To find the probability distribution for the number of jellies Jemeel chooses, we calculate the probability for each possible number of jellies (0, 1, 2, 3) out of 3 sweets chosen.

There are a total of 8 sweets (5 jellies and 3 chocolates), and Jemeel chooses 3 sweets. The total number of ways to choose 3 sweets from 8 is \(\binom{8}{3} = 56\).

Probability of choosing 0 jellies:

Choose 3 chocolates from 3: \(\binom{3}{3} = 1\)

Probability: \(\frac{1}{56}\)

Probability of choosing 1 jelly:

Choose 1 jelly from 5 and 2 chocolates from 3: \(\binom{5}{1} \times \binom{3}{2} = 5 \times 3 = 15\)

Probability: \(\frac{15}{56}\)

Probability of choosing 2 jellies:

Choose 2 jellies from 5 and 1 chocolate from 3: \(\binom{5}{2} \times \binom{3}{1} = 10 \times 3 = 30\)

Probability: \(\frac{30}{56}\)

Probability of choosing 3 jellies:

Choose 3 jellies from 5: \(\binom{5}{3} = 10\)

Probability: \(\frac{10}{56}\)

The probability distribution table is:

(a) Given \(P(X = 4) = 3P(X = 2)\), we have:

\(4k(4 + a) = 3 \times 2k(2 + a)\)

\(16k + 4ak = 12k + 6ak\)

\(4k = 2ak\)

\(a = 2\)

Using the sum of probabilities:

\(k(1 + a) + 2k(2 + a) + 3k(3 + a) + 4k(4 + a) = 1\)

\(3k + 8k + 15k + 24k = 1\)

\(50k = 1\)

\(k = \frac{1}{50}\)

(b) The probability distribution table is:

(c) To find \(\text{Var}(X)\), use:

\(\text{Var}(X) = \sum (x^2 \cdot P(X = x)) - (E(X))^2\)

\(\text{Var}(X) = \frac{3}{50} \times 1^2 + \frac{8}{50} \times 2^2 + \frac{15}{50} \times 3^2 + \frac{24}{50} \times 4^2 - 3.2^2\)

\(\text{Var}(X) = \frac{3 + 32 + 135 + 384}{50} - 10.24\)

\(\text{Var}(X) = \frac{554}{50} - 10.24\)

\(\text{Var}(X) = 11.08 - 10.24\)

\(\text{Var}(X) = 0.84\)

\(\text{Var}(X) = \frac{21}{25}\)

To solve part (b), we need to find the probability distribution for the number of times a 1 or a 6 is rolled in 3 throws of a die. The probability of rolling a 1 or a 6 in a single throw is \(\frac{1}{3}\), and the probability of not rolling a 1 or a 6 is \(\frac{2}{3}\).

Calculating each probability:

\(P(0) = \left( \frac{2}{3} \right)^3 = \frac{8}{27}\)

\(P(1) = \binom{3}{1} \left( \frac{1}{3} \right)^1 \left( \frac{2}{3} \right)^2 = 3 \times \frac{1}{3} \times \frac{4}{9} = \frac{12}{27}\)

\(P(2) = \binom{3}{2} \left( \frac{1}{3} \right)^2 \left( \frac{2}{3} \right)^1 = 3 \times \frac{1}{9} \times \frac{2}{3} = \frac{6}{27}\)

\(P(3) = \left( \frac{1}{3} \right)^3 = \frac{1}{27}\)

For part (c), the expected value \(E(X)\) is calculated as:

\(E(X) = 0 \times \frac{8}{27} + 1 \times \frac{12}{27} + 2 \times \frac{6}{27} + 3 \times \frac{1}{27}\)

\(E(X) = \frac{0}{27} + \frac{12}{27} + \frac{12}{27} + \frac{3}{27}\)

\(E(X) = \frac{27}{27} = 1\)

(i) The probability that the first ball is red is \(\frac{3}{8}\). After removing one red ball, the probability that the second ball is red is \(\frac{2}{7}\). Thus, the probability that both balls are red is \(\frac{3}{8} \times \frac{2}{7} = \frac{3}{28}\).

(ii) The probability of choosing a red then a white ball is \(\frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\). The probability of choosing a white then a red ball is \(\frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\). Therefore, the probability of different colors is \(\frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}\).

(iii) Using conditional probability, \(P(\text{first red} | \text{second red}) = \frac{P(\text{first red and second red})}{P(\text{second red})} = \frac{\frac{3}{28}}{\frac{3}{8} \times \frac{2}{7} + \frac{5}{8} \times \frac{3}{7}} = \frac{\frac{3}{28}}{\frac{21}{56}} = \frac{2}{7}\).

(iv) The probability distribution table for \(X\) is:

(v) The expected value \(E(X) = \frac{30}{56} = \frac{15}{28}\). The variance \(\text{Var}(X) = \frac{30}{56} - \left(\frac{15}{28}\right)^2 = \frac{45}{112}\) or 0.402.

(i) To find the probability distribution of \(X\), calculate \(X = \text{Red} - \text{Blue}\) for all combinations:

(ii) Calculate \(\text{E}(X)\):

\(\text{E}(X) = \frac{-1 + 0 + 3 + 4 + 9 + 8}{12} = \frac{23}{12}\)

Calculate \(\text{Var}(X)\):

\(\text{Var}(X) = \frac{1 + 0 + 3 + 8 + 27 + 32}{12} - \left(\frac{23}{12}\right)^2 = \frac{71}{12} - \left(\frac{23}{12}\right)^2\)

\(\text{Var}(X) = \frac{323}{144} \text{ or } \frac{35}{144}\)