(i) Complete the tree diagram for Bag A (3 white, 4 red, two balls chosen without replacement).

First ball: \( P(W) = \frac{3}{7} \), \( P(R) = \frac{4}{7} \).

Second ball (after white): \( P(W) = \frac{2}{6} \), \( P(R) = \frac{4}{6} \).

Second ball (after red): \( P(W) = \frac{3}{6} \), \( P(R) = \frac{3}{6} \).

Answer: Probabilities on the second branches are \( \frac{4}{6} \), \( \frac{3}{6} \), \( \frac{3}{6} \), as shown below.

(ii) Find \( P(\text{two white}) \).

\[ P(WW) = P(W_1) \cdot P(W_2 | W_1) = \frac{3}{7} \cdot \frac{2}{6} = \frac{6}{42} = \frac{1}{7} \]

Answer: \( \frac{1}{7} \).

Find \( P(\text{two white}) \) with die roll (1 or 2: Bag A, else: Bag B).

Die: \( P(\text{Bag A}) = P(1 \text{ or } 2) = \frac{2}{6} = \frac{1}{3} \), \( P(\text{Bag B}) = \frac{4}{6} = \frac{2}{3} \).

From part (a)(ii): \( P(WW | A) = \frac{1}{7} \). Given: \( P(WW | B) = \frac{2}{7} \).

\[ P(WW) = P(WW \cap A) + P(WW \cap B) = P(A) \cdot P(WW | A) + P(B) \cdot P(WW | B) \]

\[ = \frac{1}{3} \cdot \frac{1}{7} + \frac{2}{3} \cdot \frac{2}{7} = \frac{1}{21} + \frac{4}{21} = \frac{5}{21} \]

Answer:</_apps.1-2″> \( \frac{5}{21} \).

Find \( P(A | WW) \).

Using conditional probability: \[ P(A | WW) = \frac{P(A \cap WW)}{P(WW)} \]

Numerator: \( P(A \cap WW) = P(A) \cdot P(WW | A) = \frac{1}{3} \cdot \frac{1}{7} = \frac{1}{21} \).

Denominator: From part (b), \( P(WW) = \frac{5}{21} \).

\[ P(A | WW) = \frac{\frac{1}{21}}{\frac{5}{21}} = \frac{1}{5} \]

Answer: \( \frac{1}{5} \).