IB Math AA/AI — Probability Mastery Notebook

IB Mathematics — Probability

✏️ Grade 12 Self-Study Notebook ✏️

From Basics → Exam Level · 20 Questions · Multiple Choice with Explanations

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📌 ESSENTIAL FORMULA SHEET — Keep this in mind!

Addition Rule: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)

Conditional Prob: \(P(A|B) = \dfrac{P(A \cap B)}{P(B)}\)

Independent: \(P(A \cap B) = P(A) \cdot P(B)\)

Bayes' Theorem: \(P(A|B) = \dfrac{P(B|A)\cdot P(A)}{P(B)}\)

Binomial P(X=k): \(\binom{n}{k}p^k(1-p)^{n-k}\)

📘 SECTION 1 — Foundations of Probability (Q1–5)

🧠 SAMPLE SPACE · EVENT · OUTCOME · EQUALLY LIKELY

Sample Space (S) = all possible outcomes | Event (A) = a subset of S | P(A) = favourable / total (only when outcomes are equally likely!)

★ EASY

Topic: Basic Probability Definition

A fair six-sided die is rolled once. What is the probability of rolling a number greater than 4?

💡 THINK FIRST

List the sample space: S = {1, 2, 3, 4, 5, 6}. Numbers greater than 4 are: {5, 6}.
\(P(\text{event}) = \dfrac{\text{number of favourable outcomes}}{\text{total outcomes}}\)

🔍 Show Hint

Count how many numbers in {1,2,3,4,5,6} satisfy the condition "greater than 4".

★ EASY

Topic: Complement Rule

The probability that it rains tomorrow is \(0.35\). What is the probability that it does NOT rain tomorrow?

🧠 COMPLEMENT RULE: P(A') = 1 − P(A)

The complement of an event A is "A does not happen." Together they cover everything: \(P(A) + P(A') = 1\)

★ EASY

Topic: Mutually Exclusive Events

Events A and B are mutually exclusive with \(P(A) = 0.4\) and \(P(B) = 0.3\). Find \(P(A \cup B)\).

🧠 MUTUALLY EXCLUSIVE → CAN'T HAPPEN TOGETHER → P(A∩B) = 0

So: \(P(A \cup B) = P(A) + P(B)\) — just add them directly!

🔍 Show Hint

Mutually exclusive means \(P(A \cap B) = 0\), so the addition formula simplifies.

★ EASY

Topic: Venn Diagram / Addition Rule

In a class, \(P(A) = 0.5\), \(P(B) = 0.4\), and \(P(A \cap B) = 0.2\). Find \(P(A \cup B)\).

💡 KEY FORMULA

\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
The intersection is counted twice when you add P(A) and P(B), so subtract it once!

★ EASY

Topic: Tree Diagram / Sequential Events

A bag contains 3 red and 2 blue balls. Two balls are drawn with replacement. What is the probability of getting two red balls?

🧠 WITH REPLACEMENT → INDEPENDENT → MULTIPLY!

"With replacement" means the probabilities don't change between draws. Multiply each branch: \(P(\text{RR}) = P(R_1) \times P(R_2)\)

📗 SECTION 2 — Conditional Probability & Independence (Q6–10)

🧠 CONDITIONAL = "GIVEN THAT" → REDUCE THE SAMPLE SPACE

\(P(A|B)\) = "probability of A, given B already happened."
Think of it as: zoom in on B, then find A within that zoomed space.

★ EASY

Topic: Conditional Probability (basic)

Given \(P(A \cap B) = 0.12\) and \(P(B) = 0.4\), find \(P(A | B)\).

💡 FORMULA

\(P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.12}{0.4}\)

★★ EASY-MEDIUM

Topic: Independence Test — Most Commonly Confused!

\(P(A) = 0.5\), \(P(B) = 0.4\), \(P(A \cap B) = 0.2\). Are A and B independent?

🧠 INDEPENDENT TEST: P(A∩B) = P(A) × P(B) ?

