To solve this problem, we use the binomial probability formula:

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

where \(n\) is the total number of trials, \(k\) is the number of successful trials, and \(p\) is the probability of success on a single trial.

(i) For equal numbers of large and small bands, we need 10 large and 10 small bands. The probability of a band being large is 0.4, and small is 0.6. Thus,

\(P(10 \text{ large and } 10 \text{ small}) = \binom{20}{10} (0.6)^{10} (0.4)^{10}\)

Calculating this gives:

\(= 0.117\)

(ii) For more than 17 small bands, we consider 18, 19, or 20 small bands. The probability of a band being small is 0.6. Thus,

\(P(18, 19, 20 \text{ small}) = \binom{20}{18} (0.6)^{18} (0.4)^{2} + \binom{20}{19} (0.6)^{19} (0.4)^{1} + (0.6)^{20}\)

Calculating each term:

\(= 0.003087 + 0.000487 + 0.00003635\)

Summing these gives:

\(= 0.00361\)

Let the probability that an adult does not wear a watch be 0.09. We are dealing with a binomial distribution where the number of trials is 14 and the probability of success (an adult not wearing a watch) is 0.09.

We need to find the probability that more than 2 adults do not wear a watch, which is equivalent to finding:

\(P(X > 2) = 1 - P(X \leq 2)\)

Calculate \(P(X \leq 2)\) as follows:

\(P(X = 0) = (0.91)^{14}\)

\(P(X = 1) = \binom{14}{1} (0.09)^1 (0.91)^{13}\)

\(P(X = 2) = \binom{14}{2} (0.09)^2 (0.91)^{12}\)

Thus,

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

Substitute the values:

\(P(X = 0) = 0.2670\)

\(P(X = 1) = 0.3698\)

\(P(X = 2) = 0.2377\)

Therefore,

\(P(X \leq 2) = 0.2670 + 0.3698 + 0.2377 = 0.8745\)

Finally,

\(P(X > 2) = 1 - 0.8745 = 0.1255\)

Rounding to three decimal places, the probability is 0.126.

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

1. There are a fixed number of trials.
2. Each trial is independent.
3. There are only two possible outcomes for each trial.
4. The probability of success is constant for each trial.

(ii) Let the random variable \(X\) represent the number of cars made by Ford. \(X\) follows a binomial distribution with parameters \(n = 14\) and \(p = 0.28\).

We need to find \(P(X < 4)\), which is the sum of probabilities \(P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)\).

Using the binomial probability formula:

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

Calculate each probability:

\(P(X = 0) = \binom{14}{0} (0.28)^0 (0.72)^{14} = 0.72^{14} = 0.0101\)

\(P(X = 1) = \binom{14}{1} (0.28)^1 (0.72)^{13} = 14 \times 0.28 \times 0.72^{13} = 0.0548\)

\(P(X = 2) = \binom{14}{2} (0.28)^2 (0.72)^{12} = 91 \times 0.28^2 \times 0.72^{12} = 0.1385\)

\(P(X = 3) = \binom{14}{3} (0.28)^3 (0.72)^{11} = 364 \times 0.28^3 \times 0.72^{11} = 0.2154\)

Summing these probabilities gives:

\(P(X < 4) = 0.0101 + 0.0548 + 0.1385 + 0.2154 = 0.419\)

Let the random variable \(X\) represent the number of damaged tapes in a sample of 15. The probability of a tape being damaged is \(p = 0.2\), and the probability of a tape not being damaged is \(1 - p = 0.8\).

We use the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\) where \(n = 15\).

Calculate \(P(X = 0)\):

\(P(0) = (0.8)^{15} = 0.03518\)

Calculate \(P(X = 1)\):

\(P(1) = \binom{15}{1} (0.2)^1 (0.8)^{14} = 15 \times 0.2 \times (0.8)^{14} = 0.1319\)

Calculate \(P(X = 2)\):

\(P(2) = \binom{15}{2} (0.2)^2 (0.8)^{13} = 105 \times (0.2)^2 \times (0.8)^{13} = 0.2309\)

The probability that at most 2 tapes are damaged is:

\(P(X \leq 2) = P(0) + P(1) + P(2) = 0.03518 + 0.1319 + 0.2309 = 0.398\)

Let the probability of getting the required picture be 0.05 (since there is 1 desired picture out of 20).

1. The probability that Richard does not complete his collection is calculated by not getting the required picture in all 5 attempts:

   \((0.95)^5 = 0.774\)

2. The probability that he has the required picture exactly once is calculated using the binomial probability formula:

   \((0.95)^4 \times (0.05)^1 \times \binom{5}{1} = 0.204\)

3. The probability that he completes his collection with the third picture he looks at is calculated by not getting the picture in the first two attempts and getting it on the third:

   \((0.95)^2 \times (0.05) = 0.0451\)