(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:

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