IB Mathematics AA – Topic 4: Statistics & Probability

Essential Terminology:

• Experiment: A process that produces one or more outcomes
• Outcome: A possible result of an experiment
• Sample Space (S or \(\xi\)): The set of all possible outcomes
• Event (A, B, C...): A subset of the sample space (one or more outcomes)
• Probability P(A): The likelihood that event A occurs

Probability Definition

Fundamental Properties:

• Range: \(0 \leq P(A) \leq 1\) for any event A
• Impossible event: \(P(\emptyset) = 0\) (empty set)
• Certain event: \(P(S) = 1\) (entire sample space)
• Complement: \(P(A') = 1 - P(A)\) where \(A'\) is "not A"
• Sum of all outcomes: All probabilities in sample space sum to 1

⚠ Basic Probability Pitfalls:

• Probability never > 1: If you calculate P > 1, you've made an error!
• Complement rule: P(A) + P(A') = 1, always
• Equally likely assumption: Basic formula only works if all outcomes are equally likely
• Fractions vs decimals: Can express probability as fraction, decimal, or percentage

Set Operations:

Addition Rule for Probability

💡 Venn Diagram Tips:

• Always start with the intersection \(A \cap B\) when filling in values
• Work outward: subtract intersection from individual sets
• Check: all regions should sum to total (or probability 1)
• Label all regions clearly including outside both circles

Example 1: Venn Diagrams and Combined Events

Problem: In a class of 30 students:

• 18 students study French (F)
• 15 students study Spanish (S)
• 8 students study both languages

(a) Draw a Venn diagram

(b) Find P(F ∪ S)

(c) Find the probability a randomly selected student studies neither language

Solution:

(a) Venn diagram:

Step 1: Start with intersection: \(n(F \cap S) = 8\)

Step 2: Only French: \(18 - 8 = 10\)

Step 3: Only Spanish: \(15 - 8 = 7\)

Step 4: Neither: \(30 - (10 + 8 + 7) = 5\)

Venn diagram regions:
Only F: 10 students
Both F and S: 8 students
Only S: 7 students
Neither: 5 students

(b) P(F ∪ S):

Method 1: Using addition rule

\(P(F \cup S) = P(F) + P(S) - P(F \cap S)\)

\(= \frac{18}{30} + \frac{15}{30} - \frac{8}{30}\)

\(= \frac{25}{30} = \frac{5}{6}\)

Method 2: Count students studying at least one language

Students in F or S: \(10 + 8 + 7 = 25\)

\(P(F \cup S) = \frac{25}{30} = \frac{5}{6}\)

\(P(F \cup S) = \frac{5}{6}\) or 0.833

(c) Probability studying neither:

Students studying neither: 5

\(P(\text{neither}) = P((F \cup S)') = \frac{5}{30} = \frac{1}{6}\)

Or using complement: \(P((F \cup S)') = 1 - P(F \cup S) = 1 - \frac{5}{6} = \frac{1}{6}\)

P(neither) = \(\frac{1}{6}\) or 0.167

Tree Diagram Structure:

• Shows sequential events branching from left to right
• Each branch represents a possible outcome
• Probabilities written on branches
• Multiply probabilities along branches for specific outcome
• Add probabilities of different paths for combined outcomes

Multiplication Rule (AND)

⚠ Tree Diagram Pitfalls:

• Branch probabilities: All branches from same point must sum to 1
• Multiply along path: Use multiplication for single complete path
• Add different paths: Use addition for multiple paths to same outcome
• Order matters: Tree shows the sequence of events clearly

Independent Events

Mutually Exclusive Events

Important: Independent ≠ Mutually Exclusive!

• Mutually exclusive events cannot be independent (except for impossible events)
• If A and B are mutually exclusive and both have P > 0, they are dependent
• Independence: occurrence of A doesn't change probability of B
• Mutually exclusive: if A occurs, B cannot occur (P(B|A) = 0)

Conditional Probability Formula

💡 Conditional Probability Tips:

• The "given" event becomes the new sample space (denominator)
• P(A|B) can be very different from P(A)
• Use tree diagrams for sequential conditional events
• For independent events: P(A|B) = P(A)

Example 2: Conditional Probability (IB-Style)

Problem: A bag contains 5 red balls and 3 blue balls. Two balls are drawn without replacement.

(a) Find the probability both balls are red

(b) Find the probability the second ball is red given the first is red

(c) Find the probability of getting one ball of each color

Solution:

(a) Probability both balls are red:

Total balls: 8 (5 red, 3 blue)

Let R₁ = first ball is red, R₂ = second ball is red

\(P(\text{both red}) = P(R_1 \cap R_2)\)

\(= P(R_1) \times P(R_2|R_1)\)

\(= \frac{5}{8} \times \frac{4}{7}\)

(After drawing 1 red: 4 red and 3 blue remain, 7 total)

\(= \frac{20}{56} = \frac{5}{14}\)

P(both red) = \(\frac{5}{14}\) ≈ 0.357

(b) P(second red | first red):

This is \(P(R_2|R_1)\)

Given first ball was red, there are now 7 balls left (4 red, 3 blue)

\(P(R_2|R_1) = \frac{4}{7}\)

(c) Probability one of each color:

Two possible orders: Red then Blue, or Blue then Red

Path 1: Red first, then Blue

\(P(\text{RB}) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\)

Path 2: Blue first, then Red

\(P(\text{BR}) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\)

Total: Add both paths

\(P(\text{one of each}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}\)

P(one of each) = \(\frac{15}{28}\) ≈ 0.536

Common Question Types:

• "Find P(A ∪ B)": Use addition rule, subtract intersection
• "Complete Venn diagram": Start with intersection, work outward
• "Draw tree diagram": Show all outcomes, probabilities on branches
• "Are A and B independent?": Check if P(A ∩ B) = P(A) × P(B)
• "Find P(A|B)": Use conditional probability formula
• "Without replacement": Probabilities change—use conditional probability

Key Reminders:

• Always check: 0 ≤ P ≤ 1
• Venn diagrams: start with intersection
• Tree diagrams: multiply along path, add different paths
• Independent ≠ mutually exclusive
• "Given" in conditional probability means use that as denominator
• With replacement = independent; without replacement = conditional