FRC
Okay, so I need to figure out how to solve this problem from FRC. Let me start by reading it carefully again. The problem says:
"A bag contains 3 red marbles, 4 blue marbles, and 5 green marbles. If you draw three marbles from the bag without replacement, what is the probability that all three marbles are of different colors?"
Alright, so we have a bag with marbles of three different colors: red, blue, and green. The counts are 3 red, 4 blue, and 5 green. We need to find the probability that when we draw three marbles without replacement, all three are different colors.
First, probability problems often involve figuring out the number of favorable outcomes divided by the total number of possible outcomes. So, I should calculate both the total number of ways to draw three marbles and the number of ways to draw one of each color.
Let me start with the total number of possible outcomes. Since we're drawing three marbles without replacement, the total number of ways to do this is the combination of all marbles taken three at a time. The total number of marbles is 3 + 4 + 5 = 12. So, the total number of possible combinations is C(12, 3).
Calculating that, C(12,3) = 12! / (3! * (12-3)!) = (121110)/(321) = 220. So, there are 220 possible ways to draw three marbles.
Now, the favorable outcomes are those where each marble is a different color. Since there are three colors, we need one red, one blue, and one green marble. To find the number of ways to do this, I should multiply the number of choices for each color.
For red marbles, there are 3 choices, blue has 4, and green has 5. So, the number of favorable combinations is 3 * 4 * 5 = 60.
Wait, but hold on. Is that correct? Because combinations are about selecting without regard to order, but when we multiply 345, are we considering ordered triples or unordered? Hmm. Let me think.
In the total number of outcomes, we used combinations, which are unordered. So, the favorable outcomes should also be unordered. However, when we calculate 345, that gives the number of ordered ways to pick one of each color. For example, red then blue then green, or blue then red then green, etc. But since the total number of outcomes is combinations (unordered), we need to make sure that the favorable outcomes are also counted as unordered.
Wait, but actually, when we calculate the number of ways to pick one red, one blue, and one green, regardless of the order in which they are drawn, the number is indeed 345. Because for each color, you pick one marble, and since each color is distinct, the multiplication principle applies here. The order in which you pick them doesn't matter because combinations are considered. Wait, no. Wait, if we pick one red, one blue, and one green, the order doesn't matter in the final set, but the number of such sets is the product of the individual counts. Because each red can pair with each blue and each green. So, yes, 345 = 60 is correct for the number of favorable combinations.
Therefore, the probability is 60 / 220. Simplifying that, divide numerator and denominator by 20: 60 ÷ 20 = 3, 220 ÷ 20 = 11. So, 3/11.
Wait, but let me verify this another way to make sure I didn't make a mistake. Sometimes when dealing with probabilities, it's easy to mix up combinations and permutations.
Alternatively, I can calculate the probability step by step, considering the draws one by one.
The probability that the first marble is any color is 1, since we don't have any restrictions. Let's say the first marble is red. The probability of that is 3/12. Then, the second marble has to be a different color, either blue or green. There are 4 blue + 5 green = 9 marbles left. So, the probability the second marble is blue or green is 9/11. Then, the third marble has to be the remaining color. If the first was red and the second was blue, then the third has to be green, which is 5/10. If the first was red and the second was green, then the third has to be blue, which is 4/10.
But this seems more complicated because we have to consider different paths. Let's see if this approach gives the same answer.
Alternatively, since the order doesn't matter, maybe it's better to use combinations. But let's try this sequential approach.
First, the probability that the first marble is red: 3/12. Then, the second marble is blue: 4/11. Then, the third marble is green: 5/10. Multiply these together: (3/12)(4/11)(5/10).
But also, the order could be red, green, blue; blue, red, green; etc. There are 3! = 6 different orders in which the three colors can be drawn. Each of these would have the same probability. For example, blue first: 4/12, then red: 3/11, then green: 5/10. Multiply those: (4/12)(3/11)(5/10). Which is the same as (345)/(121110).
So, each permutation of the colors contributes (345)/(121110), and there are 6 permutations. So total probability is 6*(345)/(121110). Let's compute that:
6*(60)/(1320) = 360/1320 = 36/132 = 3/11. Same result as before. So that's reassuring.
Therefore, both methods give 3/11. So, the probability is 3/11.
But let me check once more. Maybe using combinations.
Total combinations: C(12,3) = 220.
Favorable combinations: C(3,1)C(4,1)C(5,1) = 345 = 60. So, 60/220 = 3/11. Yep, same answer.
Alternatively, another way: hypergeometric distribution? Not sure if that's necessary here, but since we're dealing with multiple categories, the hypergeometric distribution can be used for multivariate cases. The probability of selecting exactly one from each category is given by:
/ C(12,3) = 60 / 220 = 3/11. Same result.
So, all methods lead to 3/11. Therefore, I think that's the correct answer.
But just to be absolutely thorough, let's consider another approach. Suppose we calculate the probability of all three being different by subtracting the probabilities of other cases from 1. The other cases are: all three same color, or two of one color and one of another.
First, calculate the probability of all three being the same color. How many ways can that happen?
For red: C(3,3) = 1.
For blue: C(4,3) = 4.
For green: C(5,3) = 10.
Total same color combinations: 1 + 4 + 10 = 15.
Probability: 15 / 220.
Then, the probability of two of one color and one of another. Let's calculate that.
First, choose the color that has two marbles, and the color that has one. There are C(3,1) ways to choose the color for two marbles, and then C(2,1) ways to choose the other color. Wait, but since there are three colors, if we pick two marbles of one color and one of another, we need to consider all possible pairs.
Alternatively, for each color, compute the number of ways to choose two marbles from it and one from the remaining colors.
For red: C(3,2) * = 3*(4+5) = 3*9=27.
For blue: C(4,2) * = 6*(3+5)=6*8=48.
For green: C(5,2) * = 10*(3+4)=10*7=70.
Total two-one combinations: 27 + 48 + 70 = 145.
Therefore, total unfavorable outcomes: same color (15) + two-one (145) = 160.
Therefore, the probability of all three different colors is 1 - (160/220) = 60/220 = 3/11. Same answer again.
So, this confirms it yet again. Therefore, I'm confident the answer is 3/11.
Just to recap, the key steps are:
Calculate total combinations: C(12,3) = 220.
Calculate favorable combinations: 345 = 60.
Divide: 60/220 = 3/11.
Alternatively, using probability step-by-step considering order and multiplying by permutations, same result.
Alternatively, subtracting other probabilities from 1, same result.
Therefore, the probability is 3/11.
The probability of drawing three marbles of different colors from a bag containing 3 red, 4 blue, and 5 green marbles is calculated as follows:
Total Combinations:
The total number of ways to draw 3 marbles from 12 is:
[
\binom{12}{3} = \frac{12!}{3!(12-3)!} = 220
]
Favorable Combinations:
To get one marble of each color, multiply the choices for each color:
[
3 , (\text{red}) \times 4 , (\text{blue}) \times 5 , (\text{green}) = 60
]
Probability:
Divide favorable by total combinations:
[
https://www.18122.com/url/usdt/usdt2859.jpg
\frac{60}{220} = \frac{3}{11}
]
Answer: The probability is (\boxed{\dfrac{3}{11}}).
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