Let:

• \(M\) = event that a person is male, \(P(M) = 0.54\)

• \(F\) = event that a person is female, \(P(F) = 0.46\)

• \(C\) = event that a person is colour-blind

We know:

• \(P(C|M) = 0.05\)

• \(P(C|F) = 0.02\)

We need to find \(P(M|C)\), the probability that a person is male given that they are colour-blind.

Using Bayes' theorem:

\(P(M|C) = \frac{P(C|M) \cdot P(M)}{P(C)}\)

First, calculate \(P(C)\):

\(P(C) = P(C \cap M) + P(C \cap F)\)

\(P(C \cap M) = P(C|M) \cdot P(M) = 0.05 \times 0.54 = 0.027\)

\(P(C \cap F) = P(C|F) \cdot P(F) = 0.02 \times 0.46 = 0.0092\)

\(P(C) = 0.027 + 0.0092 = 0.0362\)

Now, substitute back into Bayes' theorem:

\(P(M|C) = \frac{0.027}{0.0362} \approx 0.746\)

Thus, the probability that the person is male given that they are colour-blind is approximately 0.746, or \(\frac{135}{181}\).

(a) The tree diagram is as follows:

1st Jump: Success (S) with probability 0.2, Failure (F) with probability 0.8.

2nd Jump: If 1st was S, then S with 0.3, F with 0.7; if 1st was F, then S with 0.1, F with 0.9.

3rd Jump: If 2nd was S, then S with 0.3, F with 0.7; if 2nd was F, then S with 0.1, F with 0.9.

(b) Calculate the probability of exactly one success:

- SFF: \(0.2 \times 0.7 \times 0.9 = 0.126\)

- FSF: \(0.8 \times 0.1 \times 0.7 = 0.056\)

- FFS: \(0.8 \times 0.9 \times 0.1 = 0.072\)

Total probability of exactly one success: \(0.126 + 0.056 + 0.072 = 0.254\)

Probability of at least one success: \(1 - 0.8 \times 0.9 \times 0.9 = 0.352\)

Probability of exactly one success given at least one success: \(\frac{0.254}{0.352} = \frac{127}{176}\)

(c) Calculate the probability for six jumps:

- Only first three are successes: \(0.2 \times 0.3 \times 0.3 \times 0.7 \times 0.9 \times 0.9 = 0.010206\)

- Only last three are successes: \(0.8 \times 0.9 \times 0.9 \times 0.1 \times 0.3 \times 0.3 = 0.005832\)

Total probability: \(0.010206 + 0.005832 = 0.016038\)

(i) To find the probability of choosing a grandparent, calculate the total number of grandparents and divide by the total number of people. There are 6 grandparents (2 in House 1, 3 in House 2, 1 in House 3) and 16 people in total (7 in House 1, 7 in House 2, 2 in House 3). Thus, the probability is \(\frac{6}{16} = \frac{3}{8}\).

(ii) To find the probability that a person chosen is a grandparent, calculate the probability for each house and sum them. The probability for House 1 is \(\frac{1}{3} \times \frac{2}{7}\), for House 2 is \(\frac{1}{3} \times \frac{3}{7}\), and for House 3 is \(\frac{1}{3} \times \frac{1}{2}\). Therefore, the total probability is \(\frac{1}{3} \times \frac{2}{7} + \frac{1}{3} \times \frac{3}{7} + \frac{1}{3} \times \frac{1}{2} = \frac{17}{42}\).

(iii) Given that the person chosen is a grandparent, calculate the probability that there is also a parent in the house. For House 1, the probability is \(\frac{2}{21}\), and for House 2, it is \(\frac{3}{21}\). Thus, the probability is \(\frac{2}{21} + \frac{3}{21} = \frac{5}{21}\). Divide by the probability from part (ii), \(\frac{17}{42}\), to get \(\frac{5}{21} \div \frac{17}{42} = \frac{10}{17}\).

(i) To find the conditional probability that Rachel wins the first game given that she loses the second, we use the formula for conditional probability:

\(P(W_1 | L_2) = \frac{P(W_1 \cap L_2)}{P(L_2)}\)

First, calculate \(P(W_1 \cap L_2)\):

\(P(W_1 \cap L_2) = P(W_1) \times P(L_2 | W_1) = 0.6 \times 0.3 = 0.18\)

Next, calculate \(P(L_2)\):

\(P(L_2) = P(W_1 \cap L_2) + P(L_1 \cap L_2) = (0.6 \times 0.3) + (0.4 \times 0.6) = 0.18 + 0.24 = 0.42\)

Thus, \(P(W_1 | L_2) = \frac{0.18}{0.42} = 0.429\).

(ii) To find the probability that Rachel wins 2 games and loses 1 game out of the first three games, consider the following scenarios:

• \(P(W_1, W_2, L_3) = 0.6 \times 0.7 \times 0.3 = 0.126\)
• \(P(W_1, L_2, W_3) = 0.6 \times 0.3 \times 0.4 = 0.072\)
• \(P(L_1, W_2, W_3) = 0.4 \times 0.4 \times 0.7 = 0.112\)

Summing these probabilities gives:

\(0.126 + 0.072 + 0.112 = 0.31\).

(a) To find the probability that all 3 eggs chosen contain the same colour sweet, consider the combinations:

- All red: \(\frac{3}{12} \times \frac{2}{11} \times \frac{1}{10} = \frac{6}{1320}\)

- All orange: \(\frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} = \frac{24}{1320}\)

- All yellow: \(\frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{60}{1320}\)

Total probability: \(\frac{6 + 24 + 60}{1320} = \frac{90}{1320} = 0.0682\)

(b) Given that all three children have the same colour sweet, the probability that all 3 eggs contain a yellow sweet is:

\(\frac{\frac{60}{1320}}{\frac{90}{1320}} = \frac{60}{90} = \frac{2}{3}\) or 0.667

(c) To find the probability that at least one child chooses an egg with an orange sweet, use the complement rule:

Probability of no orange sweets: \(\frac{8}{12} \times \frac{7}{11} \times \frac{6}{10} = \frac{336}{1320} = \frac{14}{55}\)

Probability of at least one orange sweet: \(1 - \frac{14}{55} = \frac{41}{55}\) or 0.745