Miguel's Chip Game Probability Calculation

In this article, we will delve into a probability problem presented in the context of a game played by Miguel. The game involves selecting chips with different numbers from a box, and the outcome depends on the numbers on the chosen chips. Specifically, Miguel is playing a game where a box contains four chips. Two chips are marked with the number 1, one chip has the number 3, and the last chip bears the number 5. Miguel's task is to randomly select two chips. The core of the problem lies in determining the probability of Miguel selecting two chips that both have the number 1. This problem serves as an excellent example of basic probability calculations, combining elements of combinatorics and the fundamental principles of probability theory. To accurately determine the probability, we need to consider the total number of ways Miguel can select two chips from the box, as well as the number of ways he can select two chips that both have the number 1. Understanding these concepts is crucial for solving not only this specific problem but also for tackling a wide range of probability-related scenarios. This detailed exploration will not only provide a solution to the problem but also enhance your understanding of probability calculations in general.

To fully grasp the problem, it's essential to break down the game mechanics and the possible outcomes. The game centers around a box containing four chips, each marked with a number. There are two chips with the number 1, one chip with the number 3, and one chip with the number 5. Miguel's objective is to select two chips at random. The key question we aim to answer is: What is the probability that both chips Miguel selects have the number 1? To answer this, we first need to consider the total number of ways Miguel can choose two chips from the four available. This involves using the concept of combinations, as the order in which the chips are selected does not matter. 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 this case, n = 4 (total chips), and r = 2 (chips to choose). Next, we need to determine the number of ways Miguel can select two chips that both have the number 1. Since there are only two chips with the number 1, there is only one way to choose both of them. By calculating the total number of combinations and the number of favorable outcomes, we can then determine the probability of Miguel selecting two chips with the number 1. This probability is calculated as the ratio of favorable outcomes to the total number of possible outcomes.

Before we can determine the probability of Miguel drawing two chips with the number 1, we must first calculate the total number of ways he can draw any two chips from the box. This is a classic combinatorial problem, where the order of selection does not matter. We use the combination formula, which is expressed as nCr = n! / (r! * (n-r)!), where 'n' represents the total number of items, and 'r' represents the number of items being chosen. In Miguel's game, there are four chips in total (n = 4), and he is choosing two (r = 2). Plugging these values into the combination formula gives us 4C2 = 4! / (2! * (4-2)!). To solve this, we first calculate the factorials: 4! (4 factorial) is 4 × 3 × 2 × 1 = 24, 2! (2 factorial) is 2 × 1 = 2, and (4-2)! which is 2! is also 2. Substituting these values back into the formula, we get 4C2 = 24 / (2 * 2) = 24 / 4 = 6. Therefore, there are six different ways Miguel can choose two chips from the four available. These combinations are: (1,1), (1,3), (1,5), (1,3), (1,5), and (3,5). Understanding the total possible outcomes is a crucial step in calculating the probability of the specific event we are interested in, which is Miguel drawing two chips with the number 1. This foundational calculation sets the stage for the next part of our analysis.

Now that we've established the total possible outcomes when Miguel draws two chips, our next step is to identify the favorable outcomes – the scenarios where Miguel draws two chips that both have the number 1. This is a more straightforward calculation than determining the total outcomes, as it focuses specifically on the condition we're interested in. In the box, there are two chips with the number 1. Miguel needs to choose both of these chips to achieve a favorable outcome. Since there are only two chips with the number 1, there is only one way for Miguel to select both of them. This might seem intuitive, but it's important to explicitly state this to ensure clarity in our probability calculation. The favorable outcome is the pair (1, 1). There are no other combinations that satisfy the condition of both chips having the number 1. This contrasts with the earlier calculation of total possible outcomes, where we considered all pairs of chips, regardless of their numbers. By identifying the single favorable outcome, we've isolated the specific scenario that we're interested in measuring the probability of. This step is crucial because probability is defined as the ratio of favorable outcomes to total possible outcomes. With both values now determined, we are well-positioned to calculate the probability of Miguel drawing two chips with the number 1.

With the total possible outcomes and the favorable outcomes clearly defined, we can now calculate the probability of Miguel drawing two chips with the number 1. Probability, in its simplest form, is the ratio of the number of favorable outcomes to the total number of possible outcomes. We've already determined that there is 1 favorable outcome (Miguel drawing the two chips with the number 1) and 6 total possible outcomes (all the different pairs of chips Miguel could draw). Therefore, the probability of Miguel drawing two chips with the number 1 is calculated as: Probability = (Number of Favorable Outcomes) / (Total Number of Possible Outcomes). Plugging in our values, we get: Probability = 1 / 6. This fraction represents the likelihood of Miguel drawing two chips with the number 1 in a single draw from the box. To express this probability in a more intuitive way, we can convert the fraction to a percentage. To do this, we divide 1 by 6 and then multiply by 100: (1 / 6) * 100 ≈ 16.67%. Thus, there is approximately a 16.67% chance that Miguel will draw two chips with the number 1. This calculation provides a clear and quantifiable answer to our original question, demonstrating the practical application of probability principles in a simple game scenario. The probability of 1/6 or 16.67% represents a relatively low chance, highlighting the importance of understanding probability in predicting the likelihood of events.

In conclusion, by systematically analyzing Miguel's chip game, we've successfully calculated the probability of him drawing two chips with the number 1. We began by understanding the game mechanics, identifying the total possible outcomes using combinations, and then determining the favorable outcomes. This led us to the final calculation, where we found the probability to be 1/6, or approximately 16.67%. This exercise not only provides a solution to the specific problem but also reinforces the fundamental principles of probability theory. Probability calculations are crucial in various real-world scenarios, from predicting the weather to assessing risks in financial markets. Understanding how to calculate probabilities allows us to make informed decisions and predictions based on available data. The process of breaking down a problem into smaller, manageable steps, as we did in this article, is a valuable skill in problem-solving in general. Whether you're dealing with a simple game or a complex real-world scenario, the ability to identify possible outcomes, determine favorable outcomes, and calculate probabilities is a powerful tool. This exploration of Miguel's chip game serves as a practical example of how probability concepts can be applied to everyday situations, enhancing our understanding of the world around us.