(i) To find the probability that the three peppers are all different colors, we calculate the number of ways to choose one pepper of each color and divide by the total number of ways to choose any three peppers.

The number of ways to choose one red, one green, and one yellow pepper is given by:

\(\binom{3}{1} \times \binom{4}{1} \times \binom{5}{1} = 3 \times 4 \times 5 = 60\)

The total number of ways to choose any three peppers from twelve is:

\(\binom{12}{3} = 220\)

Thus, the probability is:

\(\frac{60}{220} = \frac{3}{11}\)

(ii) To find the probability that exactly 2 of the peppers taken are green, we calculate the number of ways to choose 2 green peppers and 1 non-green pepper, and divide by the total number of ways to choose any three peppers.

The number of ways to choose 2 green peppers is:

\(\binom{4}{2} = 6\)

The number of ways to choose 1 non-green pepper (from the remaining 8 peppers) is:

\(\binom{8}{1} = 8\)

Thus, the number of favorable outcomes is:

\(6 \times 8 = 48\)

The probability is:

\(\frac{48}{220} = \frac{12}{55}\)

(i) The total probability of having burgers is given by:

\(0.8x + (0.2)(0.75) = 0.5\)

Solving for \(x\):

\(0.8x + 0.15 = 0.5\)

\(0.8x = 0.35\)

\(x = \frac{0.35}{0.8} = 0.438\)

(ii) To find the probability that Henk went swimming given that he had burgers, use Bayes' theorem:

\(P(S \mid B) = \frac{P(S \cap B)}{P(B)}\)

\(P(S \cap B) = (0.2)(0.75) = 0.15\)

\(P(S \mid B) = \frac{0.15}{0.5} = 0.3\)

(i) The probability that both sweets are toffees is calculated by considering each box separately and then summing the probabilities:

For Box A: \(\frac{1}{3} \times \frac{6}{10} \times \frac{5}{9}\)

For Box B: \(\frac{1}{3} \times \frac{5}{8} \times \frac{4}{7}\)

For Box C: \(\frac{1}{3} \times \frac{3}{10} \times \frac{2}{9}\)

Summing these probabilities gives:

\(\frac{1}{3} \times \frac{6}{10} \times \frac{5}{9} + \frac{1}{3} \times \frac{5}{8} \times \frac{4}{7} + \frac{1}{3} \times \frac{3}{10} \times \frac{2}{9} = \frac{53}{210} \approx 0.252\)

(ii) Given that both sweets are toffees, the probability that they came from Box A is calculated using conditional probability:

\(P(A \mid \text{TT}) = \frac{P(A \cap \text{TT})}{P(\text{TT})} = \frac{0.111}{0.252} = \frac{70}{159} \approx 0.440\)

(i) List all possible outcomes of drawing three balls: 123, 124, 125, 134, 135, 145, 234, 235, 245, 345. There are 10 possible outcomes.

To find the probability that the sum is odd, count the outcomes where the sum is odd: 124, 134, 145, 234. There are 4 such outcomes.

The probability is given by \(\frac{4}{10} = 0.4\).

(ii) To find the probability that the largest number \(L\) is 4, consider the outcomes where 4 is the largest: 134, 145, 234, 245. There are 3 such outcomes.

The probability is given by \(\frac{3}{10} = 0.3\).

To find the probability of getting exactly one new pen, we consider the two cases: one new pen and one old pen.

The probability of selecting one new pen first and then one old pen is:

\(P(N, \overline{N}) = \frac{3}{10} \times \frac{7}{9} = \frac{21}{90} = \frac{7}{30}\)

The probability of selecting one old pen first and then one new pen is:

\(P(\overline{N}, N) = \frac{7}{10} \times \frac{3}{9} = \frac{21}{90} = \frac{7}{30}\)

Adding these probabilities gives:

\(P(\text{exactly one new pen}) = \frac{7}{30} + \frac{7}{30} = \frac{14}{30} = \frac{7}{15}\)

Alternatively, using combinations:

Total ways to choose 2 pens from 10 is \(\binom{10}{2} = 45\).

Ways to choose 1 new pen from 3 and 1 old pen from 7 is \(\binom{3}{1} \times \binom{7}{1} = 3 \times 7 = 21\).

Thus, the probability is:

\(\frac{21}{45} = \frac{7}{15}\)