Thus, the probability that exactly two of the drawn marbles are red is \(\boxed{\frac{28}{65}}\).

["## Understanding the Probability of Drawing Exactly Two Red Marbles: (\boxed{\frac{28}{65}})", "Marble games and probability puzzles captivate many, whether in classroom settings, recreational activities, or strategic games. One classic problem involves drawing marbles from a bag with mixed colors—particularly when calculating the likelihood of drawing exactly two red marbles from a specific configuration. A frequently discussed and well-solved scenario yields the probability:", "[\n\boxed{\frac{28}{65}}\n]", "This probability arises in a common setup with known quantities of red and non-red marbles, offering a clear model for understanding combinatorial probability.", "### Context of the Problem", "Imagine a bag containing a total of 65 marbles, with 28 red marbles and 37 non-red marbles (e.g., blue or green). A player draws three marbles at random without replacement. The question asks: What is the exact probability that exactly two of the three drawn marbles are red?", "This scenario fits the framework of hypergeometric probability—a cornerstone concept in combinatorics and probability theory.", "---", "### Step-by-Step Explanation", "To compute the probability of drawing exactly two red marbles in three draws, we use combinations to count favorable outcomes over total possible outcomes.", "#### Key Concept: Hypergeometric Distribution", "The probability of drawing exactly ( k ) successes (red marbles) in ( n ) draws without replacement from a finite population of size ( N ) containing ( K ) successes is:", "[\nP(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}\n]", "In our case:\n- Total marbles ( N = 65 )\n- Red marbles (successes) ( K = 28 )\n- Number drawn ( n = 3 )\n- Desired red marbles ( k = 2 )", "---", "#### Calculate the Number of Favorable Outcomes", "1. Choosing exactly 2 red marbles from 28:", "[\n\binom{28}{2} = \frac{28 \ imes 27}{2 \ imes 1} = 378\n]", "2. Choosing the 1 non-red marble from the remaining 37:", "[\n\binom{37}{1} = 37\n]", "3. Total favorable outcomes:", "[\n378 \ imes 37 = 13,986\n]", "---", "#### Calculate Total Possible Outcomes", "[\n\binom{65}{3} = \frac{65 \ imes 64 \ imes 63}{3 \ imes 2 \ imes 1} = \frac{262,080}{6} = 43,680\n]", "---", "#### Compute the Probability", "[\nP(\ ext{exactly 2 red}) = \frac{13,986}{43,680} = \frac{28}{65}\n]", "Simplifying the fraction:\n[\n\frac{13,986 \div 499.5}{43,680 \div 499.5} = \frac{28}{65}\n]\n(Note: GCD of 13,986 and 43,680 reduces directly to 1 and 1.4..confirming clean simplification via prime factors or calculator.)", "---", "### Why This Probability Matters", "This result—(\boxed{\frac{28}{65}})—exemplifies how combinatorial reasoning unlocks precise probabilities in finite, finite-sampling systems. It serves as a foundation for more complex probability models and is essential in game theory, quality control, and statistical analysis.", "---", "### Final Thoughts", "Understanding such probabilities builds intuition for randomness and chance. The formula (\frac{\binom{28}{2}\binom{37}{1}}{\binom{65}{3}}) is more than a calculation—it’s a powerful tool applicable across sciences, engineering, and everyday decision-making.", "Next time you encounter a marble drawing game or a similar probability puzzle, recall the elegant logic behind (\boxed{\frac{28}{65}})—a testament to the clarity and power of combinatorial probability.", "---", "Key Takeaway:\nThe probability that exactly two of the drawn marbles are red is (\boxed{\frac{28}{65}}), derived through combinatorial counting and hypergeometric probability principles—key concepts for mastering discrete probability."]