The sum of all probabilities must equal 1:

\(4p + 5p^2 + 1.5p + 2.5p + 1.5p = 1\)

Simplify the equation:

\(10p^2 + 19p = 1\)

Rearrange to form a quadratic equation:

\(10p^2 + 19p - 1 = 0\)

Using the quadratic formula \(p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 10\), \(b = 19\), and \(c = -1\):

\(p = \frac{-19 \pm \sqrt{19^2 - 4 \times 10 \times (-1)}}{2 \times 10}\)

\(p = \frac{-19 \pm \sqrt{361 + 40}}{20}\)

\(p = \frac{-19 \pm \sqrt{401}}{20}\)

Calculating the roots, we find \(p = 0.1\) or \(p = -2\).

Since probability cannot be negative, \(p = 0.1\).

(i) Let \(y\) be the probability of throwing an odd number. Then the probability of throwing an even number is \(2y\). Since there are three odd and three even numbers, we have:

\(3y + 6y = 1\)

\(9y = 1\)

\(y = \frac{1}{9}\)

Thus, the probability of throwing an odd number is \(\frac{1}{3}\).

(ii) A score of 8 is obtained by throwing a 6. Since 6 is an even number, the probability \(P(X = 8) = \frac{2}{9}\).

(iii) The variance \(\text{Var}(X)\) is calculated as follows:

\(\text{Var}(X) = \frac{1}{9}(4^2 + 6^2 + 7^2 + 8^2 + 10^2) - \left(\frac{58}{9}\right)^2\)

\(= \frac{1}{9}(16 + 36 + 49 + 64 + 100) - \left(\frac{58}{9}\right)^2\)

\(= \frac{1}{9}(265) - \left(\frac{58}{9}\right)^2\)

\(= 29.44 - 41.78\)

\(= 4.02\)

(iv) The probability that the total of the scores on the two throws is 16 is:

\(P(6,10) + P(10,6) + P(8,8) = \frac{1}{81} + \frac{1}{81} + \frac{4}{81} = \frac{6}{81} = \frac{2}{27}\)

(v) Given that the total score is 16, the probability that the first score was 6 is:

\(P(6,10) = \frac{1}{81}\)

\(P(\text{first score 6 given total 16}) = \frac{\frac{1}{81}}{\frac{6}{81}} = \frac{1}{6}\)

First, use the fact that the sum of all probabilities must equal 1:

\(a + b + 0.15 + 0.4 = 1\)

\(a + b = 0.45\)

Next, use the given expected value \(E(X) = 0.75\):

\(E(X) = (-3)a + (-1)b + 0 \times 0.15 + 4 \times 0.4 = 0.75\)

\(-3a - b + 1.6 = 0.75\)

Simplify to find another equation:

\(-3a - b = -0.85\)

Now solve the system of equations:

1. \(a + b = 0.45\)

2. \(-3a - b = -0.85\)

Add the two equations to eliminate \(b\):

\(a + b - 3a - b = 0.45 - 0.85\)

\(-2a = -0.4\)

\(a = 0.2\)

Substitute \(a = 0.2\) back into the first equation:

\(0.2 + b = 0.45\)

\(b = 0.25\)

Thus, \(a = 0.2\) and \(b = 0.25\).

To find the values of \(p\) and \(q\), we use the following equations:

1. The sum of probabilities: \(0.08 + p + 0.12 + 0.16 + q + 0.22 = 1\)

2. The mean equation: \(-2(0.08) - 1(p) + 0(0.12) + 1(0.16) + 2(q) + 3(0.22) = 1.05\)

Simplifying these equations:

\(p + q = 0.42\)

\(-0.16 - p + 0.16 + 2q + 0.66 = 1.05\)

\(-p + 2q = 0.39\)

Solving these equations simultaneously:

\(p = 0.15\)

\(q = 0.27\)

To find the variance of \(X\), we use the formula:

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

\(\text{Var}(X) = 4(0.08) + p + 0.16 + 4q + 1.98 - (1.05)^2\)

Substituting \(p = 0.15\) and \(q = 0.27\):

\(\text{Var}(X) = 2.59\)

(i) To find \(P(X = 2)\), consider the cases:

- If Gohan throws a 1, she must throw again and get a 1 to score 2. The probability is \(\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}\).

- If Gohan throws a 2, her score is 2. The probability is \(\frac{1}{4}\).

Thus, \(P(X = 2) = \frac{1}{16} + \frac{1}{4} = \frac{5}{16}\).

(ii) To calculate \(E(X)\):

\(E(X) = \sum xP(X = x) = 2 \times \frac{5}{16} + 3 \times \frac{1}{16} + 4 \times \frac{3}{8} + 5 \times \frac{1}{8} + 6 \times \frac{1}{16} + 7 \times \frac{1}{16}\)

\(= \frac{10}{16} + \frac{3}{16} + \frac{12}{16} + \frac{10}{16} + \frac{6}{16} + \frac{7}{16} = \frac{60}{16} = 3.75\).

To calculate \(\text{Var}(X)\):

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

\(= 2^2 \times \frac{5}{16} + 3^2 \times \frac{1}{16} + 4^2 \times \frac{3}{8} + 5^2 \times \frac{1}{8} + 6^2 \times \frac{1}{16} + 7^2 \times \frac{1}{16} - (3.75)^2\)

\(= \frac{20}{16} + \frac{9}{16} + \frac{48}{16} + \frac{25}{8} + \frac{36}{16} + \frac{49}{16} - 14.0625\)

\(= \frac{260}{16} - \frac{225}{16} = \frac{35}{16} = 2.19\).