IB Mathematics AI – Topic 4

Key Terms:

• Experiment: An action or process that produces outcomes
• Outcome: A possible result of an experiment
• Sample Space (U or S): The set of all possible outcomes
• Event (A, B, C, ...): A subset of the sample space
• Trial: A single performance of an experiment

Basic Probability Formula:

\[ P(A) = \frac{n(A)}{n(U)} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \]

This assumes all outcomes are equally likely

Properties of Probability:

• \(0 \leq P(A) \leq 1\) for any event A
• \(P(U) = 1\) (certainty)
• \(P(\emptyset) = 0\) (impossibility)
• Complementary Events: \(P(A') = 1 - P(A)\) where \(A'\) means "not A"

Expected Number of Occurrences:

\[ \text{Expected number} = n \times P(A) \]

where n is the number of trials

⚠️ Common Pitfalls & Tips:

• Probabilities must always be between 0 and 1 (or 0% and 100%)
• Express answers as fractions, decimals, or percentages as requested
• Remember: \(P(A) + P(A') = 1\) always
• Check if outcomes are equally likely before using the basic formula
• Be careful with "at least" problems – often easier to use complement

Solution:

Setting up:

Total number of balls = 5 + 3 + 2 = 10

Sample space: \(n(U) = 10\)

(a) Probability of red:

Number of red balls = 5

\[ P(\text{Red}) = \frac{5}{10} = \frac{1}{2} = 0.5 \text{ or } 50\% \]

(b) Probability NOT blue:

Method 1 – Using complement:

\(P(\text{Blue}) = \frac{3}{10}\)

\[ P(\text{Not Blue}) = 1 - P(\text{Blue}) = 1 - \frac{3}{10} = \frac{7}{10} = 0.7 \]

Method 2 – Direct counting:

Not blue means red or green = 5 + 2 = 7 balls

\[ P(\text{Not Blue}) = \frac{7}{10} = 0.7 \text{ or } 70\% \]

(c) Expected number of green balls:

\(P(\text{Green}) = \frac{2}{10} = \frac{1}{5}\)

\[ \text{Expected number} = 50 \times \frac{1}{5} = 10 \text{ green balls} \]

Set Operations & Probability Rules:

1. Union (A ∪ B): "A or B or both"

\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

2. Intersection (A ∩ B): "Both A and B"

The overlap region in a Venn diagram

3. Complement (A'): "Not A"

\[ P(A') = 1 - P(A) \]

4. Mutually Exclusive Events:

Events that cannot occur together: \(P(A \cap B) = 0\)

For mutually exclusive events:

\[ P(A \cup B) = P(A) + P(B) \]

⚠️ Common Pitfalls & Tips:

• Always subtract the intersection when using the addition rule to avoid double-counting
• The region "only A" = \(P(A) - P(A \cap B)\)
• Check that all probabilities in your Venn diagram sum to 1
• Start with the intersection when filling in a Venn diagram from given information
• "A or B" means A ∪ B (includes both); "A or B but not both" excludes intersection

Solution:

(a) Venn Diagram:

Step 1: Start with the intersection (both languages):

\(n(F \cap S) = 8\)

Step 2: Find "only French":

Only F = Total F - Both = 18 - 8 = 10 students

Step 3: Find "only Spanish":

Only S = Total S - Both = 15 - 8 = 7 students

Step 4: Find "neither":

Neither = Total - (Only F + Both + Only S) = 30 - (10 + 8 + 7) = 5 students

(b) P(French or Spanish):

Method 1 – Using the formula:

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

\[ P(F \cup S) = \frac{18}{30} + \frac{15}{30} - \frac{8}{30} = \frac{25}{30} = \frac{5}{6} \approx 0.833 \]

Method 2 – Direct counting:

Students studying at least one language = 10 + 8 + 7 = 25

\[ P(F \cup S) = \frac{25}{30} = \frac{5}{6} \approx 0.833 \text{ or } 83.3\% \]

(c) P(Neither language):

Method 1 – Using complement:

\[ P(\text{Neither}) = 1 - P(F \cup S) = 1 - \frac{5}{6} = \frac{1}{6} \approx 0.167 \]

Method 2 – Direct:

\[ P(\text{Neither}) = \frac{5}{30} = \frac{1}{6} \text{ or } 16.7\% \]

How to Use Tree Diagrams:

1. Draw branches for each outcome at each stage
2. Label each branch with the outcome and its probability
3. Multiply along branches to find probability of a sequence
4. Add across branches to find probability of different sequences leading to same outcome

Key Rules:

• Probabilities on branches from same point must sum to 1
• For independent events, second-stage probabilities don't change
• For dependent events (without replacement), second-stage probabilities change

⚠️ Common Pitfalls & Tips:

• Remember: multiply along branches, add across different paths
• For "without replacement," probabilities change on second draw
• Check all probabilities at each stage sum to 1
• Label all branches clearly to avoid confusion
• Tree diagrams work best for 2-3 stages; become unwieldy beyond that

When to Use:

• Rolling two dice
• Selecting two items simultaneously
• Any situation with two independent events occurring together

Each cell in the table represents one outcome, making it easy to count favorable outcomes.

