Let's solve this AMC probability problem step by step. We have a bag containing 5 red marbles, 4 blue marbles, and 3 green marbles. We need to find the probability that when drawing 3 marbles simultaneously, all three are red.
First, let's count the total number of marbles. We have 5 red plus 4 blue plus 3 green, which equals 12 marbles total. To find the total number of ways to choose 3 marbles from 12, we use the combination formula: 12 choose 3, which equals 12 factorial divided by 3 factorial times 9 factorial. This simplifies to 12 times 11 times 10, divided by 3 times 2 times 1, which equals 1320 divided by 6, giving us 220 total ways.
Now we need to find how many ways we can choose 3 red marbles from the 5 red marbles available. Using the combination formula again, we calculate 5 choose 3, which equals 5 factorial divided by 3 factorial times 2 factorial. This simplifies to 5 times 4 times 3, divided by 3 times 2 times 1, which equals 60 divided by 6, giving us 10 different ways to choose 3 red marbles.
Now we can calculate the probability. The probability of drawing 3 red marbles equals the number of favorable outcomes divided by the total number of outcomes. This gives us 10 divided by 220, which simplifies to 1 over 22. Therefore, the probability that all three marbles drawn are red is 1 twenty-second, or approximately 4.5 percent.
Let's summarize our solution. We found that there are 12 total marbles, 220 total ways to choose 3 marbles, and 10 ways to choose 3 red marbles. Therefore, the probability is 10 over 220, which simplifies to 1 over 22, or approximately 4.55 percent. The final answer is one twenty-second. This type of combinatorics and probability problem is very common in AMC competitions and tests your understanding of combinations and basic probability principles.