Solution — 2023 Question 10 Probability

(a)

Tree diagram for 2023 Q10

(b)

From the tree diagram,

\begin{matrix} P (same colour) \\ = \frac{5}{12} (\frac{7}{9}) + \frac{4}{12} (\frac{3}{9}) + \frac{3}{12} (\frac{1}{9}) \\ = \frac{25}{54} ∎ \end{matrix}

(c)

\begin{matrix} P (Dan B | Lui R) \\ = \frac{P (Dan B \cap Lui R)}{P (Lui R)} \\ = \frac{\frac{4}{12} (\frac{6}{9})}{\frac{5}{12} (\frac{7}{9}) + \frac{4}{12} (\frac{6}{9}) + \frac{3}{12} (\frac{6}{9})} \\ = \frac{24}{77} ∎ \end{matrix}

(d)

tree diagram for including sophie

\begin{matrix} P (all green) \\ = \frac{3}{12} (\frac{1}{9}) (\frac{3}{12}) \\ = \frac{1}{144} ∎ \end{matrix}

(e)

For exactly two green, Dan must pick green (otherwise bag B for Lui doesn’t have green).

Case 1: Dan green, Lui green, Sophie not green.
Case 2: Dan green, Lui not green, Sophie green.

\begin{matrix} P (exactly 2 green) \\ = \frac{3}{12} (\frac{1}{9}) (\frac{9}{12}) + \frac{3}{12} (\frac{8}{9}) (\frac{2}{12}) \\ = \frac{25}{432} ∎ \end{matrix}

Remark: note how we only cared if Sophie picked green or not green, not the actual color. This reduces the number of cases we have to handle.