Solution:

(a) Tree Diagram:

Initial: 4 red, 3 blue (total = 7 marbles)

Explanation of probabilities:

• First draw: P(Red) = 4/7, P(Blue) = 3/7
• If 1st is Red: 3 red and 3 blue remain (6 total) → P(2nd Red) = 3/6
• If 1st is Blue: 4 red and 2 blue remain (6 total) → P(2nd Blue) = 2/6

(b) P(Both red):

Follow the RR path and multiply:

\[ P(\text{RR}) = \frac{4}{7} \times \frac{3}{6} = \frac{12}{42} = \frac{2}{7} \approx 0.286 \text{ or } 28.6\% \]

(c) P(Different colors):

Different colors means RB or BR. Add these paths:

\[ P(\text{RB}) = \frac{4}{7} \times \frac{3}{6} = \frac{12}{42} = \frac{2}{7} \]

\[ P(\text{BR}) = \frac{3}{7} \times \frac{4}{6} = \frac{12}{42} = \frac{2}{7} \]

\[ P(\text{Different}) = \frac{2}{7} + \frac{2}{7} = \frac{4}{7} \approx 0.571 \text{ or } 57.1\% \]

Solution:

Sample Space (Sum of two dice):

(a) P(Sum = 8):

Favorable outcomes (highlighted): (2,6), (3,5), (4,4), (5,3), (6,2)

Total outcomes = 6 × 6 = 36

\[ P(\text{Sum} = 8) = \frac{5}{36} \approx 0.139 \text{ or } 13.9\% \]

(b) P(Product > 20):

Need to find outcomes where \(\text{Die}_1 \times \text{Die}_2 > 20\)

Favorable outcomes:

• (4,6): product = 24 ✓
• (5,5): product = 25 ✓
• (5,6): product = 30 ✓
• (6,4): product = 24 ✓
• (6,5): product = 30 ✓
• (6,6): product = 36 ✓

Total favorable outcomes = 6

\[ P(\text{Product} > 20) = \frac{6}{36} = \frac{1}{6} \approx 0.167 \text{ or } 16.7\% \]

Conditional Probability Formula:

\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]

where \(P(B) > 0\)

Rearranged (useful for finding intersection):

\[ P(A \cap B) = P(A|B) \times P(B) = P(B|A) \times P(A) \]

Interpretation:

\(P(A|B)\) restricts the sample space to only those outcomes where B occurs, then finds the proportion of those where A also occurs.

Test for Independence:

Events A and B are independent if and only if:

\[ P(A \cap B) = P(A) \times P(B) \]

Alternative conditions (all equivalent):

• \(P(A|B) = P(A)\)
• \(P(B|A) = P(B)\)
• \(P(A \cap B) = P(A) \times P(B)\)

Important Distinction:

• Independent: One event doesn't affect the other (e.g., two coin flips)
• Mutually Exclusive: Events cannot both occur (\(P(A \cap B) = 0\))
• Mutually exclusive events with \(P(A) > 0\) and \(P(B) > 0\) are NOT independent

⚠️ Common Pitfalls & Tips:

• Don't confuse \(P(A|B)\) with \(P(B|A)\) – they're usually different!
• For independent events, probabilities don't change regardless of what happened before
• "With replacement" typically means independent; "without replacement" means dependent
• Use tree diagrams for conditional probability – they make the structure clear
• Check independence by testing \(P(A \cap B) = P(A) \times P(B)\)

Solution:

Given information:

\(P(S) = 0.60\)

\(P(M) = 0.40\)

\(P(S \cap M) = 0.25\)

(a) Find \(P(M|S)\):

Using the conditional probability formula:

\[ P(M|S) = \frac{P(M \cap S)}{P(S)} = \frac{0.25}{0.60} = \frac{5}{12} \approx 0.417 \]

Interpretation: 41.7% of students who play sports are also in the music club.

(b) Find \(P(S|M)\):

\[ P(S|M) = \frac{P(S \cap M)}{P(M)} = \frac{0.25}{0.40} = \frac{5}{8} = 0.625 \]

Interpretation: 62.5% of students in the music club also play sports.

(c) Test for independence:

For independence, we need \(P(S \cap M) = P(S) \times P(M)\)

Calculate \(P(S) \times P(M)\):

\[ P(S) \times P(M) = 0.60 \times 0.40 = 0.24 \]

But \(P(S \cap M) = 0.25\)

Since \(0.25 \neq 0.24\), the events are NOT independent.

Alternative check using conditional probability:

\(P(M|S) = 0.417 \neq 0.40 = P(M)\)

Since \(P(M|S) \neq P(M)\), this confirms the events are dependent. Being an athlete slightly increases the probability of being in music club.

