Probability Reyna's Coin Selection Problem Solved

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Hey guys! Let's dive into a fun probability problem involving Reyna and her coins. This is a classic example of how we can apply probability concepts to everyday situations. We'll break down the problem step by step, making sure everyone understands the logic behind it. So, buckle up, and let's get started!

The Coin Collection: Reyna's Stash

Our main focus is understanding probability in the context of Reyna's coin collection. Reyna has a total of 9 coins: 5 are worth 10 cents each (dimes), and 4 are worth 25 cents each (quarters). The core question here is: what's the likelihood that if Reyna randomly picks two coins, their combined value will be at least 35 cents? This involves a bit of combinatorics and a good grasp of probability principles. To solve this, we need to figure out all the possible combinations of coins Reyna can pick and then determine which of those combinations meet our 35-cent threshold. This problem perfectly illustrates how probability isn't just about numbers; it's about understanding the chances in real-world scenarios. We'll use combinations, which is a way of selecting items from a set where the order doesn't matter. Think of it like picking any two coins from a bag – it doesn't matter which coin you pick first, the pair is what counts. So, we'll calculate the total number of ways Reyna can pick two coins, and then we'll count the ways she can pick two coins that add up to 35 cents or more. The ratio of these two numbers will give us our probability. Probability is all about ratios – the ratio of favorable outcomes to total possible outcomes. In this case, the favorable outcomes are the coin combinations that meet our 35-cent requirement, and the total possible outcomes are all the possible two-coin combinations. By working through this problem, we'll not only find the answer but also solidify our understanding of how probability works in practical situations. So, let's jump into the calculations and see what we find!

Calculating Total Possible Outcomes

The first crucial step in solving this problem is figuring out the total number of ways Reyna can choose two coins from her collection. This is a combination problem because the order in which she selects the coins doesn't matter. Whether she picks a dime first and then a quarter, or a quarter first and then a dime, the pair of coins is the same for our purposes. To calculate combinations, we use the formula: nCr = n! / (r! * (n-r)!), where 'n' is the total number of items, 'r' is the number of items being chosen, and '!' denotes the factorial (e.g., 5! = 5 * 4 * 3 * 2 * 1). In Reyna's case, she has 9 coins in total (5 dimes and 4 quarters), and she's choosing 2. So, n = 9 and r = 2. Plugging these values into the formula, we get: 9C2 = 9! / (2! * 7!) = (9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / ((2 * 1) * (7 * 6 * 5 * 4 * 3 * 2 * 1)). We can simplify this significantly by canceling out the 7! from both the numerator and the denominator, leaving us with: 9C2 = (9 * 8) / (2 * 1) = 72 / 2 = 36. This means there are 36 different ways Reyna can choose two coins from her collection. These 36 combinations represent our total possible outcomes. To find the probability of a specific event (in this case, picking coins worth at least 35 cents), we'll compare the number of ways that event can occur to this total number of possibilities. So, we know there are 36 different pairs of coins Reyna could pick. Now, the next step is to figure out how many of those pairs add up to 35 cents or more. This will involve looking at the different combinations of dimes and quarters and seeing which ones meet our criteria. Understanding this total number of outcomes is the foundation for calculating the probability, so it's important we get this part right. With this number in hand, we can move on to the next stage of the problem and start figuring out the favorable outcomes.

Identifying Favorable Outcomes

Now that we know the total number of possible outcomes, let's figure out how many of those outcomes result in Reyna having at least 35 cents. To do this, we need to consider the different combinations of coins she could pick and their values. Remember, she has 5 dimes (10 cents each) and 4 quarters (25 cents each). There are three scenarios where the two coins Reyna picks will total 35 cents or more:

  1. One dime and one quarter: This combination gives Reyna 10 cents + 25 cents = 35 cents. So, this meets our criteria.
  2. Two quarters: This combination gives Reyna 25 cents + 25 cents = 50 cents. This also meets our criteria.

