]