Let the random variable \(X\) represent the number of buses in the sample of 11 vehicles. Since 16% of the vehicles are buses, the probability of selecting a bus is \(p = 0.16\).

We need to find \(P(X < 3) = P(0) + P(1) + P(2)\).

Using the binomial probability formula \(P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\), where \(n = 11\) and \(p = 0.16\):

\(P(0) = \binom{11}{0} (0.16)^0 (0.84)^{11} = (0.84)^{11} = 0.1469\)

\(P(1) = \binom{11}{1} (0.16)^1 (0.84)^{10} = 11 \times 0.16 \times (0.84)^{10} = 0.30782\)

\(P(2) = \binom{11}{2} (0.16)^2 (0.84)^9 = 55 \times (0.16)^2 \times (0.84)^9 = 0.2931\)

Therefore, \(P(X < 3) = 0.1469 + 0.30782 + 0.2931 = 0.748\).

The problem involves a binomial distribution where the probability of success (a resident being in favour) is 0.8, and the number of trials is 20. We need to find the probability that more than 17 residents are in favour, i.e., \(P(X > 17)\).

Using the binomial probability formula:

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

We calculate:

\(P(X = 18) = \binom{20}{18} (0.8)^{18} (0.2)^2\)

\(P(X = 19) = \binom{20}{19} (0.8)^{19} (0.2)^1\)

\(P(X = 20) = \binom{20}{20} (0.8)^{20}\)

Summing these probabilities gives:

\(P(X > 17) = P(X = 18) + P(X = 19) + P(X = 20)\)

Calculating each term:

\(P(X = 18) = \binom{20}{18} (0.8)^{18} (0.2)^2 = 0.13691\)

\(P(X = 19) = \binom{20}{19} (0.8)^{19} (0.2)^1 = 0.05765\)

\(P(X = 20) = \binom{20}{20} (0.8)^{20} = 0.01153\)

Adding these gives:

\(P(X > 17) = 0.13691 + 0.05765 + 0.01153 = 0.20609\)

Rounding to three decimal places, the probability is 0.206.

The probability of rolling an odd number (1, 3, or 5) is given by:

\(P(\text{odd}) = \frac{1}{6} + \frac{1}{6} + \frac{2}{6} = \frac{2}{3}\)

We need to find the probability of getting at least 7 odd numbers in 8 throws, which is \(P(7) + P(8)\).

Using the binomial probability formula:

\(P(7) = \binom{8}{7} \left( \frac{2}{3} \right)^7 \left( \frac{1}{3} \right)^1 = 8 \times \left( \frac{2}{3} \right)^7 \times \frac{1}{3} = 0.156\)

\(P(8) = \binom{8}{8} \left( \frac{2}{3} \right)^8 = \left( \frac{2}{3} \right)^8 = 0.0390\)

Thus, the probability of obtaining at least 7 odd numbers is:

\(P(7 \text{ or } 8) = 0.156 + 0.0390 = 0.195\)

The probability of throwing a 5 is given as 0.75. Therefore, the probability of not throwing a 5 is:

\(1 - 0.75 = 0.25\)

We need to find the probability of getting at least 8 fives in 10 throws. This is a binomial probability problem where:

\(n = 10, \ p = 0.75\)

The probability of getting exactly \(r\) fives is given by the binomial formula:

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

We need to calculate \(P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)\).

Calculating each term:

\(P(X = 8) = \binom{10}{8} (0.75)^8 (0.25)^2\)

\(P(X = 9) = \binom{10}{9} (0.75)^9 (0.25)^1\)

\(P(X = 10) = \binom{10}{10} (0.75)^{10}\)

Substituting the values:

\(P(X = 8) = 45 \times (0.75)^8 \times (0.25)^2\)

\(P(X = 9) = 10 \times (0.75)^9 \times (0.25)\)

\(P(X = 10) = 1 \times (0.75)^{10}\)

Summing these probabilities gives:

\(P(X \geq 8) = 0.526\)

Let \(X\) be the number of years out of 15 that have New Year's Day on a Saturday. \(X\) follows a binomial distribution with parameters \(n = 15\) and \(p = \frac{1}{7}\).

We need to find \(P(X \geq 3)\), which is equivalent to \(1 - P(X \leq 2)\).

Calculate \(P(X = 0)\), \(P(X = 1)\), and \(P(X = 2)\):

\(P(X = 0) = \left( \frac{6}{7} \right)^{15}\)

\(P(X = 1) = \binom{15}{1} \left( \frac{1}{7} \right) \left( \frac{6}{7} \right)^{14}\)

\(P(X = 2) = \binom{15}{2} \left( \frac{1}{7} \right)^2 \left( \frac{6}{7} \right)^{13}\)

Thus, \(P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)\).

\(P(X \leq 2) = \left( \frac{6}{7} \right)^{15} + 15 \left( \frac{1}{7} \right) \left( \frac{6}{7} \right)^{14} + \binom{15}{2} \left( \frac{1}{7} \right)^2 \left( \frac{6}{7} \right)^{13}\)

\(P(X \leq 2) \approx 0.0990 + 0.2476 + 0.2889 = 0.6355\)

Therefore, \(P(X \geq 3) = 1 - 0.6355 = 0.3645\).

Thus, the probability that at least 3 of these years have New Year's Day on a Saturday is approximately 0.365 (accept 0.364).