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IB Mathematics · Grade 11 · SL / HL

Probability
Workbook

From basics to exam level — 20 essential problems
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📐 Core Formulas — Quick Reference

Basic Probability
\( P(A) = \dfrac{\text{favourable}}{\text{total}} \)
FAVE over TOTAL
Complement
\( P(A') = 1 - P(A) \)
1 MINUS = opposite
Addition Rule
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
ADD then SUBTRACT overlap
Conditional
\( P(A|B) = \dfrac{P(A \cap B)}{P(B)} \)
GIVEN = divide by condition
Multiplication (Independent)
\( P(A \cap B) = P(A) \times P(B) \)
Independent = just MULTIPLY
Multiplication (Dependent)
\( P(A \cap B) = P(A) \times P(B|A) \)
Dependent = MULTIPLY × ADJUST
Mutually Exclusive
\( P(A \cap B) = 0 \)
CANNOT happen together
Bayes' Theorem
\( P(A|B) = \dfrac{P(B|A)\,P(A)}{P(B)} \)
FLIP the condition
Section 1

🟢 Foundation — Basic Probability

1
Easy Sample Space · Equally Likely Outcomes
FAVE ÷ TOTAL — count what you want, divide by everything possible
A bag contains 3 red, 5 blue, and 2 green marbles. One marble is chosen at random.
What is the probability of choosing a blue marble?

✏️ Solution

Total marbles = 3 + 5 + 2 = 10
Favourable outcomes (blue) = 5
\( P(\text{blue}) = \dfrac{5}{10} = \dfrac{1}{2} \)
⚠️ Watch out: both A and D/C look like ½ — always simplify your fraction!
2
Easy Complement Rule ⭐ High-frequency mistake
COMPLEMENT = 1 − P(A) — "NOT happening" = 1 minus it happening
The probability that it rains tomorrow is \(0.35\).
What is the probability that it does NOT rain?

✏️ Solution

\( P(\text{not rain}) = 1 - P(\text{rain}) = 1 - 0.35 = \mathbf{0.65} \)
⚠️ D is a trap — probability can NEVER exceed 1!
3
Easy Sample Space · Dice
LIST IT OUT — when sample space is small, write every outcome
Two fair dice are rolled. What is the probability of getting a sum of 7?

✏️ Solution

Total outcomes: \(6 \times 6 = 36\)
Pairs summing to 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) → 6 pairs
\( P(\text{sum}=7) = \dfrac{6}{36} = \dfrac{1}{6} \)
✓ Note: A and B are the same value — always express as a simplified fraction or verify both are equal.
Section 2

🟡 Core Rules — And / Or / Not

4
Easy Mutually Exclusive Events ⭐⭐ Very common error
MUTUALLY EXCLUSIVE = can't both happen → P(A∩B) = 0, so just ADD the probabilities
\( P(A) = 0.4 \) and \( P(B) = 0.3 \). Events A and B are mutually exclusive.
Find \( P(A \cup B) \).

✏️ Solution

Mutually exclusive → \(P(A \cap B) = 0\)
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.3 - 0 = \mathbf{0.7} \)
⚠️ A (0.12) is the trap: that's what you'd get if you multiplied — only multiply for independent events AND-type questions!
📖 Example — Addition Rule (Non-Mutually Exclusive)

In a class of 30 students: 18 play football, 12 play basketball, 5 play both.

\( P(F \cup B) = \dfrac{18}{30} + \dfrac{12}{30} - \dfrac{5}{30} = \dfrac{25}{30} = \dfrac{5}{6} \)

KEY: If events CAN overlap → subtract the overlap once!

5
Medium Addition Rule · Venn Diagram ⭐⭐ Exam favourite
∪ = OR → ADD both, then SUBTRACT overlap once
\( P(A) = 0.5 \), \( P(B) = 0.4 \), \( P(A \cap B) = 0.2 \).
Find \( P(A \cup B) \).

