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IB Mathematics · Probability & Combinatorics

Master the Art
of Counting & Chance

20 carefully selected questions covering Set Theory, Probability Rules, Permutations, Combinations, and Venn Diagrams — the topics most students find tricky.

Set Theory Venn Diagrams Basic Probability Complementary Events Conditional Probability Permutations nPr Combinations nCr Mutually Exclusive Independent Events Tree Diagrams

⚡ Quick Memory Points — Memorize These!

COMPLEMENT
\(P(A') = 1 - P(A)\)
"NOT A" — everything outside A. Sum always = 1.
ADDITION RULE
\(P(A \cup B) = P(A)+P(B)-P(A \cap B)\)
Subtract overlap so you don't count it twice!
MUTUALLY EXCL.
\(P(A \cap B) = 0\)
Can't happen together. No overlap in Venn.
INDEPENDENT
\(P(A \cap B) = P(A) \times P(B)\)
One event does NOT affect the other.
CONDITIONAL
\(P(A|B) = \dfrac{P(A \cap B)}{P(B)}\)
"Given B happened" — shrink the sample space!
nPr (ORDER)
\({}^nP_r = \dfrac{n!}{(n-r)!}\)
Order MATTERS. Think: arrange seats in a row.
nCr (NO ORDER)
\(\dbinom{n}{r} = \dfrac{n!}{r!(n-r)!}\)
Order does NOT matter. Think: pick a team.
SET NOTATION
\(A \cup B \text{ vs } A \cap B\)
∪ = Union (OR) · ∩ = Intersection (AND)

Set Theory & Venn Diagrams

Q 1–5
1
Set Theory Easy
A universal set \(\mathcal{U} = \{1,2,3,4,5,6,7,8,9,10\}\), \(A = \{2,4,6,8,10\}\), and \(B = \{1,2,3,4,5\}\).
Find \(A \cap B\).
💡 Key Concept \(A \cap B\) means elements that are in BOTH A and B simultaneously. List A, list B, circle what appears in both — that's your intersection.
📖 Explanation
Intersection \(A \cap B\) picks elements appearing in both sets.
A = {2,4,6,8,10} and B = {1,2,3,4,5}. Common elements: 2 (in both) and 4 (in both). So \(A \cap B = \{2,4\}\).
Note: 6,8,10 are in A but not B. 1,3,5 are in B but not A.
2
Set Theory Easy
Using the same sets: \(\mathcal{U} = \{1,...,10\}\), \(A = \{2,4,6,8,10\}\).
What is \(A'\) (the complement of A)?
💡 Memory Point: COMPLEMENT = "OUTSIDE" \(A'\) is everything in \(\mathcal{U}\) that is NOT in A. Think of it as flipping — what you didn't pick.
📖 Explanation
\(A' = \mathcal{U} \setminus A\) — remove all elements of A from the universal set.
\(\mathcal{U} = \{1,2,3,4,5,6,7,8,9,10\}\), removing \(\{2,4,6,8,10\}\) leaves \(\{1,3,5,7,9\}\).
Key check: \(P(A) + P(A') = 1\) always.
3
Venn Diagram Easy
In a class of 30 students: 18 play football (F), 12 play basketball (B), and 5 play both.
How many students play neither sport?
💡 Venn Diagram Formula \(n(F \cup B) = n(F) + n(B) - n(F \cap B)\). Then subtract from total. Subtract overlap once!
📖 Explanation
Step 1: \(n(F \cup B) = 18 + 12 - 5 = 25\) students play at least one sport.
Step 2: Neither = Total − \(n(F \cup B)\) = \(30 - 25 = \mathbf{5}\).
Common mistake: forgetting to subtract the "both" group, which gives 30 - 30 = 0. Wrong!
4
Set Theory Medium
\(A = \{a,b,c,d\}\) and \(B = \{c,d,e,f\}\).
Which statement is TRUE?
💡 Watch Out! \(A \setminus B\) (A minus B) means elements in A but NOT in B. Don't confuse with \(A \cap B\)!
📖 Explanation
• \(A \cap B = \{c,d\}\) (NOT the full union) — so option A is wrong.
• \(A \cup B = \{a,b,c,d,e,f\}\) (NOT just the intersection) — so B is wrong.
• \(A \setminus B\) = elements in A but NOT in B. Removing c,d from A: leaves \(\{a,b\}\). ✓
Option C mixed up \(\setminus\) and \(\cap\).
5
Venn Diagram Medium
A survey of 50 people found: 30 like tea (T), 25 like coffee (C), 10 like both.
A person is chosen at random. Find \(P(T \text{ only})\) — probability they like tea but NOT coffee.
💡 "Only" Trick "T only" = T minus the overlap = \(n(T) - n(T \cap C)\). Draw the Venn, shade just the left circle (not the middle)!
📖 Explanation
"T only" means like tea but NOT coffee: \(n(T \text{ only}) = 30 - 10 = 20\).
\(P(T \text{ only}) = \dfrac{20}{50} = 0.4\)
Trap: option A gives P(T) including those who also like coffee — that's NOT "T only"!