If this equation holds → independent (knowing B tells you nothing about A)
If not → dependent

★★ MEDIUM

Topic: Without Replacement (Dependent Events)

A bag has 4 red and 3 blue balls. Two are drawn without replacement. What is the probability both are red?

🧠 WITHOUT REPLACEMENT → DEPENDENT → FRACTION CHANGES!

After drawing 1 red ball: now only 3 red left out of 6 total.
\(P(R_1 \cap R_2) = P(R_1) \times P(R_2 | R_1) = \dfrac{4}{7} \times \dfrac{3}{6}\)

★★ MEDIUM

Topic: Conditional from Table

A survey of 100 students:

	Likes Math	Dislikes Math	Total
Science	35	15	50
Arts	20	30	50
Total	55	45	100

A student is selected at random. Given that they are a Science student, what is the probability they like Math?

🔍 Show Hint

Focus only on the Science row (50 students total). Of those, how many like Math?

★★ MEDIUM

Topic: Bayes' Theorem (Intro)

A medical test for a disease is 90% accurate. The disease affects 1% of the population. If a person tests positive, which formula correctly starts the calculation of the probability they actually have the disease?

🧠 BAYES = REVERSE CONDITIONAL · "FLIP IT AROUND"

You know: \(P(\text{+} | \text{disease})\). You want: \(P(\text{disease} | \text{+})\).
Bayes flips the condition! \(P(D|+) = \dfrac{P(+|D) \cdot P(D)}{P(+)}\)

📙 SECTION 3 — Discrete Random Variables & Expected Value (Q11–14)

🧠 E(X) = Σ x·P(X=x) · "WEIGHTED AVERAGE" · Var(X) = E(X²) − [E(X)]²

Expected value = long-run average. Not necessarily a value X can actually take!
For variance: compute E(X²) first (multiply each x² by its probability, then sum).

★★ MEDIUM

Topic: Expected Value E(X)

X is a discrete random variable with the distribution:

x	1	2	3	4
P(X=x)	0.1	0.3	0.4	0.2

Find \(E(X)\).

🔍 Show Hint

\(E(X) = 1(0.1) + 2(0.3) + 3(0.4) + 4(0.2)\)

★★ MEDIUM

Topic: Finding Missing Probability

For the probability distribution below, find the value of \(k\):

x	0	1	2	3
P(X=x)	\(k\)	\(2k\)	\(3k\)	\(4k\)

🧠 ALL PROBABILITIES MUST SUM TO 1

\(k + 2k + 3k + 4k = 1 \Rightarrow 10k = 1 \Rightarrow k = 0.1\)

★★★ MEDIUM-HARD

Topic: Variance Calculation

X has \(E(X) = 2\) and \(E(X^2) = 5\). Find \(\text{Var}(X)\).

🧠 Var(X) = E(X²) − [E(X)]² · "SQUARE THEN SUBTRACT SQUARE"

SD (Standard Deviation) = \(\sqrt{\text{Var}(X)}\)
Common mistake: computing E(X²) wrong — it's NOT [E(X)]²!

★★★ MEDIUM-HARD

Topic: Linear Transformations of RV

If \(E(X) = 3\) and \(\text{Var}(X) = 4\), find \(E(2X + 1)\) and \(\text{Var}(2X + 1)\).

🧠 E(aX+b) = aE(X)+b · Var(aX+b) = a²Var(X) · CONSTANTS DON'T ADD VARIANCE!

Adding a constant shifts the distribution (changes mean) but does NOT spread it out (no change to variance).
Multiplying by \(a\) scales variance by \(a^2\).

📕 SECTION 4 — Binomial Distribution (Q15–17)

🧠 BINOMIAL CHECK: FIXED n · TWO OUTCOMES · CONSTANT p · INDEPENDENT TRIALS

\(X \sim B(n, p)\) → \(P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}\)
Mean: \(E(X) = np\) | Variance: \(\text{Var}(X) = np(1-p)\)

★★ MEDIUM

Topic: Binomial Probability

A fair coin is tossed 5 times. Let X = number of heads. Find \(P(X = 3)\).

