9.3 Probability, Counting Principles & Permutations

Probability, Counting Principles & Permutations

Core Concept: Probability and combinatorics questions on scholarship and selective entry exams measure your ability to count outcomes systematically without enumerating every possibility by hand. With only about 30 to 45 seconds per item, candidates who attempt to draw full tree diagrams or write out exhaustive lists will fail. Success requires using algebraic counting principles: the Fundamental Counting Rule, factorial permutations with grouping blocks, and the complementary "at least one" probability shortcut.

Whether assessing dice combinations, marble selections from urns, seating arrangements, or survey set intersections, every problem rests on a clean partition between favourable outcomes and the total sample space.

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The Classical Probability Model

The theoretical probability of an event $E$ occurring within a finite, equally likely sample space $S$ is defined as:

$P(E) = \frac{n(E)}{n(S)} = \frac{\text{Number of Favourable Outcomes}}{\text{Total Number of Possible Outcomes}}$

The Invariant Probability Axioms

1. Bounded Range: For any event $E$, $0 \le P(E) \le 1$.
   • $P(E) = 0 \implies$ Impossible event.
   • $P(E) = 1 \implies$ Absolute certainty.
2. Exhaustive Sum: The sum of probabilities of all elementary outcomes in a sample space equals $1$: $\sum_{i} P(e_i) = 1$
3. Complementary Event Identity: The probability that event $A$ does not occur ($A'$ or $A^c$) is: $P(A') = 1 - P(A)$

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The "At Least One" Complement Shortcut

Selective school examiners frequently construct multi-stage probability items using the phrase "at least one". Calculating this directly requires computing the probability of exactly $1$ success, exactly $2$ successes, up to $n$ successes, and summing them all.

$\text{Direct Calculation (Slow): } P(\ge 1) = P(1) + P(2) + P(3) + \dots + P(n)$

The complement of "at least one success" is "zero successes" (none). Using the complementary probability rule converts a five-step addition into a single subtraction:

$\mathbf{P(\text{at least one}) = 1 - P(\text{none})}$

Worked Example: The Three-Dice Roll

When three fair standard six-sided dice are rolled simultaneously, what is the probability that at least one die lands on a 6?

• Step 1 (Identify total sample space): $n(S) = 6 \times 6 \times 6 = 216\text{ total outcomes}$
• Step 2 (Determine the complementary event): $\text{Complement: } \text{"No die lands on a 6"}$
• Step 3 (Calculate probability of non-6 on a single die): $P(\text{not 6}) = \frac{5}{6}$
• Step 4 (Calculate probability of no sixes on all three dice): $P(\text{no 6 on all 3}) = \left(\frac{5}{6}\right)^3 = \frac{125}{216}$
• Step 5 (Apply complement formula): $P(\text{at least one 6}) = 1 - \frac{125}{216} = \frac{216 - 125}{216} = \frac{91}{216}$

Exam Trap: A common error is to write $P = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$. This simple linear addition is completely invalid because rolls are independent rather than mutually exclusive; it double-counts outcomes where multiple sixes appear simultaneously.

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Compound Events: Independent vs. Dependent (Without Replacement)

The Decrementing Denominator in Selections Without Replacement

When items are removed from a container without replacement, both the target pool numerator and the total pool denominator decrement by $1$ on each subsequent draw.

Worked Example: Sequential Marble Extraction

An opaque bag contains 4 green, 6 red, and 2 blue marbles ($12$ marbles total). If two marbles are drawn at random one after another without replacement, what is the probability that exactly one marble is green and one marble is red?

• Step 1 (Recognize compound orderings): A valid pair can occur in two mutually exclusive sequences: $\text{Case 1: Green first, then Red } (G_1 \cap R_2)$ $\text{Case 2: Red first, then Green } (R_1 \cap G_2)$
• Step 2 (Compute Case 1): $P(G_1) = \frac{4}{12}, \qquad P(R_2 \mid G_1) = \frac{6}{11}$ $P(G_1 \cap R_2) = \frac{4}{12} \times \frac{6}{11} = \frac{24}{132}$
• Step 3 (Compute Case 2): $P(R_1) = \frac{6}{12}, \qquad P(G_2 \mid R_1) = \frac{4}{11}$ $P(R_1 \cap G_2) = \frac{6}{12} \times \frac{4}{11} = \frac{24}{132}$
• Step 4 (Sum mutually exclusive cases): $P(\text{One Green, One Red}) = \frac{24}{132} + \frac{24}{132} = \frac{48}{132}$
• Step 5 (Simplify fraction): $\frac{48 \div 12}{132 \div 12} = \frac{4}{11}$

Exam Trap: Candidates routinely calculate only one sequence (e.g., $G$ then $R = \frac{2}{11}$) and forget that drawing $R$ then $G$ satisfies the question criteria equally well. When order is not specified, you must multiply by the number of valid order permutations ($2! = 2$).

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The Fundamental Counting Principle

If a composite procedure can be broken down into a sequence of $k$ distinct decision stages, where Stage 1 can be completed in $n_1$ ways, Stage 2 in $n_2$ ways, and Stage $k$ in $n_k$ ways, then the total number of distinct combined outcomes is:

$\mathbf{N = n_1 \times n_2 \times n_3 \times \dots \times n_k}$

Applying Positional Restrictions

When counting outcomes under constraints, always fill the most restricted positions first.

Worked Example: Restricted Code Generation

How many 4-digit odd numbers greater than 5,000 can be formed using the digits $\{2, 3, 5, 7, 8\}$ if no digit may be repeated?

