IB Mathematics · Grade 9

Probability
& Set Theory

Master the core concepts of Sets, Combinations, and Probability — the IB way.

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Questions
4
Topics
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Topic 01

Set Theory & Notation

Union, intersection, complement, and Venn diagrams — the language of sets.

🧠
Quick Memory Keys
Union ∪ = OR = ADD
Intersect ∩ = AND = OVERLAP
Complement A' = NOT A
Subset ⊆ = INSIDE
n(A) = COUNT elements
∅ = EMPTY SET
01
Set NotationEasy
If $A = \{1, 2, 3, 4, 5\}$ and $B = \{3, 4, 5, 6, 7\}$, what is $A \cap B$?
⚠️ Tricky point: Students often confuse $\cup$ (union) with $\cap$ (intersection). Remember: $\cap$ means elements that appear in BOTH sets.
💡 Explanation

Intersection $A \cap B$ means: elements that belong to both $A$ AND $B$.

$A = \{1, 2, \mathbf{3, 4, 5}\}$ and $B = \{\mathbf{3, 4, 5}, 6, 7\}$
The shared elements are $3, 4, 5$ → $A \cap B = \{3, 4, 5\}$ ✓
$A \cup B = \{1,2,3,4,5,6,7\}$ would be the union (option A — a common wrong answer!).
02
Venn DiagramEasy
In a class of 30 students: 18 play football, 12 play basketball, and 5 play both. How many students play at least one sport?
⚠️ "At least one" = Union. Use the Inclusion-Exclusion Principle: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$
💡 Explanation

Use the Inclusion-Exclusion Principle:

$n(F \cup B) = n(F) + n(B) - n(F \cap B)$
$= 18 + 12 - 5 = \mathbf{25}$
Why subtract 5? Because those 5 students were counted twice (once in football, once in basketball).
03
ComplementMedium
Universal set $U = \{1,2,3,4,5,6,7,8,9,10\}$, $A = \{2,4,6,8,10\}$. What is $A'$ (the complement of $A$)?
💡 Explanation

$A'$ = everything in $U$ that is NOT in $A$.

$A = \{2,4,6,8,10\}$ (even numbers)
$A' = U \setminus A = \{1,3,5,7,9\}$ (odd numbers) ✓
04
Subset & Power SetMedium
How many subsets does the set $S = \{a, b, c\}$ have?
⚠️ Many students forget to count $\emptyset$ and $S$ itself. Formula: $2^n$ where $n$ = number of elements.
💡 Explanation

Number of subsets = $2^n = 2^3 = 8$

They are: $\emptyset,\ \{a\},\ \{b\},\ \{c\},\ \{a,b\},\ \{a,c\},\ \{b,c\},\ \{a,b,c\}$
Don't forget $\emptyset$ (empty set) and $S$ itself — both are valid subsets!
Topic 02

Topic 02

Combinations & Permutations

Counting without confusion — when order matters and when it doesn't.

🔢
Quick Memory Keys
Permutation = ORDER matters (arrange)
Combination = ORDER doesn't matter (choose)
$nPr$ = $\frac{n!}{(n-r)!}$
$nCr$ = $\frac{n!}{r!(n-r)!}$
Factorial = multiply DOWN to 1
$0!$ = 1 (always!)
05
FactorialEasy
What is the value of $\dfrac{6!}{4!}$?
💡 Explanation

$\dfrac{6!}{4!} = \dfrac{6 \times 5 \times \cancel{4!}}{\cancel{4!}} = 6 \times 5 = 30$

Shortcut: Cancel the common factorial — don't expand everything!
06
CombinationEasy
A team of 3 students must be chosen from 7 students. How many different teams are possible?
⚠️ A "team" has no ranking — order doesn't matter. Use $\binom{n}{r}$, not $P(n,r)$.
💡 Explanation

$\binom{7}{3} = \dfrac{7!}{3! \cdot 4!} = \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} = \dfrac{210}{6} = 35$

