To find the probability distribution of L, we consider all possible combinations of drawing 3 balls from the set {1, 2, 3, 4, 5} and determine the largest number in each combination.

The possible values for L are 3, 4, and 5. The probabilities are calculated as follows:

For the expectation, we use the formula:

\(E(L) = \sum l \cdot P(L = l) = 3 \times 0.1 + 4 \times 0.3 + 5 \times 0.6 = 4.5\)

For the variance, we use the formula:

\(\text{Var}(L) = \sum l^2 \cdot P(L = l) - (E(L))^2 = 3^2 \times 0.1 + 4^2 \times 0.3 + 5^2 \times 0.6 - 4.5^2 = 0.45\)

(i) To find the probability distribution of \(X\), we consider all possible outcomes of throwing two dice. There are a total of 36 outcomes.

- For \(x = 1\), the outcomes are \((1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (3,1), (4,1), (5,1), (6,1)\). There are 11 outcomes, so \(P(X = 1) = \frac{11}{36}\).

- For \(x = 2\), the outcomes are \((2,2), (2,3), (2,4), (2,5), (2,6), (3,2), (4,2), (5,2), (6,2)\). There are 9 outcomes, so \(P(X = 2) = \frac{9}{36}\).

- For \(x = 3\), the outcomes are \((3,3), (3,4), (3,5), (3,6), (4,3), (5,3), (6,3)\). There are 7 outcomes, so \(P(X = 3) = \frac{7}{36}\).

- For \(x = 4\), the outcomes are \((4,4), (4,5), (4,6), (5,4), (6,4)\). There are 5 outcomes, so \(P(X = 4) = \frac{5}{36}\).

- For \(x = 5\), the outcomes are \((5,5), (5,6), (6,5)\). There are 3 outcomes, so \(P(X = 5) = \frac{3}{36}\).

- For \(x = 6\), the outcome is \((6,6)\). There is 1 outcome, so \(P(X = 6) = \frac{1}{36}\).

(ii) To find \(\mathbb{E}(X)\), we calculate:

\(\mathbb{E}(X) = 1 \times \frac{11}{36} + 2 \times \frac{9}{36} + 3 \times \frac{7}{36} + 4 \times \frac{5}{36} + 5 \times \frac{3}{36} + 6 \times \frac{1}{36}\)

\(= \frac{11}{36} + \frac{18}{36} + \frac{21}{36} + \frac{20}{36} + \frac{15}{36} + \frac{6}{36}\)

\(= \frac{91}{36}\)

To solve part (ii), we need to find the probability distribution for the number of new pens in a sample of two pens.

Let \(X\) be the random variable representing the number of new pens in the sample. \(X\) can take values 0, 1, or 2.

1. Probability of 0 new pens: \(P(X = 0) = \frac{7}{10} \times \frac{6}{9} = \frac{7}{15}\).

2. Probability of 1 new pen: \(P(X = 1) = \left( \frac{3}{10} \times \frac{7}{9} \right) + \left( \frac{7}{10} \times \frac{3}{9} \right) = \frac{7}{15}\).

3. Probability of 2 new pens: \(P(X = 2) = \frac{3}{10} \times \frac{2}{9} = \frac{1}{15}\).

For part (iii), calculate the expected value \(E(X)\):

\(E(X) = 0 \times \frac{7}{15} + 1 \times \frac{7}{15} + 2 \times \frac{1}{15} = \frac{7}{15} + \frac{2}{15} = \frac{9}{15} = \frac{3}{5}\).

(i) The possible scores on the die are 1, 2, 3, and 4. The corresponding areas A are 1, 4, 9, and 16. The probability distribution is:

(ii) To find the expected value E(A), calculate:

\(E(A) = 1 \times \frac{1}{2} + 4 \times \frac{1}{6} + 9 \times \frac{1}{6} + 16 \times \frac{1}{6} = 5.33.\)

To find the variance Var(A), calculate:

\(Var(A) = (1^2 \times \frac{1}{2} + 4^2 \times \frac{1}{6} + 9^2 \times \frac{1}{6} + 16^2 \times \frac{1}{6}) - (5.33)^2 = 30.9.\)

(a) To find the probability distribution of \(X\), we calculate all possible sums of the numbers from the red and blue spinners and their probabilities:

(b) Given \(E(X) = 0.25\), we use the formula for variance:

\(\text{Var}(X) = E(X^2) - [E(X)]^2\)

Calculate \(E(X^2)\):

\(E(X^2) = \frac{1}{16}(-2)^2 + \frac{3}{16}(-1)^2 + \frac{5}{16}(0)^2 + \frac{5}{16}(1)^2 + \frac{2}{16}(2)^2\)

\(= \frac{1}{16} \times 4 + \frac{3}{16} \times 1 + \frac{5}{16} \times 0 + \frac{5}{16} \times 1 + \frac{2}{16} \times 4\)

\(= \frac{4}{16} + \frac{3}{16} + 0 + \frac{5}{16} + \frac{8}{16} = \frac{20}{16} = 1.25\)

Now, calculate \(\text{Var}(X)\):

\(\text{Var}(X) = 1.25 - (0.25)^2 = 1.25 - 0.0625 = 1.1875\)