Mathematics 9709/62 — May/June 2016

Walkthrough

A tree diagram is the natural way to represent this multi-stage probability problem. We start with the first stage: Ayman's choice of drink. There are three mutually exclusive outcomes — tea (T), coffee (C), or hot chocolate (HC) — with probabilities 0.65, 0.28, and 0.07 respectively. These must sum to 1, which they do: $0.65 + 0.28 + 0.07 = 1.00$.

From each drink choice, there is a second stage: whether milk is added or not. The problem gives conditional probabilities for milk given each drink:

• Given tea: $P(M \mid T) = 0.8$, so $P(NM \mid T) = 1 - 0.8 = 0.2$
• Given coffee: $P(M \mid C) = 0.5$, so $P(NM \mid C) = 1 - 0.5 = 0.5$
• Given hot chocolate: $P(M \mid HC) = 1$, so $P(NM \mid HC) = 1 - 1 = 0$

Each pair of second-level branches from a single node must sum to 1, which is a key check when constructing tree diagrams.

Key Takeaways

• Tree diagrams visually represent multi-stage probability problems with conditional probabilities on the branches.
• The second-level branches from each node represent conditional probabilities and must sum to 1.
• The complementary probability $P(NM) = 1 - P(M)$ is used when only one conditional probability is given.

Common Mistakes

• Forgetting to label all branches with probabilities.
• Not calculating the complementary probabilities (no milk) for each drink.
• Writing the conditional probability on the wrong branch (e.g., putting 0.8 on the NM branch from T).

Things to Be Careful About

• The probabilities on the second-level branches are conditional probabilities $P(M \mid \text{drink})$, not joint probabilities.
• All branches from a single node must sum to exactly 1.
• The problem states hot chocolate always has milk, so $P(M \mid HC) = 1$ and $P(NM \mid HC) = 0$.

Walkthrough

The question asks for the probability that Ayman's drink is coffee, given that it has milk. This is a conditional probability problem requiring the formula:

$$P(C \mid \text{milk}) = \frac{P(C \cap \text{milk})}{P(\text{milk})}$$

Step 1: Find the numerator $P(C \cap \text{milk})$

Using the multiplication law along the tree diagram path from the start through coffee to milk:

$$P(C \cap \text{milk}) = P(C) \times P(M \mid C) = 0.28 \times 0.5 = 0.14$$

Step 2: Find the denominator $P(\text{milk})$

Milk can appear with any of the three drinks, so we sum the joint probabilities along all paths that lead to milk:

• Tea and milk: $P(T \cap \text{milk}) = 0.65 \times 0.8 = 0.52$
• Coffee and milk: $P(C \cap \text{milk}) = 0.28 \times 0.5 = 0.14$
• Hot chocolate and milk: $P(HC \cap \text{milk}) = 0.07 \times 1 = 0.07$

$$P(\text{milk}) = 0.52 + 0.14 + 0.07 = 0.73$$

Step 3: Apply the conditional probability formula

$$P(C \mid \text{milk}) = \frac{0.14}{0.73} = 0.19178\ldots \approx 0.192$$

Key Takeaways

• Conditional probability $P(A \mid B) = P(A \cap B) / P(B)$ requires both a joint probability and a marginal probability.
• The total probability rule $P(B) = \sum P(A_i \cap B)$ is used when B can occur through multiple mutually exclusive paths.
• Tree diagrams make it easy to identify all relevant paths for computing joint and marginal probabilities.

Common Mistakes

• Forgetting to compute $P(\text{milk})$ correctly by summing all three joint probabilities.
• Using only one path (e.g., just coffee and milk) in the denominator instead of all paths leading to milk.
• Not showing the method (multiplication law) — unsupported answers may not receive full marks.
• Rounding too early in the calculation.

Things to Be Careful About

• The denominator $P(\text{milk})$ must include contributions from ALL drink types, not just coffee.
• $P(HC \cap \text{milk}) = 0.07 \times 1 = 0.07$ — the probability of milk given hot chocolate is 1, not 0.
• The final answer should be given to 3 significant figures as $0.192$, or as the exact fraction $14/73$.