✏️ Solution

\( P(A \cup B) = 0.5 + 0.4 - 0.2 = \mathbf{0.7} \)
⚠️ A (0.90) is the trap: forgetting to subtract the overlap!
6
Medium Venn Diagram — Finding Intersection
REARRANGE the formula: \( P(A \cap B) = P(A)+P(B)-P(A \cup B) \)
\( P(X) = 0.6 \), \( P(Y) = 0.5 \), \( P(X \cup Y) = 0.8 \).
Find \( P(X \cap Y) \).

✏️ Solution

\( P(X \cap Y) = P(X) + P(Y) - P(X \cup Y) = 0.6 + 0.5 - 0.8 = \mathbf{0.3} \)
⚠️ A (0.48) = multiplying 0.6 × 0.8 — don't multiply unless told events are independent!
Section 3

🟠 Conditional Probability

7
Medium Conditional Probability — Formula ⭐⭐⭐ #1 exam topic
P(A|B) = "A given B" → SHRINK your world to B, then find A inside it
\( P(A \cap B) = 0.12 \) and \( P(B) = 0.4 \).
Find \( P(A|B) \).

✏️ Solution

\( P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.12}{0.4} = \mathbf{0.3} \)
⚠️ A (0.048) = multiplying 0.12 × 0.4 — that's completely wrong direction!
⚠️ D = just writing the intersection — you must DIVIDE by P(B).
8
Medium Conditional Probability — Two-Way Table ⭐⭐ Table reading trap
GIVEN = row/column total — the condition restricts which total you divide by
A survey of 100 students:
Likes MathsDislikes MathsTotal
Science282250
Arts123850
Total4060100
A student is selected at random. Given the student is from Science, find the probability they like Maths.

✏️ Solution

"Given Science" → restrict to Science row → total = 50
"Like Maths AND Science" = 28
\( P(\text{Maths}|\text{Science}) = \dfrac{28}{50} = 0.56 \)
⚠️ B uses 40 (Maths column total) — that's dividing by the wrong group!
9
Hard Independence Test ⭐⭐⭐ Conceptual trap
INDEPENDENT TEST: Check if \(P(A|B) = P(A)\) — OR equivalently \(P(A \cap B) = P(A) \cdot P(B)\)
\( P(A) = 0.4 \), \( P(B) = 0.5 \), \( P(A \cap B) = 0.2 \).
Are events A and B independent?

✏️ Solution

Check: \(P(A) \times P(B) = 0.4 \times 0.5 = 0.2\)
\(P(A \cap B) = 0.2\) ✓ → They match! → Independent
⚠️ B is the classic mistake: thinking \(P(A \cap B) \neq 0\) means dependent. That's the definition of mutually exclusive, not independence!
⚠️ C: Not mutually exclusive ≠ independent. These are completely different concepts!
Section 4

🔵 Tree Diagrams & Combined Events

10
Medium Tree Diagram — Without Replacement ⭐⭐⭐ Most missed question type
WITHOUT REPLACEMENT → denominator DECREASES each draw. WITH replacement → denominator STAYS same
A box has 4 red and 6 blue balls. Two balls are drawn without replacement.
Find the probability that both are red.

✏️ Solution

1st draw: \(P(\text{red}) = \dfrac{4}{10}\)
2nd draw (red taken out): \(P(\text{red}) = \dfrac{3}{9}\)
\(P(\text{both red}) = \dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}\)
⚠️ A: \(\frac{16}{100} = \frac{4}{10} \times \frac{4}{10}\) — that's WITH replacement!
11
Medium Tree Diagram — At Least One ⭐⭐ Complement shortcut
AT LEAST ONE = use COMPLEMENT: \(1 - P(\text{none})\) — much faster!
A biased coin has \(P(\text{head}) = 0.6\). It is flipped twice.
Find the probability of getting at least one head.

✏️ Solution

Use complement: \(P(\text{at least 1 head}) = 1 - P(\text{no heads})\)
\(P(\text{no heads}) = P(\text{TT}) = 0.4 \times 0.4 = 0.16\)
\(P(\text{at least 1 head}) = 1 - 0.16 = \mathbf{0.84}\)
⚠️ D (0.48) = just doing P(HT or TH) — forgetting HH!
12
Hard Tree Diagram — Exactly One
EXACTLY ONE = list ALL paths with exactly 1 success, then ADD them
\(P(\text{Alice passes}) = 0.7\), \(P(\text{Bob passes}) = 0.6\). They take tests independently.
Find the probability that exactly one of them passes.

