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!