Probability Calculation At Most 4 Chemistry Students

Let's dive into a probability problem that Mrs. Gomes encountered at her high school. It involves figuring out the likelihood of a certain number of students having taken chemistry. So, let's break it down and solve it step by step.

Understanding the Problem

Okay, guys, so here's the scenario: Mrs. Gomes knows that 40% of the students in her high school have taken chemistry. Now, she's doing a random survey, and she picks 12 students. The big question is: what's the probability that at most 4 of these 12 students have taken chemistry? "At most 4" means we're interested in 0, 1, 2, 3, or 4 students having a chemistry background. This is a classic probability problem that we can solve using the binomial probability formula.

To really nail this problem, we need to get our heads around a few key concepts. First off, we're dealing with binomial probability. This is super important because it applies when we have a fixed number of trials (in our case, surveying 12 students), each trial is independent (one student's chemistry background doesn't affect another's), there are only two outcomes (a student has taken chemistry or hasn't), and the probability of success (a student having taken chemistry) is the same for each trial. Think of it like flipping a coin multiple times – each flip is independent, and you're either going to get heads or tails. The binomial probability formula is our trusty tool here, and it's given by: P(k) = C(n, k) * p^k * (1 - p)^(n - k). This formula helps us calculate the probability of getting exactly k successes in n trials, where p is the probability of success on a single trial. For this particular problem, we need to calculate the probabilities for 0, 1, 2, 3, and 4 students having taken chemistry and then add them up to find the overall probability of “at most 4”. Trust me, it sounds more complicated than it actually is when we put the numbers in and start crunching them.

Breaking Down the Binomial Probability Formula

Let's break down the binomial probability formula bit by bit so we really understand what's going on. The formula, as we mentioned, is: P(k) = C(n, k) * p^k * (1 - p)^(n - k). The most important component is C(n, k), which represents the number of combinations of choosing k successes from n trials. Think of it like this: if you have a group of friends and you need to pick a smaller group for a specific task, C(n, k) tells you how many different smaller groups you can form. In our chemistry problem, it tells us how many different ways we can select k students who have taken chemistry from our group of 12 surveyed students. The formula for C(n, k) is n! / (k! * (n - k)!), where "!" means factorial (e.g., 5! = 5 * 4 * 3 * 2 * 1). Don't let the factorials scare you; they're just a way to count the number of ways to arrange things. Next up, we have p^k, where p is the probability of success on a single trial and k is the number of successes we're interested in. In our case, p is the probability that a student has taken chemistry (40%, or 0.4), and k is the number of students we want to have taken chemistry (0, 1, 2, 3, or 4). So, p^k is the probability of getting k students who have taken chemistry. Lastly, we have (1 - p)^(n - k), which represents the probability of getting (n - k) failures. This is simply the probability that the remaining students haven't taken chemistry. By multiplying these three components together, we get the probability of getting exactly k successes in n trials. That's the binomial probability formula in a nutshell!

Applying the Binomial Probability Formula

Now, let's get our hands dirty and actually apply the binomial probability formula to our problem. Remember, Mrs. Gomes surveyed 12 students (n = 12), and the probability of a student having taken chemistry is 40% (p = 0.4). We want to find the probability that at most 4 students have taken chemistry, which means we need to calculate the probabilities for k = 0, 1, 2, 3, and 4, and then add them all up. Let's start with k = 0. Using the formula P(k) = C(n, k) * p^k * (1 - p)^(n - k), we get P(0) = C(12, 0) * (0.4)^0 * (0.6)^12. C(12, 0) is 1 (there's only one way to choose 0 students out of 12), (0.4)^0 is also 1 (anything to the power of 0 is 1), and (0.6)^12 is approximately 0.00217678. So, P(0) is roughly 0.00217678. Next, let's calculate P(1). We have P(1) = C(12, 1) * (0.4)^1 * (0.6)^11. C(12, 1) is 12 (there are 12 ways to choose 1 student out of 12), (0.4)^1 is 0.4, and (0.6)^11 is approximately 0.00362797. So, P(1) is roughly 0.01741425. We continue this process for k = 2, 3, and 4. For k = 2, P(2) = C(12, 2) * (0.4)^2 * (0.6)^10 ≈ 0.06385229. For k = 3, P(3) = C(12, 3) * (0.4)^3 * (0.6)^9 ≈ 0.14193953. And for k = 4, P(4) = C(12, 4) * (0.4)^4 * (0.6)^8 ≈ 0.21290930. Now, to find the probability that at most 4 students have taken chemistry, we add up these probabilities: P(0) + P(1) + P(2) + P(3) + P(4) ≈ 0.00217678 + 0.01741425 + 0.06385229 + 0.14193953 + 0.21290930 = 0.43829215. So, the probability is approximately 0.438.