Basic Probability Rules

Q 6–10
6
Basic Probability Easy
A fair die (6 sides) is rolled. What is the probability of rolling a number greater than 4?
💡 Fundamental Formula \(P(\text{event}) = \dfrac{\text{number of favorable outcomes}}{\text{total outcomes}}\). List favorable outcomes explicitly!
📖 Explanation
Numbers greater than 4 on a die: {5, 6} — that's 2 outcomes.
Total outcomes: {1,2,3,4,5,6} = 6.
\(P = \dfrac{2}{6} = \dfrac{1}{3} \approx 0.333\)
Common error: writing P = 4/6 by counting {1,2,3,4} — those are ≤ 4, not > 4!
7
Complementary Easy
The probability it rains tomorrow is 0.35. What is the probability it does NOT rain?
💡 Complement Rule: 1 Minus \(P(A') = 1 - P(A)\). If probability of rain = 0.35, then no rain = 1 − 0.35. Super simple, never forget!
📖 Explanation
\(P(\text{no rain}) = 1 - P(\text{rain}) = 1 - 0.35 = \mathbf{0.65}\)
Trap answer D (1.35) is impossible — probabilities can never exceed 1!
Remember: all probabilities must satisfy \(0 \le P \le 1\).
8
Addition Rule Medium
Events A and B are such that \(P(A) = 0.5\), \(P(B) = 0.4\), and \(P(A \cap B) = 0.2\).
Find \(P(A \cup B)\).
💡 Addition Rule — Don't Double Count! \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). The overlap gets counted twice if you just add, so subtract it once.
📖 Explanation
\(P(A \cup B) = 0.5 + 0.4 - 0.2 = \mathbf{0.7}\)
Option A (0.9) is wrong — it forgets to subtract the intersection.
Option D (1.1) is impossible since probability ≤ 1.
The intersection 0.2 is subtracted because it was counted in both P(A) and P(B).
9
Mutually Exclusive Medium
Events A and B are mutually exclusive. \(P(A) = 0.3\) and \(P(B) = 0.45\).
Which of the following is true?
💡 Mutually Exclusive = ZERO Overlap If A and B are mutually exclusive: \(P(A \cap B) = 0\), so \(P(A \cup B) = P(A) + P(B)\). They CANNOT both happen!
📖 Explanation
Mutually exclusive ⟹ \(P(A \cap B) = 0\).
So \(P(A \cup B) = 0.3 + 0.45 - 0 = \mathbf{0.75}\). ✓
Option C: \(0.3 \times 0.45 = 0.135\) — this would be \(P(A) \times P(B)\), used for independent events, not mutually exclusive!
Key distinction: Mutually exclusive ≠ Independent. Mutually exclusive events with P>0 are actually DEPENDENT!
10
Independent Events Medium
Two fair coins are flipped. What is the probability of getting exactly two heads?
💡 Independent Events → Multiply Each coin flip is independent. \(P(\text{HH}) = P(H) \times P(H) = \frac{1}{2} \times \frac{1}{2}\). List all outcomes: HH, HT, TH, TT.
📖 Explanation
Sample space: {HH, HT, TH, TT} — 4 equally likely outcomes.
Exactly 2 heads: only {HH} — 1 outcome.
\(P(HH) = \dfrac{1}{4}\)
OR using multiplication: \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\).
Trap: Option B (1/2) counts "at least one head" not "exactly two heads."

