A researcher wishes to conduct a study of the color preferences of new car buyers. Suppose that 50%
 of this population prefers the color red. If 18
 buyers are randomly selected, what is the probability that at least 3
 buyers would prefer red? Round your answer to four decimal places.

The problem describes a binomial probability scenario because we have a fixed number of trials (18 buyers), each trial results in a success (buyer prefers red) or a failure, and the probability of success is constant (50%).

Let:
- \( n = 18 \) (the number of buyers),
- \( p = 0.50 \) (the probability that a buyer prefers red),
- \( X \) be the random variable representing the number of buyers who prefer red.

We need to find \( P(X \geq 3) \), the probability that at least 3 buyers prefer red.

In this case, it's easier to first calculate \( P(X < 3) \), which is the probability that fewer than 3 buyers prefer red. Then, subtract that value from 1:

\[
P(X \geq 3) = 1 - P(X < 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2))
\]

The probability mass function for a binomial distribution is:

\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
\]

Where:
- \( \binom{n}{k} \) is the binomial coefficient \( \frac{n!}{k!(n - k)!} \).

Now, let's compute the probabilities \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \):

1. \( P(X = 0) = \binom{18}{0} (0.50)^0 (0.50)^{18} = 1 \times (0.50)^{18} \approx 0.00000381 \)
2. \( P(X = 1) = \binom{18}{1} (0.50)^1 (0.50)^{17} = 18 \times (0.50)^{18} \approx 0.00006855 \)
3. \( P(X = 2) = \binom{18}{2} (0.50)^2 (0.50)^{16} = 153 \times (0.50)^{18} \approx 0.00058267 \)

So, \( P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \approx 0.00000381 + 0.00006855 + 0.00058267 = 0.00065403 \).

Finally, \( P(X \geq 3) = 1 - 0.00065403 \approx 0.9993 \).

Thus, the probability that at least 3 buyers prefer red is approximately **0.9993**.

Do you need further details or have any questions?

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Here are five relative questions to expand on this:
1. How do you calculate the binomial coefficient \( \binom{n}{k} \)?
2. What are the properties of a binomial distribution?
3. Can the binomial distribution be approximated by a normal distribution in this case?
4. How would the probability change if the sample size or probability of success changed?
5. What is the significance of the value \( p \) in the binomial formula?

**Tip:** In binomial distributions, calculating the complement (1 - the probability of the opposite event) often simplifies problems like this one!

A researcher wishes to conduct a study of the color preferences of new car buyers. Suppose that 50% of this population prefers the color red. If 18 buyers are randomly selected, what is the probability that at least 3 buyers would prefer red? Round your answer to four decimal places.

Calculate the probability that at least 3 out of 18 randomly selected car buyers prefer the color red using a binomial distribution approach. This detailed solution covers essential probability concepts and formulas, including the binomial theorem and complement rule. Suitable for Grades 10-12.

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