A bag contains 5 red, 7 blue, and 8 green marbles. Two marbles are drawn without replacement. What is the probability both are green?
["Understanding the Probability of Drawing Two Green Marbles: A Step-by-Step Guide", "When faced with a probability question involving marbles, clarity and a methodical approach make all the difference. A classic example is determining the chance of drawing two green marbles from a bag containing marbles of different colors—specifically, a bag holding 5 red, 7 blue, and 8 green marbles—without replacement. This article breaks down the solution step-by-step, making it easy to understand how to calculate the probability of drawing two green marbles in succession.", "### The Setup: Marbles in the Bag", "Let’s start with the given quantities:", "- Red marbles: 5\n- Blue marbles: 7\n- Green marbles: 8", "The total number of marbles is:\n[\n5 + 7 + 8 = 20\n]", "### Defining the Event", "We are drawing two marbles without replacement, meaning once a marble is drawn, it is not returned to the bag. We want both marbles to be green. This is a classic problem of conditional probability, where one event affects the second.", "### Probability of First Green Marble", "The probability of drawing a green marble first is the number of green marbles divided by the total number of marbles:", "[\nP(\ ext{First green}) = \frac{8}{20}\n]", "We simplify:\n[\n\frac{8}{20} = \frac{2}{5}\n]", "### Probability of Second Green Marble (After First Green is Drawn)", "Since the first marble drawn is not replaced, the total number of marbles decreases by one, and so does the count of green marbles:", "- Green marbles left: (8 - 1 = 7)\n- Total marbles left: (20 - 1 = 19)", "So, the probability of drawing a second green marble is:", "[\nP(\ ext{Second green} \mid \ ext{First green}) = \frac{7}{19}\n]", "### Calculating Combined Probability", "Since both events must occur (first green, then green again), we multiply the probabilities using the multiplication rule for dependent events:", "[\nP(\ ext{Both green}) = P(\ ext{First green}) \ imes P(\ ext{Second green} \mid \ ext{First green}) = \frac{8}{20} \ imes \frac{7}{19}\n]", "Simplify:\n[\n\frac{8 \ imes 7}{20 \ imes 19} = \frac{56}{380}\n]", "Reduce the fraction by dividing numerator and denominator by 4:", "[\n\frac{56 \div 4}{380 \div 4} = \frac{14}{95}\n]", "### Final Answer", "[\n\boxed{\frac{14}{95} \approx 0.1474} \quad \ ext{or about } 14.74%\n]", "This means there’s a 14.74% chance that both marbles drawn are green when six marbles are randomly selected without replacement from the bag.", "### Why This Matters", "Understanding such probabilities helps in fields ranging from statistics and game theory to quality control and risk assessment. It reflects the principle that earlier draws reduce the pool, altering likelihoods—making absence significant in sequence-based draws.", "By applying step-by-step reasoning and conditional probability, even complex marble probability problems become approachable and instructive. Whether for study, math competitions, or everyday curiosity, mastering these fundamentals builds strong analytical habits."]