Conditional Probability

Q 11–14
11
Conditional P Medium
\(P(A) = 0.6\), \(P(B) = 0.5\), \(P(A \cap B) = 0.3\).
Find \(P(A \mid B)\) — the probability of A given B.
💡 Conditional Probability Formula \(P(A|B) = \dfrac{P(A \cap B)}{P(B)}\). "Given B" means B is now your new universe — divide by P(B)!
📖 Explanation
\(P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.3}{0.5} = \mathbf{0.6}\)
Note: \(P(A|B) = P(A) = 0.6\). This means A and B are actually INDEPENDENT! (Knowing B happened doesn't change P(A).)
Option D (0.18) = 0.6 × 0.3 — wrong formula entirely.
12
Conditional P Medium
A bag contains 3 red, 4 blue, and 3 green balls. One ball is drawn and not replaced. Then a second is drawn.
What is \(P(\text{both red})\)?
💡 Without Replacement → Dependent! First draw changes what's left. Use: \(P(\text{R then R}) = P(R_1) \times P(R_2 | R_1)\). The second P uses 9 balls (one gone).
📖 Explanation
There are 3 red balls, 10 total.
\(P(R_1) = \dfrac{3}{10}\)
After 1 red is removed: 2 red remain, 9 total.
\(P(R_2|R_1) = \dfrac{2}{9}\)
\(P(\text{both red}) = \dfrac{3}{10} \times \dfrac{2}{9} = \dfrac{6}{90} = \dfrac{1}{15}\)
Option B uses replacement (wrong — the problem says "not replaced"). That's the most common mistake here!
13
Tree Diagram Medium
A student passes a test with probability 0.7. If they pass, probability of getting a scholarship is 0.8. If they fail, probability of scholarship is 0.1.
What is the probability of getting a scholarship?
💡 Total Probability — Two Paths Draw a tree! Path 1: Pass → Scholarship. Path 2: Fail → Scholarship. Add both P(path) × P(outcome on path).
📖 Explanation
Two paths to scholarship:
• Pass AND Scholarship: \(0.7 \times 0.8 = 0.56\)
• Fail AND Scholarship: \(0.3 \times 0.1 = 0.03\)
\(P(\text{scholarship}) = 0.56 + 0.03 = \mathbf{0.59}\)
Option A (0.56) only counts the "pass" path — forgot the second branch!
14
Independence Test Hard
\(P(A) = 0.4\), \(P(B) = 0.5\), \(P(A \cup B) = 0.7\).
Are events A and B independent?
💡 Independence Check First find \(P(A \cap B)\) from the addition rule. Then check: does \(P(A \cap B) = P(A) \times P(B)\)?
📖 Explanation
Step 1 — Find \(P(A \cap B)\) using addition rule:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(0.7 = 0.4 + 0.5 - P(A \cap B)\)
\(P(A \cap B) = 0.2\)
Step 2 — Independence check:
\(P(A) \times P(B) = 0.4 \times 0.5 = 0.2 = P(A \cap B)\) ✓
So YES, A and B are independent.
Option C is wrong: \(P(A \cap B) \ne 0\) does NOT mean dependent — it means not mutually exclusive!

