Final answer:
The probability of selecting 3 Democrats and 2 Republicans from a city council of 6 Democrats and 8 Republicans for a 5-person committee is 280/1001.
Step-by-step explanation:
To solve this problem, we'll use the concept of combinations since the order of selection does not matter. The formula for combinations is C(n, k) = n! / (k! (n-k)!), where n is the total number of items to choose from, k is the number we are selecting, and ! denotes factorial.
The total number of ways to select a committee of 5 people from 14 (6 Democrats and 8 Republicans) is C(14, 5). The number of ways to select 3 Democrats from 6 is C(6, 3), and the number of ways to select 2 Republicans from 8 is C(8, 2).
The probability of selecting 3 Democrats and 2 Republicans is therefore calculated by multiplying C(6, 3) with C(8, 2) and dividing by C(14, 5):
P(3 Democrats and 2 Republicans) = (C(6, 3) × C(8, 2)) / C(14, 5).
Simplifying the answer gives us:
P(3 Democrats and 2 Republicans) = (20 × 28) / 2002 = 560 / 2002 = 280/1001, which cannot be simplified further.