A box contains 4 red, 5 blue, and 6 green marbles. Two marbles are drawn without replacement. What is the probability that both are green?
["Title: Probability of Drawing Two Green Marbles: A Complete Guide", "---", "When it comes to probability problems involving marbles, understanding combinations is key—especially when drawing without replacement. This article explores a classic probability question: What is the probability of drawing two green marbles in succession from a box containing 4 red, 5 blue, and 6 green marbles?", "---", "### Understanding the Problem", "You start with a total number of marbles:", "- Red: 4\n- Blue: 5\n- Green: 6", "Total marbles = 4 + 5 + 6 = 15 marbles", "We are drawing two marbles without replacement—meaning once a marble is drawn, it is not replaced. The goal is to compute the probability that both marbles are green.", "---", "### Step-by-Step Probability Calculation", "Step 1: Probability of drawing a green marble first\nThere are 6 green marbles out of 15 total marbles.\n[\nP(\ ext{First green}) = \frac{6}{15}\n]", "Step 2: Probability of drawing a second green marble\nAfter drawing one green marble, one green marble is removed. Now, only 5 green marbles remain, and total marbles decrease to 14.\n[\nP(\ ext{Second green | First green}) = \frac{5}{14}\n]", "Step 3: Multiply the probabilities\nSince these are dependent events (drawing without replacement), multiply the probabilities:\n[\nP(\ ext{Both green}) = \frac{6}{15} \ imes \frac{5}{14}\n]", "Simplify:\n[\n\frac{6}{15} = \frac{2}{5}, \quad \frac{5}{14} = \frac{5}{14}\n]\n[\nP = \frac{2}{5} \ imes \frac{5}{14} = \frac{10}{70} = \frac{1}{7}\n]", "---", "### Final Answer", "The probability that both marbles drawn without replacement are green is (\frac{1}{7}), or approximately 0.1429 (14.29%).", "---", "### Bonus: Development via Combinations", "Alternatively, you can compute this using combinations (combinations count unordered selections).", "- Total ways to choose any 2 marbles from 15:\n[\n\binom{15}{2} = \frac{15 \ imes 14}{2} = 105\n]", "- Ways to choose 2 green marbles from 6:\n[\n\binom{6}{2} = \frac{6 \ imes 5}{2} = 15\n]", "- Probability both green:\n[\nP = \frac{15}{105} = \frac{1}{7}\n]", "This confirms the earlier result.", "---", "### Why This Matters", "This type of probability reasoning is essential in statistics, gaming, quality control, and decision-making under uncertainty. Whether you’re analyzing dice rolls, card draws, or defective parts in manufacturing, understanding sampling without replacement helps build intuition and solve real-world problems.", "---", "Keywords: probability two green marbles draw without replacement, probability marble problem, marbles probability calculation, 4 red 5 blue 6 green marbles, conditional probability, how to calculate combined probability", "---", "Summary:\nDrawing two green marbles from a bag with 4 red, 5 blue, and 6 green marbles has a probability of (\frac{1}{7}) when drawn without replacement, calculated through sequential probability or combinations.", "---", "Start mastering probability today—one marble at a time!"]