✏️ Solution

Path 1: Alice passes, Bob fails: \(0.7 \times 0.4 = 0.28\)
Path 2: Alice fails, Bob passes: \(0.3 \times 0.6 = 0.18\)
\(P(\text{exactly one}) = 0.28 + 0.18 = \mathbf{0.46}\)
⚠️ A (0.42) = \(0.7 \times 0.6\) = P(both pass), which is NOT what was asked!
Section 5

🔴 Advanced — Bayes & Exam Level

13
Hard Bayes' Theorem — Intro ⭐⭐⭐ Exam classic
BAYES = FLIP conditional probability. Draw a tree → find the branch → divide by total probability of outcome
A factory has two machines. Machine A produces 60% of items; Machine B produces 40%. Machine A has a 5% defect rate; Machine B has a 10% defect rate.

An item is chosen at random and found to be defective. What is the probability it came from Machine B?

✏️ Solution — Tree Method

\(P(\text{defective from A}) = 0.60 \times 0.05 = 0.030\)
\(P(\text{defective from B}) = 0.40 \times 0.10 = 0.040\)
\(P(\text{defective}) = 0.030 + 0.040 = 0.070\)
\(P(B|\text{defective}) = \dfrac{0.040}{0.070} = \dfrac{4}{7} \approx 0.571\)
⚠️ D (0.40) = just the proportion from B — doesn't account for defect rate at all!
14
Hard Combined Conditional — Multi-step ⭐⭐⭐ Tricky wording
TOTAL PROBABILITY: \(P(B) = P(B|A_1)P(A_1) + P(B|A_2)P(A_2)\) — sum all paths leading to B
A medical test for a disease has:
• \(P(\text{positive} | \text{has disease}) = 0.95\) (sensitivity)
• \(P(\text{positive} | \text{no disease}) = 0.08\) (false positive)
• \(P(\text{has disease}) = 0.02\)

Find \(P(\text{positive test})\).

✏️ Solution — Law of Total Probability

\(P(+) = P(+|\text{disease}) \cdot P(\text{disease}) + P(+|\text{no disease}) \cdot P(\text{no disease})\)
\(= 0.95 \times 0.02 + 0.08 \times 0.98\)
\(= 0.019 + 0.0784 = \mathbf{0.0970} \approx 0.097\)
⚠️ C (0.08) = only counting false positives. A = only counting true positives. Easy to miss that you need BOTH paths!
15
Hard Permutations & Probability ⭐⭐ Counting + probability combo
COUNTING METHODS: Permutation = order matters \(\frac{n!}{(n-r)!}\) · Combination = order doesn't \(\binom{n}{r}\)
5 students (including Alice and Bob) stand in a line at random.
What is the probability that Alice and Bob stand next to each other?

✏️ Solution — Treat pair as a block

Total arrangements of 5 people: \(5! = 120\)
Treat Alice+Bob as one block → 4 units → \(4! = 24\) arrangements
Alice and Bob can swap within the block: \(\times 2\) → \(24 \times 2 = 48\)
\(P = \dfrac{48}{120} = \dfrac{2}{5}\)
⚠️ A (1/5): forgot to multiply by 2 for the internal swap of Alice/Bob!
Section 6

🟣 Exam Level — IB Style

16
Exam Conditional Independence — Proof type ⭐⭐⭐ Proof-style IB question
PROVE INDEPENDENCE: Show \(P(A \cap B) = P(A) \cdot P(B)\) — calculate both sides separately
\(P(A) = 0.3\), \(P(B) = 0.4\). A and B are independent.
Find \( P(A' \cap B') \) (probability that neither A nor B occurs).

