Question: A transportation analyst is analyzing a network with 6 red buses, 4 green buses, and 5 blue buses. If she randomly selects 5 buses for maintenance, what is the probability that she selects at least two buses of each color?

["Title: Probability of Selecting at Least Two Buses of Each Color: A Combinatorial Analysis on a Mixed Fleet Network", "Meta Description:\nExplore the probability of selecting at least two red, green, and blue buses when randomly choosing 5 vehicles from a fleet of 6 red, 4 green, and 5 blue buses. Learn the combinatorial method to solve this real-world transportation challenge.", "---", "### Introduction", "In urban transit systems, maintaining vehicle fleets efficiently is crucial for reliability and safety. Imagine a transportation analyst tasked with selecting 5 buses for scheduled maintenance from a mixed fleet: 6 red buses, 4 green buses, and 5 blue buses. The key question arises: What is the probability that among the selected 5 buses, at least two are red, at least two are green, and at least two are blue?", "At first glance, the condition “at least two of each color” appears impossible with only 5 buses selected—since having at least two buses of three different colors totals at least 6 buses (2+2+2). This immédiatly raises a red flag: Is it even possible to select 5 buses and have at least two of each color?", "This article explores the combinatorics behind this question, clarifies the constraints, and explains why the desired outcome is combinatorially impossible—but also educates on how such probabilities are calculated when valid conditions hold.", "---", "### Understanding the Fleet Composition", "Let’s summarize the fleet:", "- Red buses: 6\n- Green buses: 4\n- Blue buses: 5\n- Total buses: 6 + 4 + 5 = 15 buses", "The analyst picks 5 buses at random.", "---", "### Analyzing the Core Question: At Least Two of Each Color", "To select at least two buses of each color among 5 buses requires a minimum of:", "- 2 red\n- 2 green\n- 2 blue\n- Total: at least 6 buses", "But only 5 buses are selected. Therefore, it is impossible to satisfy the condition of having at least two buses of each color in a sample of only 5.", "Thus, the event “selecting at least two buses of each color” has zero favorable outcomes.", "👉 Conclusion:\nThe probability is 0.", "---", "### When Does the Condition “At Least Two of Each” Make Sense?", "This scenario becomes meaningful only when selecting at least 6 buses—so that the minimum requirement of 2 per color fits within 5 selections (which it doesn’t). Alternatively, if the analyst selected 6 or more buses, the probability of having at least two of each color could be computed using hypergeometric distribution.", "For completeness, let’s briefly describe how this would be calculated with a valid example:", "Suppose the analyst selected 6 buses instead of 5. Then, we compute the probability of choosing at least 2 red, at least 2 green, and at least 2 blue buses.", "This involves summing the probabilities of valid combinations such as:", "- 2 red, 2 green, 2 blue\n- 3 red, 2 green, 1 blue — but only if all counts ≥ 2\n- 2 red, 3 green, 1 blue — again invalid\n- etc.", "Only combinations satisfying ≥2 for each color within the total sample size are counted.", "---", "### Mathematical Model: Hypergeometric Probability", "For a realistic version of the problem, suppose the selection is 6 buses from the 15. The probability of selecting at least two red, two green, and two blue buses is:", "[\nP = \sum \frac{\binom{R}{r} \binom{G}{g} \binom{B}{b}}{\binom{15}{6}}\n]", "where:\n- ( R \geq 2, G \geq 2, B \geq 2 ), and\n- ( r + g + b = 6 ), ( r \leq 6, g \leq 4, b \leq 5 )", "Only combinations like ( (2,2,2) ), or ( (3,2,1) ) are invalid if any count is less than 2.", "Valid case: exactly (2,2,2)", "[\n\binom{6}{2} \binom{4}{2} \binom{5}{2} = 15 \ imes 6 \ imes 10 = 900\n]", "Total ways:\n[\n\binom{15}{6} = 5005\n]", "So,\n[\nP = \frac{900}{5005} \approx 0.1798 \ ext{ or } 17.98%\n]", "But this is for 6 buses — not 5.", "---", "### Why 5 Buses Make the Event Impossible", "Given only 5 buses selected, maximum colors possible is 3 — but no combination of 5 buses can include two from each of three colors due to the arithmetic constraint.", "Mathematically, the problem violates a fundamental pigeonhole principle: requiring at least 2 from each of 3 groups within only 5 items is classically infeasible.", "---", "### Final Answer and Takeaway", "The probability of selecting at least two red, two green, and two blue buses when choosing only 5 buses from a fleet of 6 red, 4 green, and 5 blue buses is 0.", "This makes intuitive and mathematical sense — the event cannot occur due to insufficient total selections.", "However, this problem exemplifies core principles in transportation planning and statistical analysis: combinatorics underpinning risk assessment, maintenance scheduling, and resource allocation. Understanding feasibility is as important as calculating probability.", "For planners and analysts, verifying constraints before performing probability calculations prevents misleading conclusions — especially when real-world operations demand precise feasibility checks.", "---", "### Key Takeaways", "- The condition “at least two of each color” requires 6 or more buses.\n- Selecting only 5 buses makes the event impossible.\n- Combinatorial analysis confirms zero favorable outcomes.\n- Real-world applications benefit from validating selection constraints before computation.", "---", "Related Keywords:\nprobability of selecting buses, transportation maintenance scheduling, combinatorics in logistics, hypergeometric distribution bus selection, fleet maintenance analysis, identifying impossible events in analytics", "---", "Want to calculate realistic probabilities for bus maintenance?\nUse computational combinatorics or software (e.g., Python, R) with hypergeometric functions to simulate valid sample spaces.", "---", "By mastering such foundational probability concepts, transportation analysts ensure efficient, reliable, and data-driven fleet management."]