Probability Puzzle: Balls, Sums, And Multiples Of 3
Hey guys! Let's dive into a fun probability problem that's sure to get your brains buzzing. We're going to break down the steps to solve this puzzle involving drawing balls, summing numbers, and figuring out those tricky multiples of 3. So, grab your thinking caps, and let's get started!
Understanding the Problem and Setting Up
Alright, so here's the deal: we have a box with nine balls. Each ball is uniquely numbered from 1 to 9. We're going to randomly pick out three of these balls. The real challenge comes with the conditions: first, the sum of the numbers on the first two balls cannot be a multiple of 3. Second, the sum of all three balls must be a multiple of 3. Seems a bit complex, right? But don't worry, we'll take it step by step.
Our goal is to figure out the probability of this happening. Probability, as you probably know, is all about the chances of something occurring. It's calculated as: (Number of favorable outcomes) / (Total number of possible outcomes). So, we need to find out how many different sets of three balls meet our criteria and then divide that by all the possible sets of three balls we could pick.
To make this easier to grasp, we will categorize the numbers 1 to 9 based on their remainders when divided by 3. This approach is key to understanding the problem. Let's see how:
- Remainder 0: 3, 6, 9 (These numbers are perfectly divisible by 3)
- Remainder 1: 1, 4, 7 (These leave a remainder of 1 when divided by 3)
- Remainder 2: 2, 5, 8 (These leave a remainder of 2 when divided by 3)
This categorization will be super useful in figuring out the sums. Now we can see how the sums relate to multiples of 3. Remember, a number is a multiple of 3 if its remainder is 0 when divided by 3.
Finding Favorable Outcomes
Okay, now let's think about the different ball combinations that satisfy our conditions. Here's where the remainder categories come into play. We need the first two balls' sum NOT to be a multiple of 3, and the sum of all three balls TO be a multiple of 3.
Consider the possible combinations that result in a multiple of 3 when summing three numbers. The remainders must either be (0, 0, 0), (1, 1, 1), or (0, 1, 2). Think about it. If you add up the remainders, they have to equal 0, 3, or 6, which are all multiples of 3. Let's break down the combinations that give us these results, and also fulfill the condition that the first two balls don't sum to a multiple of 3.
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Case 1: (0, 1, 2) in any order. If we pick a number from each remainder group (0, 1, and 2), the sum of the three balls is always a multiple of 3. This meets the second condition. The combinations for the remainders of the first two balls can be (1, 2) or (2, 1). Why? Because the sum of numbers with remainders 1 and 2 will never be a multiple of 3. The first ball is from group 1 and the second ball is from group 2, or vice versa. The sum is not divisible by 3. This satisfies our first condition.
So, let's calculate the combinations: There are 3 options for a ball with remainder 0, 3 options for a ball with remainder 1, and 3 options for a ball with remainder 2. The order matters for the first two balls, so the number of possibilities is 3 * 3 * 3 = 27 combinations. Because the first two balls can be the numbers with the remainders (1, 2) or (2, 1) and these balls can be in either order we need to multiply our result by 2. Thus 2 * 27 = 54 favorable outcomes.
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Case 2: (1, 1, 1). The sum of the numbers in the first two balls cannot be a multiple of 3. But the total sum must be. The only way is that the remainders of the three balls are 1, 1, and 1. The numbers are all from the group with remainder 1. In this case, the sum of the three balls is a multiple of 3. However, we need to ensure the first two balls do not sum to a multiple of 3. But that is impossible. When taking the numbers from the same group (remainder 1), the sum of the first two balls will be a multiple of 3.
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Case 3: (0, 0, 0) This case can be handled in a way that the total sum of the numbers is a multiple of 3. If we select 3 balls with the remainder 0, the total sum is 0. But the sum of the first two balls is a multiple of 3, because they are both multiple of 3. So we cannot use this case.
So, the only viable scenario is Case 1. 54 favorable outcomes meet the conditions.
Calculating Total Possible Outcomes
Now, we need to figure out the total number of ways we can choose any three balls out of nine. The order of selection doesn't matter, so we use combinations. The formula for combinations is: nCr = n! / (r! * (n-r)!), where n is the total number of items, and r is the number of items to choose.
In our case, n = 9 (total balls) and r = 3 (balls we're selecting). Let's do the math:
- 9C3 = 9! / (3! * 6!)
- 9C3 = (9 * 8 * 7) / (3 * 2 * 1)
- 9C3 = 84
So, there are 84 total possible combinations of picking three balls from the box.
Determining the Probability
We're almost there, guys! We have the number of favorable outcomes (54) and the total number of possible outcomes (84). Now, we just divide the favorable outcomes by the total outcomes to get the probability.
- Probability = (Favorable Outcomes) / (Total Outcomes)
- Probability = 54 / 84
- Probability = 9 / 14
So, the probability is 9/14. That means, that if we do this random ball-drawing experiment many, many times, we can expect that the conditions will be met in roughly 64.3% of the cases. Pretty cool, huh?
Conclusion and Final Thoughts
And there you have it! We've successfully navigated this probability puzzle. By breaking down the problem into smaller parts, categorizing the numbers, and using the right formulas, we've found the probability of this somewhat tricky scenario. Remember that the key is careful consideration of all possible cases, understanding the divisibility rules, and not being afraid to break down the problem into more manageable chunks.
Key Takeaways: This problem highlights the importance of methodical thinking, understanding combinatorics, and how patterns in numbers (like multiples of 3) can help solve probability questions. It’s also a reminder that probability can be super interesting and is relevant in so many different areas.
Hope you had fun with this. Keep practicing, keep learning, and keep asking questions. Until next time, happy calculating, and keep those problem-solving skills sharp! Let me know in the comments if you have any questions, or if you want me to tackle another brain teaser.