Solution (a)
Step 1 (Formula / Governing Law & Given Data): [1 Mark]
Let $E_1$ be the event of choosing Bag I and $E_2$ be the event of choosing Bag II. Let $R$ be the event of drawing a red ball. According to the Law of Total Probability:
$$P(R) = P(E_1) \cdot P(R|E_1) + P(E_2) \cdot P(R|E_2)$$Given data:
- Die shows < 3 (i.e., 1 or 2): $P(E_1) = \frac{2}{6} = \frac{1}{3}$
- Die shows ≥ 3 (i.e., 3, 4, 5, or 6): $P(E_2) = \frac{4}{6} = \frac{2}{3}$
- Bag I (3 Red, 4 White): $P(R|E_1) = \frac{3}{7}$
- Bag II (8 Red, 6 White): $P(R|E_2) = \frac{8}{14} = \frac{4}{7}$
Step 2 (Step-by-Step Mathematical Working): [1.5 Marks]
Substituting the values into the formula:
$$P(R) = \left( \frac{1}{3} \times \frac{3}{7} \right) + \left( \frac{2}{3} \times \frac{4}{7} \right)$$ $$P(R) = \frac{3}{21} + \frac{8}{21}$$ $$P(R) = \frac{3 + 8}{21}$$Step 3 (Final Concluded Answer): [0.5 Mark]
$$P(R) = \frac{11}{21}$$💡 EL10 Concept Insight: Think of this like a branching path: you first flip a coin (the die) to decide which room to enter, and then you calculate the chance of picking a red ball in that specific room; we just add up the chances from both possible paths.
Solution (b)
Step 1 (Formula / Governing Law & Given Data): [1 Mark]
Given:
- $P(X \cup Y) = a$
- $P(X \cup Y) - P(X \cap Y) = b$ (Probability of exactly one event)
We know $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$.
Step 2 (Step-by-Step Mathematical Working): [1.5 Marks]
From the given equations:
$$P(X \cap Y) = P(X \cup Y) - b = a - b$$Now, consider the expression $P(X') + P(Y')$:
$$P(X') + P(Y') = (1 - P(X)) + (1 - P(Y)) = 2 - (P(X) + P(Y))$$Since $P(X) + P(Y) = P(X \cup Y) + P(X \cap Y)$, substitute the known values:
$$P(X) + P(Y) = a + (a - b) = 2a - b$$