Question: A box contains 7 green educational badges and 9 digital certificates. If three items are selected at random, what is the probability that exactly 2 are badges?
["Title: Probability Calculation: Selecting Badges and Certificates – A Step-by-Step Guide", "Meta Description:\nDiscover how to calculate the probability of selecting exactly 2 green educational badges and 1 digital certificate from a box containing 7 badges and 9 digital certificates when drawing 3 items at random.", "---", "When managing educational resources, understanding probability can greatly assist in planning distribution, inventory, or random selection processes. A common question arises: What is the probability that exactly 2 out of 3 randomly selected items from a box are green educational badges, given there are 7 badges and 9 digital certificates?", "In this SEO-optimized article, we break down the probability calculation using combinatorics, offering a clear and practical guide for educators, administrators, and data enthusiasts.", "Understanding the Problem", "We have:\n- Total green badges: 7\n- Total digital certificates: 9\n- Total items in the box: 7 + 9 = 16\n- We randomly select 3 items without replacement\n- We want the probability that exactly 2 are badges and 1 is a digital certificate", "Step 1: Total Ways to Choose 3 Items", "The total number of ways to choose any 3 items from 16 is given by the combination formula:", "[\n\binom{16}{3} = \frac{16!}{3!(16-3)!} = \frac{16 \ imes 15 \ imes 14}{3 \ imes 2 \ imes 1} = 560\n]", "Step 2: Ways to Choose Exactly 2 Badges and 1 Certificate", "- Ways to choose 2 badges from 7:\n[\n\binom{7}{2} = \frac{7 \ imes 6}{2 \ imes 1} = 21\n]", "- Ways to choose 1 certificate from 9:\n[\n\binom{9}{1} = 9\n]", "- Total favorable outcomes (2 badges + 1 certificate):\n[\n21 \ imes 9 = 189\n]", "Step 3: Calculating the Probability", "The probability ( P ) is the ratio of favorable outcomes to total possible outcomes:", "[\nP = \frac{189}{560}\n]", "This fraction can be simplified by dividing numerator and denominator by 7:", "[\nP = \frac{27}{80}\n]", "So, the probability that exactly 2 of the 3 selected items are badges is ( \frac{27}{80} ), or approximately 0.3375 (33.75%).", "Why This Matters", "Understanding such probabilities helps institutions:\n- Allocate resources efficiently\n- Assess fairness in random sampling\n- Model educational distribution and certification tracking", "Conclusion", "By applying basic combinatorial principles, we’ve calculated the precise probability of selecting exactly 2 badges and 1 digital certificate when drawing 3 items from the box. This method applies broadly to educational management and statistics, reinforcing the role of probability in data-driven decision-making.", "Keywords: probability probability calculation educational badges digital certificates chance selection combinatorics, educational resources management, statistical probability, classroom resource planning", "Read more: Best practices for distributing educational badges and certificates\nCompare: Badges vs Certificates — Choosing the best credential for your program", "---", "Note: Optimizing with relevant keywords ensures visibility for educators and data analysts searching for probability problems in educational settings."]
