10.2 Multi-Stage Events, Tree Diagrams, Replacement vs. Non-Replacement, and Conditional Probability

Two-Stage Sample Spaces and Two-Dimensional Grids

A multi-stage (compound) experiment consists of two or more sequential steps—or simultaneous actions treated in order, such as tossing two coins or rolling two standard six-sided dice. If Stage 1 has $m$ equally likely outcomes and Stage 2 has $n$ equally likely outcomes, a two-dimensional sample space grid (array) displays all $m \times n$ ordered pairs $(x, y)$.

Listing ordered pairs prevents a classic probability error: treating distinct ordered pathways as a single outcome.

• Tossing two fair coins: The $2 \times 2 = 4$ equally likely ordered outcomes are $(H, H)$, $(H, T)$, $(T, H)$, and $(T, T)$. Getting "one Head and one Tail in any order" corresponds to two outcomes—$(H, T)$ and $(T, H)$—so its probability is $\frac{2}{4} = \frac{1}{2} = 0.50$, not $\frac{1}{3}$.
• Rolling two fair six-sided dice: There are $6 \times 6 = 36$ equally likely ordered pairs $(d_1, d_2)$. A sum of $7$ can occur in $6$ ordered ways—$(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$—so:

$$P(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}$$

By contrast, a sum of $12$ occurs in only $1$ way, $(6,6)$, giving $P(\text{sum} = 12) = \frac{1}{36}$, while a sum of at least $10$ occurs in $3 + 2 + 1 = 6$ ways (sums of $10$, $11$, and $12$), giving $P(\text{sum} \ge 10) = \frac{6}{36} = \frac{1}{6}$.

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Probability Tree Diagrams: Multiply Along Branches, Add Across Paths

When stage outcomes have unequal probabilities or involve more than two stages, a probability tree diagram organises every sequential path from left to right. Every group of branches emerging from a single node must sum to $1$.

Two calculation rules govern every probability tree:

1. Multiply Along Branches (AND Rule): To find the probability of following one specific path from start to finish (Event $A$ in Stage 1 and Event $B$ in Stage 2), multiply the branch probabilities along that path:

$$P(A \text{ and } B) = P(A) \times P(B \mid A)$$

1. Add Across End-Paths (OR Rule): Because distinct end-paths of a tree are mutually exclusive, if multiple paths satisfy the question (for example, choosing "one girl and one boy in either order"), add the probabilities of those favourable end-paths together.
2. The "At Least One" Tree Shortcut: If a two-stage tree asks for the probability that an event occurs at least once across the two stages, three of the four end-paths are favourable. Rather than calculating and adding all three paths, calculate the single path where the event occurs neither time and subtract from $1$: $P(\text{at least one}) = 1 - P(\text{none})$.

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Independent Events (With Replacement) vs. Dependent Events (Without Replacement)

Before writing the second-stage branch fractions on a tree diagram, always check whether the first selection is replaced:

1. Sampling With Replacement (Independent Events)

When a drawn item is returned to the pool before the second draw—or when stages are physically independent, such as spinning a wheel twice or forecasting rain on two separate weeks—the outcome of Stage 1 does not alter the probabilities for Stage 2. Both the numerator and the denominator remain unchanged:

$$P(A \text{ then } B) = P(A) \times P(B)$$

For example, if the probability of rain on a Tuesday sports afternoon is $0.30$ and the probability of rain on the following Tuesday is independently $0.30$, the probability of rain on neither Tuesday is $0.70 \times 0.70 = 0.49$, so the probability of rain on at least one of the two Tuesdays is $1 - 0.49 = 0.51$ ($51\%$). Simply adding $0.30 + 0.30 = 0.60$ would be incorrect because it double-counts the $0.30 \times 0.30 = 0.09$ chance that it rains on both Tuesdays.

2. Sampling Without Replacement (Dependent Events)

Whenever you select two student representatives, two raffle prize-winners, or two library books from a finite group of $N$ without returning the first selection, the second stage depends directly on what happened first:

• The second-stage denominator always decreases by $1$, from $N$ to $N - 1$, because one item or person has already been removed.
• If the second draw is from the same category as the first (which originally had $k$ members), its numerator also decreases by $1$, from $k$ to $k - 1$:

$$P(\text{both from same category}) = \frac{k}{N} \times \frac{k - 1}{N - 1}$$

• If the second draw is from a different category (which originally had $m$ members), that category still has all $m$ members available, so its numerator stays $m$ while the denominator is $N - 1$:

$$P(\text{first category } A \text{ then category } B) = \frac{k}{N} \times \frac{m}{N - 1}$$

Worked Example: Selecting Two Student Leaders Without Replacement

Suppose a shortlist of $10$ Year 6 leadership finalists consists of $6$ girls ($G$) and $4$ boys ($B$). Two co-captains are drawn at random without replacement:

• Path 1 (Girl then Girl): $P(G_1 \cap G_2) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$
• Path 2 (Girl then Boy): $P(G_1 \cap B_2) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90} = \frac{4}{15}$
• Path 3 (Boy then Girl): $P(B_1 \cap G_2) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90} = \frac{4}{15}$
• Path 4 (Boy then Boy): $P(B_1 \cap B_2) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$

Notice that all four end-paths sum to $\frac{30 + 24 + 24 + 12}{90} = \frac{90}{90} = 1$. If a question asks for the probability of selecting one girl and one boy in any order, you must add Path 2 and Path 3: $\frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}$.

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Conditional Probability from Two-Way Tables and Screening

Conditional probability, written $P(A \mid B)$ and read "the probability of $A$ given that $B$ has occurred", measures the chance of $A$ within the restricted subgroup where $B$ is already known to be true:

$$P(A \mid B) = \frac{n(A \text{ and } B)}{n(B)} = \frac{P(A \text{ and } B)}{P(B)}$$

In LANTITE two-way table and Venn diagram questions, look closely at the wording that defines the denominator:

• "If a student is chosen at random from the whole survey..." uses the grand total as the denominator.
• "Given that a student is in Year 12..." or "Of the students who catch the bus..." restricts the denominator strictly to the row total or column total for that condition, never the grand total.

Worked Example: Senior School Transport Two-Way Table

Compare how three slightly different phrases change the denominator:

1. Joint probability (out of all $250$ students): "What is the probability that a randomly selected senior student is in Year 12 and catches the bus?" $\rightarrow \frac{66}{250} = 0.264 = 26.4\%$.
2. Row-conditioned probability ($P(\text{Bus} \mid \text{Year 12})$): "Given that a randomly selected student is in Year 12, what is the probability they catch the bus?" $\rightarrow \frac{66}{120} = \frac{11}{20} = 0.55 = 55\%$.
3. Column-conditioned probability ($P(\text{Year 12} \mid \text{Bus})$): "Of the students who catch the bus, what is the probability that a randomly selected student is in Year 12?" $\rightarrow \frac{66}{150} = \frac{11}{25} = 0.44 = 44\%$.

The same conditional principle applies to educational screening and diagnostic tools. Suppose a literacy screening tool is administered to $400$ Year 1 students, of whom $60$ genuinely require intensive reading intervention and $340$ do not. If the screening tool flags $54$ of the $60$ struggling readers as "at risk" (true positives) but also flags $36$ of the $340$ proficient readers as "at risk" (false positives), then a total of $54 + 36 = 90$ students receive an "at risk" flag. Given that a student was flagged "at risk", the conditional probability that the student genuinely requires intensive intervention is $\frac{54}{90} = \frac{3}{5} = 0.60 = 60\%$.