• Stage 1 (Analyse Constraints):
  • To be $> 5,000$, the thousands digit (first position) must be $5, 7,$ or $8$.
  • To be odd, the units digit (last position) must be $3, 5,$ or $7$.
  • Because $5$ and $7$ appear in both constraint sets, we must split into mutually exclusive cases:
• Case A: First digit is even (must be $8$):
  • Position 1 (Thousands): $1$ choice ($8$).
  • Position 4 (Units): $3$ choices ($3, 5, 7$).
  • Position 2 (Hundreds): $3$ remaining digits available.
  • Position 3 (Tens): $2$ remaining digits available. $\text{Total for Case A} = 1 \times 3 \times 2 \times 3 = 18\text{ numbers}$
• Case B: First digit is odd (must be $5$ or $7$):
  • Position 1 (Thousands): $2$ choices ($5$ or $7$).
  • Position 4 (Units): $2$ remaining odd choices ($3$ plus whichever of $5, 7$ was not used).
  • Position 2 (Hundreds): $3$ remaining digits available.
  • Position 3 (Tens): $2$ remaining digits available. $\text{Total for Case B} = 2 \times 3 \times 2 \times 2 = 24\text{ numbers}$
• Total Count: $18 + 24 = 42\text{ valid 4-digit numbers}$.

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Factorials & Permutations with Grouping Constraints

The number of distinct arrangements of $n$ unique objects placed in a straight line is given by $n$ factorial ($n!$):

$n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$

Memorise standard factorials to preserve exam momentum:

$$1! = 1 \qquad 2! = 2 \qquad 3! = 6 \qquad 4! = 24 \qquad 5! = 120 \qquad 6! = 720 \qquad 7! = 5,040$$

The "Super-Item" (Block) Grouping Method

When specific items must always sit together, bind them into a single consolidated element.

Original Entities: A, B, C, D, E  (where A and B must sit together)
Step 1: Fuse [A, B] into a single block: [AB]
Step 2: Count available entities to arrange: [AB], C, D, E  ──► 4 elements
Step 3: Arrange external entities: 4! = 24 ways
Step 4: Multiply by internal permutations inside block: 2! = 2 ways [AB or BA]
Total Arrangements = 4! × 2! = 24 × 2 = 48 ways

$\mathbf{\text{Arrangements with Grouping} = (N - k + 1)! \times k!}$

Where $N$ is total objects, and $k$ is the number of bound objects.

Worked Example: Line Arrangements with Grouping

Six students—4 boys and 2 girls—stand in a straight line for a selective school induction photograph. If the 2 girls must stand next to each other, how many distinct line arrangements are possible?

• Step 1 (Create the girl-block): Fuse the 2 girls into a single entity: $(G_1 G_2)$.
• Step 2 (Count total movable units): We have 4 boys plus 1 girl-block $= 5$ total units.
• Step 3 (Arrange the 5 units): $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\text{ ways}$
• Step 4 (Account for internal permutations): The 2 girls can be ordered as $(G_1, G_2)$ or $(G_2, G_1)$: $2! = 2\text{ ways}$
• Step 5 (Compute total arrangements): $\text{Total} = 5! \times 2! = 120 \times 2 = 240\text{ arrangements}$

Exam Trap: A common error is to calculate $5! = 120$ and forget to multiply by $2!$, landing on a planted distractor.

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Two-Way Tables, Venn Diagrams & Inclusion-Exclusion

When probability involves overlapping categories, the Principle of Inclusion-Exclusion prevents double-counting the intersection:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $n(A \cup B) = n(A) + n(B) - n(A \cap B)$

          ┌───────────────────────────┐
          │ Universal Set (U)         │
          │   ┌───────┬───┬───────┐   │
          │   │ Only  │ A │ Only  │   │
          │   │   A   │ ∩ │   B   │   │
          │   │       │ B │       │   │
          │   └───────┴───┴───────┘   │
          │         Neither           │
          └───────────────────────────┘

Worked Example: Survey Set Operations

In a cohort of 36 Year 8 students surveyed at a selective entry seminar, 22 students play tennis, 18 play badminton, and 4 play neither sport. How many students play both tennis and badminton?

• Step 1 (Determine total students playing at least one sport): $n(\text{Tennis} \cup \text{Badminton}) = 36 - 4 = 32\text{ students}$
• Step 2 (Apply Inclusion-Exclusion formula): $n(T \cup B) = n(T) + n(B) - n(T \cap B)$ $32 = 22 + 18 - n(T \cap B)$ $32 = 40 - n(T \cap B)$
• Step 3 (Isolate the intersection): $n(T \cap B) = 40 - 32 = 8\text{ students}$

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Probability & Counting Reference Table

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High-Yield Traps in Probability & Permutations

1. The Denominator Invariance Fallacy in Sampling

In rapid-fire tests, students frequently treat dependent events as independent, keeping the denominator constant (e.g., computing two card draws as $\frac{4}{52} \times \frac{3}{52}$ instead of $\frac{4}{52} \times \frac{3}{51}$). Whenever the words "without replacement" or "drawn together" appear, the denominator must decrement on every subsequent draw.

2. Overlooking Order Permutations in Distinct Selections

When asked for the probability of selecting one blue and one red marble, students often calculate $P(B_1 \cap R_2)$ and stop. Unless the question explicitly specifies "the first is blue and the second is red", the selection can occur in multiple orders. For two items, multiply by $2!$; for three distinct items, multiply by $3! = 6$.

3. The Union Without Intersection Trap

When students are asked for the probability that a randomly chosen card from a standard 52-card deck is a Heart or a King, they often add $\frac{13}{52} + \frac{4}{52} = \frac{17}{52}$. This erroneously double-counts the King of Hearts. The true probability is $\frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$.