(i) To find the probability distribution of \(X\), calculate \(X = R - B\) for each combination of red (R) and blue (B) spinner outcomes. The possible values of \(X\) are -2, -1, 0, 1, 2, 3. The probability distribution table is:

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

\(E(X) = \frac{-2 \times 1 + (-1) \times 2 + 0 \times 3 + 1 \times 3 + 2 \times 2 + 3 \times 1}{12} = 0.5\)

\(\text{Var}(X) = \frac{(-2)^2 \times 1 + (-1)^2 \times 2 + 0^2 \times 3 + 1^2 \times 3 + 2^2 \times 2 + 3^2 \times 1}{12} - (0.5)^2\)

\(= \frac{26}{12} - \frac{1}{4} = \frac{23}{12}\)

(iii) The probability that \(X = 1\) given \(X\) is non-zero is:

\(P(X \neq 0) = \frac{9}{12}\)

\(P(X = 1 \mid X \neq 0) = \frac{P(X = 1 \cap X \neq 0)}{P(X \neq 0)} = \frac{\frac{3}{12}}{\frac{9}{12}} = \frac{1}{3}\)

(i) The sum of all probabilities must equal 1:

\(6p + 0.1 = 1\)

Solving for \(p\):

\(6p = 0.9\)

\(p = 0.15\)

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

\(\text{Var}(X) = (-1)^2 \times p + 0^2 \times p + 1^2 \times 2p + 2^2 \times 2p + 4^2 \times 0.1 - (1.15)^2\)

Substitute \(p = 0.15\):

\(\text{Var}(X) = 1 \times 0.15 + 0 + 1 \times 0.3 + 4 \times 0.3 + 16 \times 0.1 - 1.15^2\)

\(= 0.15 + 0 + 0.3 + 1.2 + 1.6 - 1.3225\)

\(= 1.9275\)

Rounded to three significant figures:

\(\text{Var}(X) = 1.93\)

(a) To find the probabilities, we use \(P(X = x) = k(x^2 - 1)\) for each value of \(x\):

• For \(x = -2\), \(P(X = -2) = k((-2)^2 - 1) = 3k\).
• For \(x = 2\), \(P(X = 2) = k(2^2 - 1) = 3k\).
• For \(x = 3\), \(P(X = 3) = k(3^2 - 1) = 8k\).

Since the total probability must be 1, we have:

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

\(14k = 1\)

\(k = \frac{1}{14}\)

Thus, the probability distribution table is:

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

\(E(X) = (-2) \times \frac{3}{14} + 2 \times \frac{3}{14} + 3 \times \frac{8}{14}\)

\(= \frac{-6}{14} + \frac{6}{14} + \frac{24}{14}\)

\(= \frac{12}{14} = \frac{12}{7}\)

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

\(\text{Var}(X) = (-2)^2 \times \frac{3}{14} + 2^2 \times \frac{3}{14} + 3^2 \times \frac{8}{14} - (E(X))^2\)

\(= 4 \times \frac{3}{14} + 4 \times \frac{3}{14} + 9 \times \frac{8}{14} - \left(\frac{12}{7}\right)^2\)

\(= \frac{12 + 12 + 72}{14} - \frac{144}{49}\)

\(= \frac{96}{14} - \frac{144}{49}\)

\(= \frac{192}{49}\)

To solve this problem, we need to calculate the probability distribution for the number of heads obtained when the three coins are tossed.

Let the random variable X represent the number of heads obtained. The possible values of X are 0, 1, 2, and 3.

1. Probability of 0 heads, P(0):
\(P(0) = (1 - 0.4) \times (1 - 0.75) \times (1 - 0.5) = 0.6 \times 0.25 \times 0.5 = 0.075\)

2. Probability of 1 head, P(1):
\(P(1) = 0.4 \times 0.25 \times 0.5 + 0.6 \times 0.75 \times 0.5 + 0.6 \times 0.25 \times 0.5 = 0.35\)

3. Probability of 2 heads, P(2):
\(P(2) = 0.4 \times 0.75 \times 0.5 + 0.4 \times 0.25 \times 0.5 + 0.6 \times 0.75 \times 0.5 = 0.425\)

4. Probability of 3 heads, P(3):
\(P(3) = 0.4 \times 0.75 \times 0.5 = 0.15\)

The probability distribution table is:

To calculate the mean, E(X):
\(E(X) = 0 \times 0.075 + 1 \times 0.35 + 2 \times 0.425 + 3 \times 0.15 = 1.65\)

To calculate the variance, Var(X):
\(Var(X) = (0^2 \times 0.075) + (1^2 \times 0.35) + (2^2 \times 0.425) + (3^2 \times 0.15) - (1.65)^2 = 0.678\)

To solve this problem, we first need to determine the probability distribution of the random variable \(X\), which represents the number of cats chosen when 3 animals are selected from 5 dogs and 2 cats.

We can have 0, 1, or 2 cats chosen. The probabilities are calculated as follows:

• \(P(X = 0) = \frac{\binom{5}{3} \cdot \binom{2}{0}}{\binom{7}{3}} = \frac{\frac{5 \times 4 \times 3}{3 \times 2 \times 1}}{\frac{7 \times 6 \times 5}{3 \times 2 \times 1}} = \frac{2}{7}\)
• \(P(X = 1) = \frac{\binom{5}{2} \cdot \binom{2}{1}}{\binom{7}{3}} = \frac{\frac{5 \times 4}{2 \times 1} \cdot 2}{\frac{7 \times 6 \times 5}{3 \times 2 \times 1}} = \frac{4}{7}\)
• \(P(X = 2) = \frac{\binom{5}{1} \cdot \binom{2}{2}}{\binom{7}{3}} = \frac{5}{\frac{7 \times 6 \times 5}{3 \times 2 \times 1}} = \frac{1}{7}\)

The probability distribution table is:

Given \(E(X) = \frac{6}{7}\), we find \(\text{Var}(X)\) using the formula:

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

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

\(E(X^2) = 0^2 \cdot \frac{2}{7} + 1^2 \cdot \frac{4}{7} + 2^2 \cdot \frac{1}{7} = 0 + \frac{4}{7} + \frac{4}{7} = \frac{8}{7}\)

Then, \(\text{Var}(X) = \frac{8}{7} - \left(\frac{6}{7}\right)^2 = \frac{8}{7} - \frac{36}{49} = \frac{56}{49} - \frac{36}{49} = \frac{20}{49}\)

Thus, \(\text{Var}(X) = \frac{20}{49}\) or approximately 0.408.