Let's calculate the number of ways each of these scenarios can occur:

  • One dime and one quarter: Reyna has 5 dimes, so she has 5 choices for picking a dime. She has 4 quarters, so she has 4 choices for picking a quarter. Therefore, there are 5 * 4 = 20 ways to pick one dime and one quarter.
  • Two quarters: This is a combination problem similar to what we did earlier. Reyna has 4 quarters, and she's choosing 2. So, we use the combination formula: 4C2 = 4! / (2! * 2!) = (4 * 3 * 2 * 1) / ((2 * 1) * (2 * 1)) = (4 * 3) / (2 * 1) = 12 / 2 = 6. There are 6 ways to pick two quarters.

So, in total, there are 20 ways to pick one dime and one quarter, and 6 ways to pick two quarters. This means there are 20 + 6 = 26 favorable outcomes. These are the combinations of coins that will give Reyna at least 35 cents. We've now identified all the ways Reyna can get the outcome we're interested in. The next step is to use this information, along with the total number of possible outcomes we calculated earlier, to determine the probability. By figuring out the favorable outcomes, we're getting closer to solving the problem and understanding the chances of Reyna getting at least 35 cents when she picks two coins.

Calculating the Probability

Alright, we're in the home stretch! We've done the groundwork, and now it's time to calculate the probability. Remember, probability is the ratio of favorable outcomes to total possible outcomes. We've already figured out both of these numbers:

  • Total possible outcomes: 36 (the number of ways Reyna can pick any two coins)
  • Favorable outcomes: 26 (the number of ways Reyna can pick two coins worth at least 35 cents)

So, the probability of Reyna picking two coins worth at least 35 cents is 26/36. But we can simplify this fraction to make it easier to understand. Both 26 and 36 are divisible by 2, so we can divide both the numerator and the denominator by 2: 26 / 2 = 13 36 / 2 = 18. This gives us a simplified fraction of 13/18. So, the probability that Reyna chooses two coins worth at least 35 cents is 13/18. This means that if Reyna were to randomly pick two coins many, many times, we would expect that about 13 out of every 18 times, the coins would be worth 35 cents or more. Probability is often expressed as a fraction, but it can also be expressed as a decimal or a percentage. To convert 13/18 to a decimal, we simply divide 13 by 18, which gives us approximately 0.722. To convert this to a percentage, we multiply by 100, giving us about 72.2%. So, we can also say that there is a 72.2% chance that Reyna will pick two coins worth at least 35 cents. This problem demonstrates how we can use probability to quantify the likelihood of an event occurring. By breaking down the problem into smaller steps, like calculating total outcomes and favorable outcomes, we can tackle complex situations and arrive at a clear answer. So, there you have it! The probability of Reyna picking two coins worth at least 35 cents is 13/18, or approximately 72.2%.

Final Answer

So, to recap, we've successfully navigated Reyna's coin conundrum! We started by understanding the problem, which involved figuring out the probability of Reyna picking two coins worth at least 35 cents from her collection of 5 dimes and 4 quarters. Then, we broke the problem down into manageable steps. First, we calculated the total number of possible outcomes, which was the number of ways Reyna could pick any two coins from her collection. We used the combination formula to find that there were 36 different ways she could do this. Next, we identified the favorable outcomes – the combinations of coins that would result in a total value of 35 cents or more. We found that there were two scenarios that met this criteria: picking one dime and one quarter, or picking two quarters. We calculated the number of ways each of these scenarios could occur and found that there were 20 ways to pick one dime and one quarter, and 6 ways to pick two quarters, for a total of 26 favorable outcomes. Finally, we calculated the probability by dividing the number of favorable outcomes by the total number of possible outcomes. This gave us a probability of 26/36, which we simplified to 13/18. We also expressed this probability as a decimal (approximately 0.722) and as a percentage (approximately 72.2%). Therefore, the final answer is that the probability of Reyna picking two coins worth at least 35 cents is 13/18, or about 72.2%. This problem is a great example of how probability can be applied to real-world scenarios. By understanding the basic principles of probability and breaking down problems into smaller steps, we can solve complex questions and make informed decisions. And that's the power of probability – it helps us understand and quantify the chances of different events occurring, which is a valuable skill in many areas of life.