Order does NOT matter (a team {A,B,C} is the same as {B,C,A}) → use Combination.
07
PermutationMedium
In how many ways can 4 different books be arranged on a shelf?
💡 Explanation

Arranging all 4 books = $4! = 4 \times 3 \times 2 \times 1 = 24$

Position 1: 4 choices → Position 2: 3 choices → Position 3: 2 → Position 4: 1
Order matters when arranging on a shelf → Permutation!
08
Permutation vs CombinationHard
A password is made from 3 digits chosen from $\{1,2,3,4,5\}$ with no repetition. How many passwords are possible?
⚠️ Password "123" ≠ "321" — order MATTERS here! Use Permutation.
💡 Explanation

$P(5,3) = \dfrac{5!}{(5-3)!} = \dfrac{5!}{2!} = \dfrac{120}{2} = 60$

Or: $5 \times 4 \times 3 = 60$ (5 choices for 1st digit, 4 for 2nd, 3 for 3rd)
$\binom{5}{3} = 10$ would be wrong here — that ignores order.
09
Combinations — ContextHard
From a standard deck of 52 cards, 5 cards are dealt. How many different 5-card hands are possible? (Leave in $\binom{n}{r}$ form.)
💡 Explanation

A "hand" of cards has no order → Combination: $\binom{52}{5} = 2{,}598{,}960$

Note: $\binom{52}{5} = \binom{52}{47}$ (they are equal — a useful identity!)
$P(52,5)$ would over-count by a factor of $5! = 120$.
Topic 03

Topic 03

Basic Probability

Single events, sample spaces, and the fundamental rules of chance.

🎲
Quick Memory Keys
$P(A)$ = favorable / total
Range = $0 \leq P(A) \leq 1$
Complement = $P(A') = 1 - P(A)$
Certain = $P = 1$
Impossible = $P = 0$
Sum rule = all probs add to 1
10
Basic ProbabilityEasy
A fair die is rolled. What is the probability of getting a number greater than 4?
💡 Explanation

Numbers greater than 4 on a die: $\{5, 6\}$ → 2 outcomes.

$P(\text{>4}) = \dfrac{2}{6} = \dfrac{1}{3}$
A common mistake: counting 4 itself. "Greater than 4" excludes 4!
11
Complement RuleEasy
The probability that it rains tomorrow is $0.35$. What is the probability that it does NOT rain?
💡 Explanation

$P(A') = 1 - P(A) = 1 - 0.35 = 0.65$

This always works because $P(A) + P(A') = 1$ (either it rains or it doesn't!).
12
Mutually ExclusiveMedium
Events $A$ and $B$ are mutually exclusive. $P(A) = 0.4$ and $P(B) = 0.3$. What is $P(A \cup B)$?
⚠️ Mutually exclusive = cannot happen at the same time → $P(A \cap B) = 0$
💡 Explanation

For mutually exclusive events: $P(A \cup B) = P(A) + P(B)$

$= 0.4 + 0.3 = 0.7$
Why? Because $P(A \cap B) = 0$, the general formula simplifies: $P(A \cup B) = P(A) + P(B) - \underbrace{P(A \cap B)}_{=0}$
13
Sample SpaceMedium
Two fair coins are flipped. What is the probability of getting exactly one Head?
⚠️ Always write out the sample space first: {HH, HT, TH, TT}
💡 Explanation

Sample space: $\{HH, HT, TH, TT\}$ — 4 equally likely outcomes.

Exactly one Head: $\{HT, TH\}$ → 2 outcomes
$P(\text{exactly 1H}) = \dfrac{2}{4} = \dfrac{1}{2}$
Topic 04

Topic 04

Conditional & Combined Probability

Independent events, conditional probability, and probability trees.

