To find the probability distribution, we first determine \(k\) by ensuring the probabilities sum to 1:

\(k(-2)^2 + k(1)^2 + k(2)^2 + k(3)^2 = 1\)

\(4k + k + 4k + 9k = 18k = 1\)

\(k = \frac{1}{18}\)

Thus, the probability distribution table is:

To find \(E(X)\):

\(E(X) = \frac{4 \times (-2) + 1 \times 1 + 4 \times 2 + 9 \times 3}{18}\)

\(= \frac{-8 + 1 + 8 + 27}{18}\)

\(= \frac{28}{18} = \frac{14}{9}\)

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

\(\text{Var}(X) = \frac{4 \times (-2)^2 + 1 \times 1^2 + 4 \times 2^2 + 9 \times 3^2}{18} - \left(\frac{14}{9}\right)^2\)

\(= \frac{16 + 1 + 16 + 81}{18} - \left(\frac{14}{9}\right)^2\)

\(= \frac{114}{18} - \frac{196}{81}\)

\(= \frac{317}{81}\)

Let the probability of choosing a rope of length 3 m be denoted as \(P(3)\) and the probability of choosing a rope of length 5 m be \(P(5)\).

Given that there are four times as many ropes of length 3 m as there are ropes of length 5 m, we have:

\(P(3) = \frac{4}{5} \quad \text{and} \quad P(5) = \frac{1}{5}\)

The expectation \(E(X)\) is calculated as:

\(E(X) = 3 \times P(3) + 5 \times P(5) = 3 \times \frac{4}{5} + 5 \times \frac{1}{5} = \frac{12}{5} + \frac{5}{5} = \frac{17}{5} = 3.4\)

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

First, find \(E(X^2)\):

\(E(X^2) = 3^2 \times P(3) + 5^2 \times P(5) = 9 \times \frac{4}{5} + 25 \times \frac{1}{5} = \frac{36}{5} + \frac{25}{5} = \frac{61}{5}\)

Then, use the formula for variance:

\(\text{Var}(X) = E(X^2) - (E(X))^2 = \frac{61}{5} - \left(\frac{17}{5}\right)^2 = \frac{61}{5} - \frac{289}{25}\)

Convert \(\frac{61}{5}\) to \(\frac{305}{25}\) and subtract:

\(\text{Var}(X) = \frac{305}{25} - \frac{289}{25} = \frac{16}{25} = 0.64\)

To solve this problem, we first need to determine the probability distribution of \(X\), the number of girls in the team.

We have 3 boys and 5 girls, and we are choosing a team of 4. The possible values for \(X\) are 1, 2, 3, and 4.

For \(X = 1\):

We choose 1 girl and 3 boys. The probability is given by:

\(P(X=1) = \frac{\binom{5}{1} \cdot \binom{3}{3}}{\binom{8}{4}} = \frac{5 \cdot 1}{70} = \frac{1}{14}\)

For \(X = 2\):

We choose 2 girls and 2 boys. The probability is given by:

\(P(X=2) = \frac{\binom{5}{2} \cdot \binom{3}{2}}{\binom{8}{4}} = \frac{10 \cdot 3}{70} = \frac{3}{7}\)

For \(X = 3\):

We choose 3 girls and 1 boy. The probability is given by:

\(P(X=3) = \frac{\binom{5}{3} \cdot \binom{3}{1}}{\binom{8}{4}} = \frac{10 \cdot 3}{70} = \frac{3}{7}\)

For \(X = 4\):

We choose 4 girls. The probability is given by:

\(P(X=4) = \frac{\binom{5}{4} \cdot \binom{3}{0}}{\binom{8}{4}} = \frac{5 \cdot 1}{70} = \frac{1}{14}\)

Thus, the probability distribution table is:

Given \(E(X) = \frac{5}{2}\), we calculate \(\text{Var}(X)\) using the formula:

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

First, calculate \(E(X^2)\):

\(E(X^2) = 1^2 \cdot \frac{1}{14} + 2^2 \cdot \frac{3}{7} + 3^2 \cdot \frac{3}{7} + 4^2 \cdot \frac{1}{14}\)

\(= \frac{1}{14} + \frac{12}{7} + \frac{27}{7} + \frac{16}{14}\)

\(= \frac{1}{14} + \frac{24}{14} + \frac{54}{14} + \frac{16}{14}\)

\(= \frac{95}{14}\)

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

\(\text{Var}(X) = \frac{95}{14} - \left(\frac{5}{2}\right)^2\)

\(= \frac{95}{14} - \frac{25}{4}\)

\(= \frac{95}{14} - \frac{87.5}{14}\)

\(= \frac{15}{28}\)

Thus, \(\text{Var}(X) = \frac{15}{28}\) (0.536).

(i) The score is 6 if the cards drawn are (3, 9) or (9, 3). The probability of each pair is \(\frac{1}{5} \times \frac{1}{5} = \frac{1}{25}\). Therefore, the probability of a score of 6 is \(2 \times \frac{1}{25} = \frac{2}{25} = 0.08\).

(ii) The probability distribution table is:

(iii) The mean score is calculated as \(\sum x p = 0 \times 0.2 + 1 \times 0.24 + 2 \times 0.08 + 3 \times 0.08 + 4 \times 0.16 + 5 \times 0.16 + 6 \times 0.08 = 2.56\).

(iv) Judy wins if the score is 4, 5, or 6. The probability of these scores is \(0.16 + 0.16 + 0.08 = 0.4\). The probability of a draw (score 0) is 0.2. Therefore, the probability that Judy wins with the second choice is \(0.2 \times 0.4 = 0.08\).

(v) The probability that Judy wins with the \(n\)th choice is \((0.2)^{n-1} \times 0.4\).

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

Since the sum of all probabilities must equal 1, we have:

\(k + 2k + 3k + 4k + 5k = 1\)

\(15k = 1\)

\(k = \frac{1}{15}\) (0.0667)

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

\(E(X) = 1 \cdot k + 2 \cdot 2k + 3 \cdot 3k + 4 \cdot 4k + 5 \cdot 5k\)

\(= k + 4k + 9k + 16k + 25k\)

\(= 55k\)

Substituting \(k = \frac{1}{15}\), we get:

\(E(X) = 55 \times \frac{1}{15} = \frac{55}{15} = \frac{11}{3}\) (3.67)