This is a binomial probability problem, which can be approached using the binomial distribution.

Problem Breakdown:

• There are 8 people being polled, so the number of trials (n) is 8.
• We are concerned with the probability of at least 6 Democrats being selected, so we are interested in the number of Democrats (denoted as $X$) being greater than or equal to 6.
• The probability of selecting a Democrat (success) is p = 0.35, since 35% of the voters are Democrats.
• This follows a binomial distribution where $X \sim B(n = 8, p = 0.35)$.

The binomial probability mass function is given by:

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

Where:

• $n$ is the total number of trials (8),
• $k$ is the number of successes (Democrats in this case),
• $p$ is the probability of success (0.35),
• $\binom{n}{k}$ is the binomial coefficient, representing the number of ways to choose $k$ successes out of $n$ trials.

To find the probability that at least 6 are Democrats (i.e., $P(X \geq 6)$):

We will calculate the probability for $X = 6, 7, 8$ and add them together:

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8)$

Let's calculate each of these probabilities:

1. Probability for $X = 6$:

$P(X = 6) = \binom{8}{6} (0.35)^6 (0.65)^2$

$P(X = 6) = \frac{8!}{6!(8-6)!} \times (0.35)^6 \times (0.65)^2$

$P(X = 6) = 28 \times (0.35)^6 \times (0.65)^2$

2. Probability for $X = 7$:

$P(X = 7) = \binom{8}{7} (0.35)^7 (0.65)^1$

$P(X = 7) = 8 \times (0.35)^7 \times (0.65)$

3. Probability for $X = 8$:

$P(X = 8) = \binom{8}{8} (0.35)^8 (0.65)^0$

$P(X = 8) = 1 \times (0.35)^8$

Now, you can compute these values to get the final probability. Would you like me to calculate the exact values for these probabilities?