To solve this problem, we can use the binomial probability formula because we are dealing with a situation where there are two possible outcomes (either a buyer prefers red or they do not), and we have a fixed number of trials (14 buyers).

The formula for binomial probability is:

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

Where:

• $n = 14$ (the number of trials, or buyers),
• $k = 2$ (the number of successes, or buyers who prefer red),
• $p = 0.40$ (the probability of a buyer preferring red),
• $\binom{n}{k}$ is the binomial coefficient, calculated as $\frac{n!}{k!(n - k)!}$.

Now, plugging the values into the formula:

$P(X = 2) = \binom{14}{2} (0.40)^2 (0.60)^{12}$

First, calculate the binomial coefficient:

$\binom{14}{2} = \frac{14 \times 13}{2 \times 1} = 91$

Now, calculate the probabilities:

$(0.40)^2 = 0.16$ $(0.60)^{12} \approx 0.00217678$

Now, multiply everything together:

$P(X = 2) = 91 \times 0.16 \times 0.00217678 \approx 0.0318$

Thus, the probability that exactly 2 buyers prefer red is approximately 0.0318 (rounded to four decimal places).

Would you like further details on how this formula works, or have any other questions about this problem?

Here are 5 related questions:

1. What is the probability that no buyers prefer red?
2. What is the probability that at least 3 buyers prefer red?
3. How would the probability change if the proportion of buyers who prefer red were 0.50 instead of 0.40?
4. How do we calculate the mean and variance for a binomial distribution?
5. What are the assumptions behind using the binomial distribution in this problem?

Tip: In binomial problems, always ensure that the trials are independent and the probability of success remains constant throughout the experiment.