Permutations — Order Matters

Q 15–17
15
Permutations Easy
In how many ways can 5 students be arranged in a row of 5 chairs?
💡 Factorial Rule for All n Items When ALL n items are arranged in n spots: answer = \(n!\). Here: \(5! = 5 \times 4 \times 3 \times 2 \times 1\). Think: 5 choices for seat 1, 4 for seat 2, ...
📖 Explanation
\(5! = 5 \times 4 \times 3 \times 2 \times 1 = \mathbf{120}\)
This is a permutation of all 5 items: \({}^5P_5 = 5!\)
Option D (60) = \(5!/2\) — would be for arranging 5 people where 2 are identical (wrong here, all different students).
Memorize: \(5! = 120\), \(4! = 24\), \(3! = 6\).
16
Permutations Medium
How many different 3-letter codes can be made from the letters {A, B, C, D, E} if no letter is repeated?
💡 nPr: Arrange r from n (Order Matters) \({}^nP_r = \dfrac{n!}{(n-r)!}\). Here: 5 letters, choose 3 positions, order matters (ABC ≠ BAC).
📖 Explanation
\({}^5P_3 = \dfrac{5!}{(5-3)!} = \dfrac{5!}{2!} = \dfrac{120}{2} = \mathbf{60}\)
Or use the slot method: 5 choices × 4 choices × 3 choices = 60. ✓
Option A (10) = \({}^5C_3\) — that's combinations (order doesn't matter). Wrong here because ABC ≠ CBA.
Option C (125) = \(5^3\) — that's WITH repetition allowed.
17
Permutations Hard
6 people (including Alice and Bob) sit in a row. In how many arrangements are Alice and Bob always next to each other?
💡 "Glue Together" Trick Treat Alice+Bob as ONE unit → now you have 5 units. Arrange 5 units: \(5!\). But Alice and Bob can swap within their unit: ×2!
📖 Explanation
Step 1: Glue Alice+Bob into one "super-person" → 5 units to arrange.
Step 2: Arrangements of 5 units = \(5! = 120\)
Step 3: Inside the unit, Alice-Bob or Bob-Alice → \(\times 2! = \times 2\)
Total = \(120 \times 2 = \mathbf{240}\)
Option D (720) = \(6!\) = all arrangements with NO restriction. Divide by the constraint gives 240.

Combinations — Order Doesn't Matter

Q 18–20
18
Combinations Easy
A committee of 3 people is to be chosen from a group of 8. How many different committees are possible?
💡 Committee = Combination (Order Doesn't Matter) \(\dbinom{n}{r} = \dfrac{n!}{r!(n-r)!}\). {Alice, Bob, Carol} is the SAME committee as {Carol, Bob, Alice}. That's why we use C, not P!
📖 Explanation
\(\dbinom{8}{3} = \dfrac{8!}{3! \cdot 5!} = \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = \dfrac{336}{6} = \mathbf{56}\)
Option C (336) = \({}^8P_3\) = permutation (treating order as important). But a committee {A,B,C} = {C,A,B}, so divide by \(3! = 6\).
Key: "Committee," "team," "group," "selection" → always Combination!
19
Combinations Medium
A hand of 5 cards is dealt from a standard deck of 52 cards. What is the probability of getting exactly 3 aces?
💡 Probability with Combinations \(P = \dfrac{\text{favorable combinations}}{\text{total combinations}}\). Favorable: choose 3 aces from 4 AND 2 non-aces from 48.
📖 Explanation
Total 5-card hands: \(\dbinom{52}{5}\)
Favorable: choose exactly 3 aces from 4 available: \(\dbinom{4}{3} = 4\) ways, AND choose 2 non-aces from remaining 48: \(\dbinom{48}{2} = 1128\) ways.
\(P = \dfrac{4 \times 1128}{\dbinom{52}{5}} = \dfrac{4512}{2598960} \approx 0.00174\)
Option C: only counts aces, forgets you must also choose the other 2 cards!
Option D uses permutation logic (ordered drawing), incorrect for card hands.
20
P vs C Challenge Hard
A class has 10 students. In how many ways can a president, secretary, and treasurer be chosen (one person per role), versus simply choosing a group of 3?
Which pair of values is correct?
💡 The Classic P vs C Distinction Roles = Order matters (President ≠ Secretary) → use \({}^{10}P_3\).
Group = Order doesn't matter → use \(\dbinom{10}{3}\). The ratio between them is always \(r! = 3! = 6\).
📖 Explanation
Roles (Permutation):
\({}^{10}P_3 = \dfrac{10!}{7!} = 10 \times 9 \times 8 = \mathbf{720}\)

Group of 3 (Combination):
\(\dbinom{10}{3} = \dfrac{10!}{3! \cdot 7!} = \dfrac{720}{6} = \mathbf{120}\)

Note: \({}^{10}P_3 \div 3! = {}^{10}C_3\) → \(720 \div 6 = 120\). ✓
The key insight: every combination corresponds to \(3! = 6\) permutations (6 ways to assign the 3 different roles).