(a) Let the probability of a person being a Notes supporter be 0.3. We use the binomial distribution with parameters 6 (trials) and 0.3 (success probability).

The probability that no more than 2 people are Notes supporters is:

\(P(0, 1, 2) = \binom{6}{0} (0.3)^0 (0.7)^6 + \binom{6}{1} (0.3)^1 (0.7)^5 + \binom{6}{2} (0.3)^2 (0.7)^4\)

Calculating each term:

\(\binom{6}{0} (0.3)^0 (0.7)^6 = 0.1176\)

\(\binom{6}{1} (0.3)^1 (0.7)^5 = 0.3025\)

\(\binom{6}{2} (0.3)^2 (0.7)^4 = 0.3241\)

Adding these probabilities gives:

\(0.1176 + 0.3025 + 0.3241 = 0.744\)

(b) The probability that a person supports neither choir is:

\(1 - (0.3 + 0.45) = 0.25\)

The probability that none of the 6 people support either choir is:

\(P(6 \text{ support neither choir}) = (0.25)^6\)

Calculating this gives:

\((0.25)^6 = 0.000244\)

The problem involves a binomial distribution where the probability of success (a household being satisfied) is 0.66, and the number of trials is 10. We need to find the probability that at least 8 households are satisfied, which means finding the probability of 8, 9, or 10 successes.

The probability of exactly 8 successes is given by:

\(P(X = 8) = \binom{10}{8} (0.66)^8 (0.34)^2\)

The probability of exactly 9 successes is given by:

\(P(X = 9) = \binom{10}{9} (0.66)^9 (0.34)^1\)

The probability of exactly 10 successes is given by:

\(P(X = 10) = \binom{10}{10} (0.66)^{10}\)

Thus, the probability of at least 8 successes is:

\(P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)\)

Calculating each term:

\(P(X = 8) = \binom{10}{8} (0.66)^8 (0.34)^2 = 45 \times 0.66^8 \times 0.34^2\)

\(P(X = 9) = \binom{10}{9} (0.66)^9 (0.34)^1 = 10 \times 0.66^9 \times 0.34\)

\(P(X = 10) = \binom{10}{10} (0.66)^{10} = 1 \times 0.66^{10}\)

Summing these probabilities gives:

\(P(X \geq 8) = 0.284\)

The probability of obtaining a score of 8 or more with two dice is calculated first. The possible outcomes for scores of 8 or more are: (2,6), (3,5), (4,4), (5,3), (6,2), (3,6), (4,5), (5,4), (6,3), (4,6), (5,5), (6,4), (5,6), (6,5), (6,6). There are 15 favorable outcomes out of 36 total outcomes, so the probability is:

\(P(\text{score} \geq 8) = \frac{15}{36} = \frac{5}{12}\)

We need to find the probability of obtaining a score of 8 or more on fewer than 3 occasions in 8 throws. This is a binomial distribution problem where:

\(n = 8, \quad p = \frac{5}{12}\)

We calculate the probability of obtaining a score of 8 or more on 0, 1, or 2 occasions:

\(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\)

Using the binomial probability formula:

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

Calculate each term:

\(P(X = 0) = \binom{8}{0} \left(\frac{5}{12}\right)^0 \left(\frac{7}{12}\right)^8 = 0.01341\)

\(P(X = 1) = \binom{8}{1} \left(\frac{5}{12}\right)^1 \left(\frac{7}{12}\right)^7 = 0.07661\)

\(P(X = 2) = \binom{8}{2} \left(\frac{5}{12}\right)^2 \left(\frac{7}{12}\right)^6 = 0.1915\)

Summing these probabilities gives:

\(P(X < 3) = 0.01341 + 0.07661 + 0.1915 = 0.28152\)

Rounding to three decimal places, the probability is:

\(\boxed{0.282}\)

Let the probability that a customer rates the logo as good be \(p = 0.42\). The number of customers rating the logo as good follows a binomial distribution \(B(n, p)\) where \(n = 10\).

We need to find the probability that fewer than 8 customers rate the logo as good, i.e., \(P(X < 8)\).

This is equivalent to \(1 - P(X \geq 8)\).

Using the binomial probability formula:

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

Calculate \(P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)\).

\(P(X = 8) = \binom{10}{8} (0.42)^8 (0.58)^2\)

\(P(X = 9) = \binom{10}{9} (0.42)^9 (0.58)^1\)

\(P(X = 10) = \binom{10}{10} (0.42)^{10} (0.58)^0\)

Thus, \(P(X \geq 8) = 10C_8 (0.42)^8 (0.58)^2 + 10C_9 (0.42)^9 (0.58)^1 + 10C_{10} (0.42)^{10}\).

Therefore, \(P(X < 8) = 1 - (10C_8 (0.42)^8 (0.58)^2 + 10C_9 (0.42)^9 (0.58)^1 + 10C_{10} (0.42)^{10})\).

Calculating this gives \(P(X < 8) = 0.983\).

Let the probability that a person is a man be \(p = 0.34\). The number of men in a sample of 14 people follows a binomial distribution \(B(14, 0.34)\).

We need to find \(P(X \leq 2)\), where \(X\) is the number of men.

\(P(X = 0) = \binom{14}{0} (0.34)^0 (0.66)^{14} = (0.66)^{14}\)

\(P(X = 1) = \binom{14}{1} (0.34)^1 (0.66)^{13} = 14 \times 0.34 \times (0.66)^{13}\)

\(P(X = 2) = \binom{14}{2} (0.34)^2 (0.66)^{12} = \frac{14 \times 13}{2} \times (0.34)^2 \times (0.66)^{12}\)

Calculating each term:

\(P(X = 0) = 0.0029758\)

\(P(X = 1) = 0.02146239\)

\(P(X = 2) = 0.071866\)

Adding these probabilities gives:

\(P(X \leq 2) = 0.0029758 + 0.02146239 + 0.071866 = 0.0963\)