💡 SETUP

\(X \sim B(5,\ 0.5)\)
\(P(X=3) = \binom{5}{3}(0.5)^3(0.5)^2 = 10 \times \dfrac{1}{32}\)

★★★ HARD

Topic: Binomial — "At Least" / Cumulative

\(X \sim B(6,\ 0.3)\). Find \(P(X \geq 2)\).

🧠 "AT LEAST" → USE COMPLEMENT: P(X≥2) = 1 − P(X≤1) = 1 − P(X=0) − P(X=1)

Always easier to compute what you DON'T want, then subtract from 1!
\(P(X=0) = 0.7^6 \approx 0.1176\) \(P(X=1) = 6 \times 0.3 \times 0.7^5 \approx 0.3025\)

★★★ HARD

Topic: Binomial Mean & Variance

A binomial distribution has mean \(\mu = 6\) and variance \(\sigma^2 = 2.4\). Find the values of \(n\) and \(p\).

💡 SOLVE SIMULTANEOUSLY

\(np = 6\) and \(np(1-p) = 2.4\)
Divide second by first: \(1-p = 0.4\), so \(p = 0.6\), then \(n = 10\)

📓 SECTION 5 — Normal Distribution & Exam Level (Q18–20)

🧠 NORMAL: SYMMETRIC · BELL-SHAPED · X~N(μ,σ²) · Z-SCORE = (x−μ)/σ

Standardize with Z-score → use GDC or Z-table.
68-95-99.7 Rule: within 1σ: 68%, within 2σ: 95%, within 3σ: 99.7%

★★★ HARD

Topic: Normal Distribution — Finding Probability

Heights are normally distributed: \(X \sim N(170, 64)\) (in cm, variance = 64).
Find \(P(166 \leq X \leq 178)\) using the standardised Z-score approach.

💡 Z-SCORE SETUP

\(\sigma = \sqrt{64} = 8\)
\(Z_1 = \dfrac{166-170}{8} = -0.5\) \(Z_2 = \dfrac{178-170}{8} = 1\)
\(P(-0.5 \leq Z \leq 1) = \Phi(1) - \Phi(-0.5) \approx 0.8413 - 0.3085\)

★★★★ EXAM LEVEL

Topic: Inverse Normal — Find the Value

Scores in an exam follow \(X \sim N(60, 100)\). The top 10% of students receive an A grade.
What is the minimum score required to get an A? (Use \(\Phi^{-1}(0.9) \approx 1.282\))

🧠 INVERSE NORMAL: FIND x FROM P · "WORK BACKWARDS" · x = μ + Zσ

Top 10% means \(P(X > x) = 0.1\), so \(P(X \leq x) = 0.9\).
Find Z first from the table, then reverse the Z-formula: \(x = \mu + Z\sigma\)

🔍 Show Hint

\(\sigma = \sqrt{100} = 10\). \(x = 60 + 1.282 \times 10\)

★★★★★ IB EXAM LEVEL

Topic: Combined — Bayes + Conditional + Tree Diagram

A factory has two machines: Machine A produces 60% of output, Machine B produces 40%.
Machine A has a defect rate of 5%, Machine B has a defect rate of 10%.
A randomly selected item is found to be defective.
What is the probability it was produced by Machine A?

💡 BAYES' THEOREM — STEP BY STEP

Step 1: Find total \(P(\text{defect})\) using the Law of Total Probability:
\(P(D) = P(D|A)\cdot P(A) + P(D|B)\cdot P(B) = (0.05)(0.6) + (0.10)(0.4)\)
\(= 0.03 + 0.04 = 0.07\)

Step 2: Apply Bayes:
\(P(A|D) = \dfrac{P(D|A)\cdot P(A)}{P(D)} = \dfrac{0.03}{0.07}\)

🧠 TOTAL PROBABILITY → THEN BAYES → "BRANCH × BRANCH ÷ TOTAL"

Draw the tree! Multiply along branches. Bayes = one branch ÷ sum of all matching branches.

🎓 End of Probability Notebook · Keep Practicing! · Good Luck on Your IB Exam! 🍀