Calculating Individual Probabilities

Let's break down the calculation of individual probabilities a bit more to make sure we've got it all covered. We've already touched on the binomial probability formula, P(k) = C(n, k) * p^k * (1 - p)^(n - k), but let's really dig into how each part contributes to the final result. Take the case of k = 2, for example. We're trying to find the probability that exactly 2 out of the 12 surveyed students have taken chemistry. The first part we need to calculate is C(12, 2), which, as we discussed earlier, represents the number of ways to choose 2 students out of 12. Using the formula n! / (k! * (n - k)!), we get 12! / (2! * 10!) = (12 * 11) / (2 * 1) = 66. This tells us there are 66 different ways to select 2 students from a group of 12. Next, we need to calculate (0.4)^2, which is the probability that both of the chosen students have taken chemistry. This is simply 0.4 * 0.4 = 0.16. Then, we calculate (0.6)^10, which is the probability that the remaining 10 students have not taken chemistry. This is approximately 0.00604662. Finally, we multiply these three values together: 66 * 0.16 * 0.00604662 ≈ 0.06385229. This gives us the probability of exactly 2 students having taken chemistry. We repeat this process for k = 0, 1, 3, and 4, carefully calculating each part of the formula and then multiplying them together. It’s a bit tedious, but breaking it down step by step ensures we get the correct result. Remember, accuracy is key in probability problems!

Summing the Probabilities

Alright, guys, we've done the heavy lifting by calculating the individual probabilities for 0, 1, 2, 3, and 4 students having taken chemistry. Now comes the final step: summing these probabilities to find the overall probability that at most 4 students have taken chemistry. This step is actually pretty straightforward – we simply add up the probabilities we calculated earlier. So, we have: P(at most 4) = P(0) + P(1) + P(2) + P(3) + P(4). We found that P(0) ≈ 0.00217678, P(1) ≈ 0.01741425, P(2) ≈ 0.06385229, P(3) ≈ 0.14193953, and P(4) ≈ 0.21290930. Adding these values together, we get: P(at most 4) ≈ 0.00217678 + 0.01741425 + 0.06385229 + 0.14193953 + 0.21290930 = 0.43829215. So, the probability that at most 4 students have taken chemistry is approximately 0.43829215. The question asks us to round the answer to the nearest thousandth, which means we look at the fourth decimal place. In this case, it's a 2, so we don't need to round up. Therefore, our final answer is 0.438. That's it – we've solved the problem! By breaking it down into smaller steps and carefully applying the binomial probability formula, we were able to find the probability that at most 4 students in Mrs. Gomes' survey have taken chemistry.

Final Answer

So, after all the calculations and steps, we've arrived at the final answer. The probability that at most 4 students out of the 12 surveyed have taken chemistry is approximately 0.438 when rounded to the nearest thousandth. We got here by understanding the binomial probability formula, breaking down the problem into manageable parts, calculating individual probabilities for 0, 1, 2, 3, and 4 students, and then summing those probabilities up. It's a bit of a journey, but hopefully, you've seen how each step makes sense and contributes to the final solution. This kind of problem is a great example of how probability works in the real world. Mrs. Gomes could use this information to get a better understanding of the distribution of chemistry knowledge among her students. And for us, it's a good reminder that even seemingly complex problems can be solved by breaking them down and applying the right tools and formulas. Keep practicing, and you'll become a probability pro in no time!