Transition Matrix Structure:

A transition matrix T is a square matrix where:

• Entry \(t_{ij}\) = probability of moving FROM state i TO state j
• Each row represents starting state
• Each column represents ending state
• All entries are between 0 and 1
• Each row sums to 1 (must go somewhere)

State Vector:

Current probabilities of being in each state, written as a column vector:

\[ \mathbf{s} = \begin{pmatrix} p_1 \\ p_2 \\ \vdots \\ p_n \end{pmatrix} \]

where \(p_i\) is probability of being in state i, and \(\sum p_i = 1\)

Finding Future States:

\[ \mathbf{s}_{n+1} = T \mathbf{s}_n \]

After k steps: \(\mathbf{s}_k = T^k \mathbf{s}_0\)

Steady State (Long-term Behavior):

A steady state vector \(\mathbf{s}_{\infty}\) satisfies:

\[ T \mathbf{s}_{\infty} = \mathbf{s}_{\infty} \]

This represents the long-term equilibrium probabilities.

Finding Steady State:

1. Set up equation \(T\mathbf{s} = \mathbf{s}\)
2. Solve the system of equations
3. Use constraint that probabilities sum to 1
4. Alternatively: Calculate high powers of T (e.g., \(T^{20}\)) and read steady state from any column

⚠️ Common Pitfalls & Tips:

• Check row sums: Every row of a transition matrix must sum to 1
• Multiply on the LEFT: \(T\mathbf{s}\), not \(\mathbf{s}T\)
• State vectors must sum to 1 (represent 100% of probability)
• Use your GDC for matrix multiplication and finding powers
• Steady state is found when \(T\mathbf{s} = \mathbf{s}\), not \(\mathbf{s}T = \mathbf{s}\)
• Not all Markov chains have a unique steady state

Solution:

(a) Transition Matrix:

Rows: FROM (current state); Columns: TO (next state)

\[ T = \begin{pmatrix} 0.8 & 0.4 \\ 0.2 & 0.6 \end{pmatrix} \begin{matrix} \leftarrow \text{TO Active} \\ \leftarrow \text{TO Inactive} \end{matrix} \]

\[ \begin{matrix} \uparrow \\ \text{FROM A} \end{matrix} \quad \begin{matrix} \uparrow \\ \text{FROM I} \end{matrix} \]

Initial state vector:

\[ \mathbf{s}_0 = \begin{pmatrix} 0.7 \\ 0.3 \end{pmatrix} \]

(b) After 1 month:

\[ \mathbf{s}_1 = T\mathbf{s}_0 = \begin{pmatrix} 0.8 & 0.4 \\ 0.2 & 0.6 \end{pmatrix} \begin{pmatrix} 0.7 \\ 0.3 \end{pmatrix} \]

Calculate:

Active after 1 month: \(0.8(0.7) + 0.4(0.3) = 0.56 + 0.12 = 0.68\)

Inactive after 1 month: \(0.2(0.7) + 0.6(0.3) = 0.14 + 0.18 = 0.32\)

\[ \mathbf{s}_1 = \begin{pmatrix} 0.68 \\ 0.32 \end{pmatrix} \]

After 1 month: 68% Active, 32% Inactive

(c) Steady State:

Let \(\mathbf{s}_{\infty} = \begin{pmatrix} a \\ i \end{pmatrix}\) where \(a + i = 1\)

We need \(T\mathbf{s}_{\infty} = \mathbf{s}_{\infty}\):

\[ \begin{pmatrix} 0.8 & 0.4 \\ 0.2 & 0.6 \end{pmatrix} \begin{pmatrix} a \\ i \end{pmatrix} = \begin{pmatrix} a \\ i \end{pmatrix} \]

This gives us:

Equation 1: \(0.8a + 0.4i = a\) → \(0.4i = 0.2a\) → \(2i = a\)

Equation 2 (Constraint): \(a + i = 1\)

Solving:

Substitute \(a = 2i\) into the constraint:

\( (2i) + i = 1 \) → \(3i = 1\) → \(i = 1/3\)

Then \(a = 2i = 2(1/3) = 2/3\)

\[ \mathbf{s}_{\infty} = \begin{pmatrix} 2/3 \\ 1/3 \end{pmatrix} \approx \begin{pmatrix} 0.667 \\ 0.333 \end{pmatrix} \]

Steady State: 66.7% Active, 33.3% Inactive

In the long run, regardless of starting distribution, the system stabilizes at 2/3 Active and 1/3 Inactive users.

Basic Probability

• \(P(A) = \frac{n(A)}{n(U)}\)
• \(P(A') = 1 - P(A)\)
• \(0 \leq P(A) \leq 1\)

Combined Events

• \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
• Mutually exclusive: \(P(A \cap B) = 0\)

Conditional Probability

• \(P(A|B) = \frac{P(A \cap B)}{P(B)}\)
• \(P(A \cap B) = P(A|B) \times P(B)\)

Independence

• \(P(A \cap B) = P(A) \times P(B)\)
• \(P(A|B) = P(A)\)