Let the probability that a water pistol does not work properly be denoted by \(p = 0.08\). The probability that a water pistol works properly is \(1 - p = 0.92\).

We are dealing with a binomial distribution where \(n = 19\) and \(p = 0.08\). We need to find the probability that at most 2 do not work properly, i.e., \(P(X \leq 2)\).

The probability \(P(X \leq 2)\) is given by:

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

Using the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), we calculate:

\(P(X = 0) = \binom{19}{0} (0.08)^0 (0.92)^{19} = (0.92)^{19}\)

\(P(X = 1) = \binom{19}{1} (0.08)^1 (0.92)^{18} = 19 \times 0.08 \times (0.92)^{18}\)

\(P(X = 2) = \binom{19}{2} (0.08)^2 (0.92)^{17} = \frac{19 \times 18}{2} \times (0.08)^2 \times (0.92)^{17}\)

Summing these probabilities gives:

\(P(X \leq 2) = (0.92)^{19} + 19 \times 0.08 \times (0.92)^{18} + \frac{19 \times 18}{2} \times (0.08)^2 \times (0.92)^{17}\)

Calculating this expression yields \(P(X \leq 2) = 0.809\).

Let the probability that a car is fitted with satellite navigation equipment be denoted by \(p = 0.76\).

The number of cars fitted with the equipment follows a binomial distribution with parameters \(n = 11\) and \(p = 0.76\).

We need to find \(P(X < 10)\), which is equivalent to \(1 - P(X \geq 10)\).

Calculate \(P(X \geq 10) = P(X = 10) + P(X = 11)\).

Using the binomial probability formula:

\(P(X = 10) = \binom{11}{10} (0.76)^{10} (0.24)^{1}\)

\(P(X = 11) = \binom{11}{11} (0.76)^{11} (0.24)^{0}\)

\(P(X = 10) = 11 \times (0.76)^{10} \times (0.24)\)

\(P(X = 11) = (0.76)^{11}\)

\(P(X \geq 10) = 11 \times (0.76)^{10} \times (0.24) + (0.76)^{11}\)

\(P(X \geq 10) = 0.219\)

Thus, \(P(X < 10) = 1 - 0.219 = 0.781\).

The problem involves a binomial distribution where the number of trials is 10, and the probability of success (rolling a six) is \(\frac{1}{6}\).

We need to find the probability of getting between 3 and 5 sixes inclusive, i.e., \(P(3 \leq X \leq 5)\).

This can be calculated as:

\(P(3, 4, 5) = \binom{10}{3} \left( \frac{1}{6} \right)^3 \left( \frac{5}{6} \right)^7 + \binom{10}{4} \left( \frac{1}{6} \right)^4 \left( \frac{5}{6} \right)^6 + \binom{10}{5} \left( \frac{1}{6} \right)^5 \left( \frac{5}{6} \right)^5\)

Calculating each term:

\(\binom{10}{3} \left( \frac{1}{6} \right)^3 \left( \frac{5}{6} \right)^7 = 0.155045\)

\(\binom{10}{4} \left( \frac{1}{6} \right)^4 \left( \frac{5}{6} \right)^6 = 0.054253\)

\(\binom{10}{5} \left( \frac{1}{6} \right)^5 \left( \frac{5}{6} \right)^5 = 0.012924\)

Adding these probabilities gives:

\(0.155045 + 0.054253 + 0.012924 = 0.222\)

The problem involves a binomial distribution where the probability of success (a household having a printer) is 0.68, and the number of trials is 8. We need to find the probability that 5, 6, or 7 households have a printer.

The probability of exactly k successes in a binomial distribution is given by:

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

where \(n = 8\), \(p = 0.68\), and \(1-p = 0.32\).

Calculate the probabilities for 5, 6, and 7 successes:

\(P(X = 5) = \binom{8}{5} (0.68)^5 (0.32)^3\)

\(P(X = 6) = \binom{8}{6} (0.68)^6 (0.32)^2\)

\(P(X = 7) = \binom{8}{7} (0.68)^7 (0.32)^1\)

Sum these probabilities:

\(P(5, 6, 7) = \binom{8}{5} (0.68)^5 (0.32)^3 + \binom{8}{6} (0.68)^6 (0.32)^2 + \binom{8}{7} (0.68)^7 (0.32)\)

Using the binomial coefficients and calculations:

\(P(5, 6, 7) = 0.722\)

(i) The greatest number of books that could be read is 12. The probability that a member reads this number of books is given by \(P(12) = (0.7)^{12}\).

Calculating this gives \(P(12) = 0.0138\).

(ii) To find the probability that a member reads fewer than 10 books, we calculate \(1 - P(10, 11, 12)\).

Using the binomial probability formula, we have:

\(P(10) = \binom{12}{10} (0.7)^{10} (0.3)^2\)

\(P(11) = \binom{12}{11} (0.7)^{11} (0.3)^1\)

\(P(12) = (0.7)^{12}\)

Thus, \(P(10, 11, 12) = \binom{12}{10} (0.7)^{10} (0.3)^2 + \binom{12}{11} (0.7)^{11} (0.3) + (0.7)^{12}\).

Calculating these probabilities gives \(P(10, 11, 12) = 0.2528\).

Therefore, the probability that a member reads fewer than 10 books is \(1 - 0.2528 = 0.747\).