🌳
Quick Memory Keys
Independent = $P(A \cap B) = P(A) \times P(B)$
Conditional = $P(A|B) = \frac{P(A \cap B)}{P(B)}$
With replacement = Independent
Without replacement = Dependent
Tree diagram = multiply ALONG branches
OR on tree = ADD the ends
14
Independent EventsEasy
A coin is flipped and a die is rolled. What is the probability of getting Heads AND a 6?
💡 Explanation

Coin and die are independent (one doesn't affect the other).

$P(\text{H} \cap \text{6}) = P(\text{H}) \times P(\text{6}) = \dfrac{1}{2} \times \dfrac{1}{6} = \dfrac{1}{12}$
15
Without ReplacementMedium
A bag has 4 red and 6 blue balls. Two balls are drawn without replacement. What is the probability both are red?
⚠️ "Without replacement" means the second draw changes — the denominator decreases!
💡 Explanation

1st draw: $\dfrac{4}{10}$ (4 red out of 10 total)

2nd draw (no replacement): $\dfrac{3}{9}$ (3 red left out of 9 total)
$P(\text{both red}) = \dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}$
Option B is the "with replacement" answer — a very common mistake!
16
Conditional ProbabilityHard
$P(A) = 0.5$, $P(B) = 0.4$, $P(A \cap B) = 0.2$. Find $P(A \mid B)$.
⚠️ $P(A|B)$ means: "probability of A, GIVEN that B has already occurred." Denominator is $P(B)$, not 1.
💡 Explanation

$P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.2}{0.4} = 0.5$

Notice: $P(A|B) = P(A) = 0.5$ here — this means A and B are independent!
Independence test: $P(A \cap B) = P(A) \times P(B)$? $0.2 = 0.5 \times 0.4$ ✓
17
Probability TreeHard
A test is 90% accurate. A disease affects 1% of the population. If a person tests positive, what is closest to the probability they actually have the disease?
⚠️ This is Bayes' Theorem territory — very counter-intuitive! Use a tree or a table of 1000 people.
💡 Explanation (Imagining 1000 people)

Out of 1000 people: 10 have the disease, 990 don't.

Of 10 sick: $10 \times 0.9 = 9$ test positive (true positives)
Of 990 healthy: $990 \times 0.1 = 99$ test positive (false positives!)
Total positives = $9 + 99 = 108$. Of those, only 9 are actually sick.
$P(\text{sick} \mid \text{positive}) = \dfrac{9}{108} \approx 8.3\%$ — surprisingly low!
18
At Least OneMedium
A die is rolled twice. What is the probability of getting at least one 6?
⚠️ "At least one" → use complement: $P(\text{at least one 6}) = 1 - P(\text{no 6 at all})$
💡 Explanation

$P(\text{no 6 on either roll}) = \dfrac{5}{6} \times \dfrac{5}{6} = \dfrac{25}{36}$

$P(\text{at least one 6}) = 1 - \dfrac{25}{36} = \dfrac{11}{36}$
Don't just double $\frac{1}{6}$! That would double-count rolling 6 on both dice.
19
Sets + ProbabilityHard
In a group of 50 students: 30 study Maths ($M$), 25 study Science ($S$), and 10 study both. A student is chosen at random. Find $P(M \cup S)$.
💡 Explanation

$n(M \cup S) = n(M) + n(S) - n(M \cap S) = 30 + 25 - 10 = 45$

$P(M \cup S) = \dfrac{45}{50} = 0.9$
Option A gives $> 1$, which is impossible for a probability — always check!
20
Combinations + ProbabilityHard ★
A committee of 4 is chosen from 6 men and 4 women. What is the probability the committee has exactly 2 women?
⚠️ Combine $\binom{n}{r}$ with probability. Numerator = (ways to choose 2 women from 4) × (ways to choose 2 men from 6).
💡 Explanation

Total ways to choose 4 from 10: $\binom{10}{4} = \dfrac{10!}{4! \cdot 6!} = 210$

Ways to choose exactly 2 women AND 2 men:
$\binom{4}{2} \times \binom{6}{2} = 6 \times 15 = 90$
$P = \dfrac{90}{210} = \dfrac{3}{7} \approx 0.429$
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