Probability Of Chemistry Students: A Step-by-Step Solution

Hey guys! Let's dive into a probability problem together. We've got Mrs. Gomes, who's found that 40% of the students at her high school take chemistry. She's conducting a random survey of 12 students, and we need to figure out the probability that at most 4 of these students have taken chemistry. Sounds interesting, right? Let's break it down step by step.

Understanding the Problem

Okay, so the first thing we need to do is really understand what the problem is asking. We know that the probability of a student taking chemistry is 40%, or 0.4. This is our success rate. We also know that Mrs. Gomes is surveying 12 students, which means we have 12 trials. The question asks for the probability that at most 4 students have taken chemistry. This means we need to consider the probabilities of 0, 1, 2, 3, and 4 students having taken chemistry and then add them all up. This "at most" phrasing is super important because it tells us we're dealing with a cumulative probability. Ignoring this detail can totally throw off your calculations, so always double-check what the question is really asking!

Why is This a Binomial Probability Problem?

This problem is a classic example of a binomial probability problem. Why? Because it meets the four conditions of a binomial setting, often remembered by the acronym BINS:

• Binary: Each student either has taken chemistry (success) or hasn't (failure). There are only two outcomes.
• Independent: We assume that one student's choice to take chemistry doesn't affect another student's choice. This is the "randomly surveys" part coming into play – it suggests independence.
• Number of trials: There's a fixed number of trials, which is 12 students in this case.
• Success: The probability of success (a student taking chemistry) is the same for each trial, which is 40% or 0.4.

Recognizing that this is a binomial situation is key because it allows us to use the binomial probability formula, which we'll get to in a bit.

The Binomial Probability Formula

Alright, now that we've nailed down the type of problem, let's introduce the star of the show: the binomial probability formula. This formula helps us calculate the probability of getting exactly k successes in n trials. Here it is:

$$P(X = k) = \binom{n}{k} * p^k * (1 - p)^{(n - k)}$$

Whoa, looks a bit scary, right? Don't worry, we'll break it down. Each part has a specific meaning:

• $P(X = k)$: This is the probability of getting exactly k successes. It's what we're trying to find.
• $\binom{n}{k}$: This is the binomial coefficient, often read as "n choose k." It represents the number of ways to choose k successes from n trials. You can calculate it using the formula: $\binom{n}{k} = \frac{n!}{k!(n - k)!}$, where "!" means factorial (e.g., 5! = 5 * 4 * 3 * 2 * 1).
• $p$: This is the probability of success on a single trial (0.4 in our case).
• $k$: This is the number of successes we want (0, 1, 2, 3, or 4 in our case).
• $n$: This is the total number of trials (12 students).
• $(1 - p)$: This is the probability of failure on a single trial (1 - 0.4 = 0.6).
• $(n - k)$: This is the number of failures.

So, this formula essentially calculates the probability by considering the number of ways to get k successes, the probability of those k successes happening, and the probability of the remaining failures happening. Pretty neat, huh?

Calculating the Probabilities

Okay, formula in hand, let's get to the calculations! Remember, we need to find the probability of at most 4 students having taken chemistry. This means we need to calculate the probabilities for 0, 1, 2, 3, and 4 students and then add them together. Buckle up, because we're going to be using that formula a few times!

Probability of 0 Students

First up, let's find the probability that no students (0) have taken chemistry. Plugging the values into our formula:

$$P(X = 0) = \binom{12}{0} * (0.4)^0 * (0.6)^{12}$$

Let's break it down:

• $\binom{12}{0} = 1$ (There's only one way to choose 0 items from 12).
• $(0.4)^0 = 1$ (Anything to the power of 0 is 1).
• $(0.6)^{12} \approx 0.002176$

So, $P(X = 0) = 1 * 1 * 0.002176 \approx 0.002176$

Probability of 1 Student

Next, let's calculate the probability that exactly 1 student has taken chemistry:

$$P(X = 1) = \binom{12}{1} * (0.4)^1 * (0.6)^{11}$$

• $\binom{12}{1} = 12$ (There are 12 ways to choose 1 student from 12).
• $(0.4)^1 = 0.4$
• $(0.6)^{11} \approx 0.0036279$

So, $P(X = 1) = 12 * 0.4 * 0.0036279 \approx 0.017414$

Probability of 2 Students

Now, let's find the probability for 2 students:

$$P(X = 2) = \binom{12}{2} * (0.4)^2 * (0.6)^{10}$$

• $\binom{12}{2} = \frac{12!}{2!10!} = \frac{12 * 11}{2 * 1} = 66$
• $(0.4)^2 = 0.16$
• $(0.6)^{10} \approx 0.0060466$

So, $P(X = 2) = 66 * 0.16 * 0.0060466 \approx 0.063796$

Probability of 3 Students

Time for 3 students:

$$P(X = 3) = \binom{12}{3} * (0.4)^3 * (0.6)^9$$

• $\binom{12}{3} = \frac{12!}{3!9!} = \frac{12 * 11 * 10}{3 * 2 * 1} = 220$
• $(0.4)^3 = 0.064$
• $(0.6)^9 \approx 0.0100777$

So, $P(X = 3) = 220 * 0.064 * 0.0100777 \approx 0.141861$

Probability of 4 Students

Last but not least, let's calculate the probability for 4 students:

$$P(X = 4) = \binom{12}{4} * (0.4)^4 * (0.6)^8$$

• $\binom{12}{4} = \frac{12!}{4!8!} = \frac{12 * 11 * 10 * 9}{4 * 3 * 2 * 1} = 495$
• $(0.4)^4 = 0.0256$
• $(0.6)^8 \approx 0.016796$

So, $P(X = 4) = 495 * 0.0256 * 0.016796 \approx 0.212840$

Adding the Probabilities

Phew! We've calculated the individual probabilities. Now, remember, we need the probability of at most 4 students. So, we need to add up all those probabilities we just calculated:

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X \leq 4) \approx 0.002176 + 0.017414 + 0.063796 + 0.141861 + 0.212840$

$P(X \leq 4) \approx 0.438087$

Rounding the Answer

Finally, the question asks us to round the answer to the nearest thousandth. So, 0.438087 rounded to the nearest thousandth is 0.438.

The Final Answer

Therefore, the probability that at most 4 students have taken chemistry is approximately 0.438. Awesome! We did it!

Key Takeaways

• Understanding the Question: Always, always read the question carefully and make sure you understand what it's asking. Phrases like "at most" and "at least" are super important.
• Binomial Settings: Recognize when a problem fits the binomial setting (BINS). It makes your life so much easier!
• The Formula: The binomial probability formula might look intimidating, but break it down piece by piece, and you'll be a pro in no time.
• Cumulative Probability: When dealing with "at most" or "at least," remember to add up the relevant probabilities.
• Calculators and Tools: Don't be afraid to use a calculator or statistical software to help with the calculations, especially those factorials and binomial coefficients!

I hope this breakdown was helpful, guys! Probability problems can seem tough at first, but with a little practice and a good understanding of the concepts, you'll be solving them like a boss. Keep up the great work!