In a group of 100 people, 70% prefer coffee over tea, and 30% prefer tea over coffee. If you randomly select 5 people from this group, what is the probability that exactly 3 of them prefer coffee?
To solve this problem, we'll use the binomial probability formula:

P(x) = C(n, x) * p^x * q^(n-x)

where n is the number of trials (in this case, selecting 5 people), x is the number of successful outcomes (selecting 3 who prefer coffee), p is the probability of success (0.7 for preferring coffee), and q is the probability of failure (0.3 for not preferring coffee).

1. First, calculate C(5, 3) using the combination formula: C(5, 3) = 5! / (3!(5-3)!) = 10.
2. Next, calculate p^x * q^(n-x): (0.7)^3 * (0.3)^2 ≈ 0.03087.
3. Finally, multiply these values together: 10 * 0.03087 ≈ 0.3087.

So, the probability that exactly 3 out of the 5 selected people prefer coffee is approximately 0.3087 or 30.87%.