Let X be the random variable representing the number of green sweets taken. The possible values of X are 0, 1, and 2.

Calculate the probabilities:

\(P(X = 0) = P(B, B) = \frac{5}{7} \times \frac{4}{6} = \frac{10}{21}\)

\(P(X = 1) = P(G, B) + P(B, G) = \frac{2}{7} \times \frac{5}{6} + \frac{5}{7} \times \frac{2}{6} = \frac{10}{21}\)

\(P(X = 2) = P(G, G) = \frac{2}{7} \times \frac{1}{6} = \frac{1}{21}\)

Calculate the expected value:

\(E(X) = 0 \times \frac{10}{21} + 1 \times \frac{10}{21} + 2 \times \frac{1}{21} = \frac{4}{7}\)

Calculate the variance:

\(\text{Var}(X) = 0^2 \times \frac{10}{21} + 1^2 \times \frac{10}{21} + 2^2 \times \frac{1}{21} - \left(\frac{4}{7}\right)^2 = \frac{50}{147}\)

To find the probability distribution for the number of white roses chosen, we calculate the probabilities for choosing 0, 1, or 2 white roses.

There are a total of 10 roses (5 yellow, 3 red, 2 white).

Probability of choosing 0 white roses:

The probability of choosing 3 non-white roses is:

\(P(0) = \frac{8}{10} \times \frac{7}{9} \times \frac{6}{8} = \frac{42}{90}\)

Probability of choosing 1 white rose:

The probability of choosing 1 white and 2 non-white roses is:

\(P(1) = \left( \frac{2}{10} \times \frac{8}{9} \times \frac{7}{8} \right) \times 3 = \frac{42}{90}\)

The factor of 3 accounts for the different orders in which the roses can be chosen.

Probability of choosing 2 white roses:

The probability of choosing 2 white and 1 non-white rose is:

\(P(2) = \left( \frac{2}{10} \times \frac{1}{9} \times \frac{8}{8} \right) \times 3 = \frac{6}{90}\)

The factor of 3 accounts for the different orders in which the roses can be chosen.

Thus, the probability distribution table is:

(i) To complete the table, calculate the sum of the numbers from Spinner A and Spinner B for each combination. For example, if Spinner A shows 1 and Spinner B shows -3, then \(X = 1 + (-3) = -2\). Complete the table similarly for all combinations.

(ii) To find the probability distribution of \(X\), count the frequency of each sum from the completed table and divide by the total number of outcomes (16). The probabilities are: \(-2: \frac{1}{16}, -1: \frac{2}{16}, 0: \frac{4}{16}, 1: \frac{3}{16}, 2: \frac{2}{16}, 3: \frac{3}{16}, 4: \frac{1}{16}\).

(iii) Calculate \(\text{Var}(X)\) using the formula \(\text{Var}(X) = E(X^2) - [E(X)]^2\). First, find \(E(X) = \sum x \cdot P(x) = 1\). Then, \(E(X^2) = \sum x^2 \cdot P(x) = \frac{62}{16}\). Thus, \(\text{Var}(X) = \frac{62}{16} - 1^2 = \frac{23}{8} = 2.875\).

(iv) To find the probability that \(X\) is even given \(X\) is positive, use conditional probability: \(P(\text{even} \mid \text{positive}) = \frac{P(\text{even and positive})}{P(\text{positive})}\). The even positive values are 2 and 4, with probabilities \(\frac{2}{16} + \frac{1}{16} = \frac{3}{16}\). The positive values are 1, 2, 3, and 4, with probabilities \(\frac{3}{16} + \frac{2}{16} + \frac{3}{16} + \frac{1}{16} = \frac{9}{16}\). Thus, \(P(\text{even} \mid \text{positive}) = \frac{3/16}{9/16} = \frac{5}{9}\).

(i) The tree diagram shows three stages, each with a probability of success \(S = 0.4\) and failure \(F = 0.6\). Each branch represents a login attempt.

(ii) The probability distribution is calculated as follows:

• \(P(X = 0) = 0.4\)
• \(P(X = 1) = 0.6 \times 0.4 = 0.24\)
• \(P(X = 2) = 0.6^2 \times 0.4 = 0.144\)
• \(P(X = 3) = 0.6^3 = 0.216\)

(iii) The expected number of unsuccessful attempts is calculated using \(E(X) = \sum x P(X = x)\):

\(E(X) = 0 \times 0.4 + 1 \times 0.24 + 2 \times 0.144 + 3 \times 0.216 = 1.176\)

(i) The probability that exactly one rabbit is white can be calculated by considering two cases: one white and one non-white rabbit.

Case 1: First rabbit is white, second is non-white: \(\frac{6}{9} \times \frac{3}{8}\).

Case 2: First rabbit is non-white, second is white: \(\frac{3}{9} \times \frac{6}{8}\).

Adding these probabilities gives: \(\frac{6}{9} \times \frac{3}{8} + \frac{3}{9} \times \frac{6}{8} = \frac{1}{2}\).

(ii) The probability distribution table is constructed as follows:

(iii) The expected value \(E(X)\) is calculated as:

\(E(X) = 0 \times \frac{1}{12} + 1 \times \frac{1}{2} + 2 \times \frac{5}{12} = \frac{16}{12} = \frac{4}{3}\).