✏️ Solution

If A and B are independent, so are A' and B'!
\(P(A') = 1 - 0.3 = 0.7\), \(P(B') = 1 - 0.4 = 0.6\)
\(P(A' \cap B') = 0.7 \times 0.6 = \mathbf{0.42}\)
⚠️ D (0.58) = \(P(A \cup B) = 0.3+0.4-0.12 = 0.58\), which is the probability of at least one — not neither!
17
Exam Bayes — Medical Testing (Advanced) ⭐⭐⭐ IB HL favourite
POSITIVE PREDICTIVE VALUE: Even a good test has lots of false positives if disease is RARE — always compute \(P(\text{disease}|\text{positive})\)
Using data from Q14: \(P(\text{disease}) = 0.02\), sensitivity \(= 0.95\), false positive rate \(= 0.08\), \(P(\text{positive}) = 0.097\).

Find \(P(\text{has disease} \mid \text{positive test})\). Round to 3 s.f.

✏️ Solution — Bayes' Theorem

\( P(\text{disease}|+) = \dfrac{P(+|\text{disease}) \cdot P(\text{disease})}{P(+)} \)
\( = \dfrac{0.95 \times 0.02}{0.097} = \dfrac{0.019}{0.097} \approx \mathbf{0.196} \)
🤯 Only ~20% of positive tests are actually diseased! This is why medical statistics is so counterintuitive — rare diseases + even small false positive rates = most positives are false.
⚠️ A (0.95): that's the sensitivity — totally different thing!
18
Exam Combinations & Probability ⭐⭐⭐ Combinatorics-probability
nCr PROBABILITY = \(\dfrac{\text{ways to get what you want (using }^nC_r\text{)}}{\text{total ways to choose}}\)
A committee of 3 is chosen from 5 men and 4 women.
What is the probability that the committee contains exactly 2 women?

✏️ Solution

Total ways to choose 3 from 9: \(\binom{9}{3} = 84\)
Ways to choose 2 women from 4: \(\binom{4}{2} = 6\)
Ways to choose 1 man from 5: \(\binom{5}{1} = 5\)
Favourable = \(6 \times 5 = 30\)
\(P = \dfrac{30}{84} = \dfrac{5}{14} \approx 0.357\)
⚠️ A (6/84) = only counting the women, forgetting to multiply by the ways to choose the man!
19
Exam Conditional + Independence Combined
INDEPENDENT ↔ CONDITIONAL EQUAL: A indep. B iff \(P(A|B) = P(A)\) iff \(P(B|A) = P(B)\)
\(P(C) = 0.5\), \(P(D) = 0.6\), \(P(C|D) = 0.5\).
Which statement is definitely true?

✏️ Solution

Check independence: \(P(C|D) = 0.5 = P(C)\) ✓ → C and D are independent!
Also: \(P(C \cap D) = P(C) \times P(D) = 0.5 \times 0.6 = 0.3 \neq 0\) → NOT mutually exclusive
For D: since independent, \(P(D|C) = P(D) = 0.6\) ✓ — D is also true!
⚠️ But the question asks which is definitely true: B captures the core fact. D follows from it but is a consequence. In IB exam context, choose B as the fundamental statement.
20
Exam Full IB-Style Multi-part Problem ⭐⭐⭐⭐ Boss question
IB EXAM STRATEGY: DRAW TREE → LABEL → MULTIPLY branches → ADD final outcomes
Three archers each shoot once. Their probabilities of hitting the target are:
\(P(A) = 0.8\), \(P(B) = 0.7\), \(P(C) = 0.6\). Shots are independent.

Find the probability that the target is hit by at least two archers.

✏️ Solution — Systematic Listing

Exactly 3 hit: \(0.8 \times 0.7 \times 0.6 = 0.336\)
Exactly 2 hit (3 cases):
A,B miss C: \(0.8 \times 0.7 \times 0.4 = 0.224\)
A,C miss B: \(0.8 \times 0.3 \times 0.6 = 0.144\)
B,C miss A: \(0.2 \times 0.7 \times 0.6 = 0.084\)
Exactly 2 total: \(0.224 + 0.144 + 0.084 = 0.452\)
\(P(\text{at least 2}) = 0.452 + 0.336 = \mathbf{0.692} \approx 0.692\)
⚠️ A (0.336) = only counting all-three hit. B (0.788) = close but missing some cases.
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