The problem can be modeled using a binomial distribution where the probability of success (owning a cell phone) is given by \(p = 0.75\), and the number of trials \(n = 8\).

We need to find the probability that the number of successes (adults owning a cell phone) is between 4 and 6 inclusive, i.e., \(P(4 \leq X \leq 6)\).

This can be calculated as:

\(P(4 \leq X \leq 6) = P(X = 4) + P(X = 5) + P(X = 6)\)

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

\(P(X = 4) = \binom{8}{4} (0.75)^4 (0.25)^4\)

\(P(X = 5) = \binom{8}{5} (0.75)^5 (0.25)^3\)

\(P(X = 6) = \binom{8}{6} (0.75)^6 (0.25)^2\)

Calculating each:

\(P(X = 4) = 70 \times (0.75)^4 \times (0.25)^4\)

\(P(X = 5) = 56 \times (0.75)^5 \times (0.25)^3\)

\(P(X = 6) = 28 \times (0.75)^6 \times (0.25)^2\)

Summing these probabilities gives:

\(P(4 \leq X \leq 6) = 0.606\)

(i) To find the probability that the numbers on four dice add up to 5, consider the combinations: (1, 1, 1, 2), (1, 1, 2, 1), (1, 2, 1, 1), (2, 1, 1, 1). Each die has a probability of \(\frac{1}{6}\) for a specific number. The probability for one combination is \(\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{1296}\). There are 4 such combinations, so the total probability is \(4 \times \frac{1}{1296} = \frac{1}{324}\).

(ii) Let \(p = \frac{1}{324}\) be the probability from part (i) and \(q = 1 - p = \frac{323}{324}\). We use the binomial probability formula for exactly 1 or 2 successes in 7 trials: \(P(1,2) = \binom{7}{1} p^1 q^6 + \binom{7}{2} p^2 q^5\). Calculate each term: \(\binom{7}{1} \times \frac{1}{324} \times \left(\frac{323}{324}\right)^6 + \binom{7}{2} \times \left(\frac{1}{324}\right)^2 \times \left(\frac{323}{324}\right)^5 = 0.0214\).

Let the random variable \(X\) represent the number of houses with solar heating. \(X\) follows a binomial distribution with parameters \(n = 19\) and \(p = 0.12\), i.e., \(X \sim B(19, 0.12)\).

We need to find \(P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)\).

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

\(P(X = 0) = \binom{19}{0} (0.12)^0 (0.88)^{19} = (0.88)^{19}\)

\(P(X = 1) = \binom{19}{1} (0.12)^1 (0.88)^{18} = 19 \times (0.12) \times (0.88)^{18}\)

\(P(X = 2) = \binom{19}{2} (0.12)^2 (0.88)^{17} = \frac{19 \times 18}{2} \times (0.12)^2 \times (0.88)^{17}\)

\(P(X = 3) = \binom{19}{3} (0.12)^3 (0.88)^{16} = \frac{19 \times 18 \times 17}{6} \times (0.12)^3 \times (0.88)^{16}\)

Summing these probabilities gives:

\(P(X < 4) = (0.88)^{19} + 19 \times (0.12) \times (0.88)^{18} + \frac{19 \times 18}{2} \times (0.12)^2 \times (0.88)^{17} + \frac{19 \times 18 \times 17}{6} \times (0.12)^3 \times (0.88)^{16}\)

Calculating these values, we find \(P(X < 4) = 0.813\).

(i) The three conditions for a binomial distribution are:

1. The probability of success, denoted as \(p\), is constant for each trial.
2. The trials are independent.
3. There are a fixed number of trials, and each trial has only two possible outcomes (success or failure).

(ii) Let \(X\) be the number of months George buys shares in a small company. \(X\) follows a binomial distribution with parameters \(n = 18\) and \(p = 0.15\).

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 \(P(X = 2)\) using the binomial probability formula:

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

\(P(X = 0) = \binom{18}{0} (0.15)^0 (0.85)^{18} = (0.85)^{18}\)

\(P(X = 1) = \binom{18}{1} (0.15)^1 (0.85)^{17} = 18 \times (0.15) \times (0.85)^{17}\)

\(P(X = 2) = \binom{18}{2} (0.15)^2 (0.85)^{16} = \frac{18 \times 17}{2} \times (0.15)^2 \times (0.85)^{16}\)

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

\(P(X \geq 3) = 1 - P(X \leq 2)\)

Substitute the values:

\(P(X \geq 3) = 1 - [(0.85)^{18} + 18 \times (0.15) \times (0.85)^{17} + \frac{18 \times 17}{2} \times (0.15)^2 \times (0.85)^{16}]\)

\(P(X \geq 3) = 0.520\)

The probability that a mango is medium or large is given by:

\(p = 0.70 + 0.15 = 0.85\)

We need to find the probability that fewer than 12 mangoes are medium or large. This is equivalent to finding:

\(P(X < 12) = 1 - P(X \geq 12)\)

Where \(X\) is the number of medium or large mangoes out of 14. We use the binomial distribution:

\(P(X \geq 12) = P(X = 12) + P(X = 13) + P(X = 14)\)

Using the binomial probability formula:

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

Calculate each probability:

\(P(X = 12) = \binom{14}{12} (0.85)^{12} (0.15)^2\)

\(P(X = 13) = \binom{14}{13} (0.85)^{13} (0.15)^1\)

\(P(X = 14) = \binom{14}{14} (0.85)^{14} (0.15)^0\)

Summing these gives:

\(P(X \geq 12) = 0.6479\)

Thus,

\(P(X < 12) = 1 - 0.6479 = 0.352\)