A box contains 5 red, 4 blue, and 3 green balls. Two balls are drawn at random without replacement. What is the probability both are red?
["Understanding the Probability of Drawing Two Red Balls Without Replacement", "In combinatorics and probability theory, calculating the chance of drawing red balls from a set is a classic example of conditional probability. Consider a box containing 5 red, 4 blue, and 3 green balls—totaling 12 balls. If two balls are drawn at random without replacement, what is the probability that both balls drawn are red?", "### Total Balls and Basic Setup", "The total number of balls in the box is:\n5 (red) + 4 (blue) + 3 (green) = 12 balls", "We want to calculate the probability of drawing two red balls consecutively, without putting any ball back into the box after the first draw.", "### Step-by-step Probability Calculation", "Step 1: Probability the first ball is red", "There are 5 red balls out of 12 total balls.\nSo,\n[\nP(\ ext{First ball red}) = \frac{5}{12}\n]", "Step 2: Probability the second ball is red (given the first was red)", "After removing one red ball, only 4 red balls remain, and total balls drop to 11.\nSo,\n[\nP(\ ext{Second ball red} \mid \ ext{First red}) = \frac{4}{11}\n]", "Step 3: Multiply the probabilities", "Since the draws are dependent (without replacement), the joint probability is the product:\n[\nP(\ ext{Both red}) = \frac{5}{12} \ imes \frac{4}{11} = \frac{20}{132} = \frac{5}{33}\n]", "### Final Answer", "The probability that both balls drawn at random without replacement are red is:\n[\n\boxed{\frac{5}{33}}\n]", "This result illustrates how conditional probability works in sampling without replacement—each draw affects the next, reducing the chance of repeated successes.", "---", "Whether you’re studying probability fundamentals or applying it in game theory, this classic problem reinforces key concepts essential for statistical reasoning and risk assessment in various fields."]
