Solution: The diagonal of the rectangle is $ \sqrt{6^2 + 8^2} = 10 $ cm. The circumference is $ \pi \times \text{diameter} = 10\pi $ cm. The circumference of the throughfall is $\boxed{10\pi}$ cm.**Question:** A bag contains 7 red marbles and 8 blue marbles. If four marbles are drawn at random, what is the probability that exactly two are red?

["Understanding the Probability of Drawing Two Red and Two Blue Marbles", "When drawing marbles from a bag, calculating probability involves combinatorics—the science of counting combinations. In this scenario, we have a bag containing 7 red marbles and 8 blue marbles, for a total of 15 marbles. We randomly draw four marbles, and we want to find the probability that exactly two are red and two are blue.", "### Step 1: Total Possible Ways to Draw 4 Marbles\nThe total number of ways to choose any 4 marbles from 15 is given by the combination formula:\n[\n\binom{15}{4} = \frac{15!}{4!(15-4)!} = \frac{15 \ imes 14 \ imes 13 \ imes 12}{4 \ imes 3 \ imes 2 \ imes 1} = 1365\n]", "### Step 2: Favorable Outcomes: Exactly Two Red and Two Blue Marbles\nTo get exactly two red and two blue marbles:\n- Choose 2 red marbles from 7:\n[\n\binom{7}{2} = \frac{7 \ imes 6}{2} = 21\n]\n- Choose 2 blue marbles from 8:\n[\n\binom{8}{2} = \frac{8 \ imes 7}{2} = 28\n]\n- Multiply these to get the total favorable combinations:\n[\n21 \ imes 28 = 588\n]", "### Step 3: Compute the Probability\nProbability is favorable outcomes divided by total outcomes:\n[\nP(\ ext{exactly 2 red}) = \frac{588}{1365}\n]\nSimplify the fraction by dividing numerator and denominator by 21:\n[\n\frac{588 \div 21}{1365 \div 21} = \frac{28}{65}\n]", "### Final Answer\nThe probability that exactly two of the four drawn marbles are red is $\boxed{\frac{28}{65}}$.", "---", "This classic probability problem combines counting combinations and understanding conditional selection—essential tools for mastering data analysis and risk assessment in real-world decision-making. Whether in games, surveys, or operational planning, precise probability calculations ensure informed, evidence-based choices."]