Self-Study Edition
Probability
& Set Theory
20 carefully curated questions covering the most commonly missed topics. Choose wisely — every answer reveals a full explanation.
🧠
⚡ Quick Memory Points
∪ = Union = OR = everything in either set |
∩ = Intersection = AND = only what's shared
A' = Complement = NOT A = everything outside A
A ⊆ B = A is a subset of B (A fits inside B)
n(A∪B) = n(A)+n(B)−n(A∩B) ← golden formula!
Q 01
Sets · Basic
Let \(U = \{1,2,3,4,5,6,7,8\}\), \(A = \{1,2,3,4\}\), \(B = \{3,4,5,6\}\).
Find \(A \cup B\).
💡 Explanation
Union (∪) means ALL elements that appear in A
or B (or both). List everything: {1,2,3,4} from A plus {5,6} new from B = {1,2,3,4,5,6}. Don't double-count {3,4}!
Q 02
⚠ Tricky · Complement
Using the same sets above, find \((A \cap B)'\).
Hint: Do the intersection first, then take the complement with respect to U.
💡 Explanation
Step 1: A ∩ B = {3,4} (elements in BOTH sets).
Step 2: (A ∩ B)' = U minus {3,4} = {1,2,5,6,7,8}.
Common mistake: forgetting to include {7,8} which are in U but in neither A nor B!
Q 03
⚠ Tricky · Venn Diagram
In a class of 30 students, 18 study French, 15 study Spanish, and 7 study both.
How many students study neither language?
Apply the inclusion-exclusion formula: \(n(F \cup S) = n(F) + n(S) - n(F \cap S)\)
💡 Explanation
n(F ∪ S) = 18 + 15 − 7 = 26. This is the number who study at least one language.
Neither = Total − n(F ∪ S) = 30 − 26 =
4.
Trap: many students just subtract the overlap (7) from 30, getting 23. Always find the union first!
Q 04
Subsets
How many subsets does the set \(S = \{a, b, c\}\) have?
💡 Explanation
Number of subsets = 2ⁿ where n is the number of elements.
Here n = 3, so 2³ = 8.
The subsets are: {}, {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c}. Don't forget the empty set ∅!
Chapter 2 · Basic Probability
🎲
⚡ Quick Memory Points
P(A) = favourable / total | always between 0 and 1
P(A') = 1 − P(A) ← complement rule
P(A∪B) = P(A)+P(B)−P(A∩B) ← addition rule
Mutually exclusive: P(A∩B)=0, so P(A∪B)=P(A)+P(B)
Q 05
Basic Probability
A bag contains 4 red, 3 blue, and 5 green marbles. A marble is picked at random.
What is the probability of not picking a blue marble?
💡 Explanation
P(blue) = 3/12 = 1/4.
P(not blue) = 1 − 1/4 = 3/4.
Note: 9/12 and 3/4 are equal — but option C is the simplified form. Always simplify fractions in IB!
Q 06
★ Hard · Addition Rule
\(P(A) = 0.5\), \(P(B) = 0.4\), \(P(A \cap B) = 0.2\).
Find \(P(A \cup B)\).
💡 Explanation
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.5 + 0.4 − 0.2 =
0.7.
The most common error: adding 0.5 + 0.4 = 0.9 without subtracting the overlap. You'd be double-counting the intersection!
Q 07
⚠ Tricky · Mutually Exclusive
Events A and B are mutually exclusive with \(P(A)=0.3\) and \(P(B)=0.5\).
Which of the following is impossible?
💡 Explanation
Mutually exclusive means A and B
cannot happen at the same time, so P(A ∩ B) = 0 by definition.
Therefore P(A ∩ B) = 0.15 is
impossible for mutually exclusive events.
A, B, C are all consistent with the given info.
Q 08
Probability · Dice
A fair die is rolled once. What is the probability of getting a number greater than 4?
💡 Explanation
"Greater than 4" on a standard die = {5, 6} → 2 outcomes out of 6.
P = 2/6 =
1/3.
Watch the language: "greater than 4" does NOT include 4. "At least 4" would give {4,5,6} = 3/6 = 1/2.
Chapter 3 · Conditional Probability
🔑
⚡ Quick Memory Points
P(A|B) = P(A∩B) / P(B) ← "given B has happened"
Independent: P(A|B) = P(A) → knowing B tells you nothing about A
Independent test: P(A∩B) = P(A)×P(B) ?
Dependent (without replacement): pool shrinks each time!
Q 09
⚠ Tricky · Conditional
\(P(A) = 0.6\), \(P(B) = 0.5\), \(P(A \cap B) = 0.3\).
Find \(P(A \mid B)\).
💡 Explanation
P(A|B) = P(A∩B) / P(B) = 0.3 / 0.5 =
0.6.
Notice P(A|B) = P(A) = 0.6 — this means A and B are
independent!
Also check: P(A)×P(B) = 0.6×0.5 = 0.3 = P(A∩B) ✓ (independence confirmed)
Q 10
★ Hard · Without Replacement
A box has 5 red and 3 blue balls. Two balls are drawn without replacement.
What is the probability that both balls are red?
💡 Explanation
P(1st red) = 5/8.
P(2nd red | 1st was red) = 4/7 ← only 4 red remain, total is now 7!
P(both red) = 5/8 × 4/7 = 20/56 =
5/14.
Option A (25/64) is the wrong answer for "with replacement". Without replacement changes the denominator!
Chapter 4 · Combinations & Permutations
🔢
⚡ Quick Memory Points
Permutation = order matters → nPr = n!/(n−r)!
Combination = order does NOT matter → nCr = n!/r!(n−r)!
Trick: "Choose" = Combination. "Arrange" = Permutation
nCr = nC(n−r) ← e.g. 10C3 = 10C7 (symmetry!)
Q 11
Combinations
A team of 3 is chosen from 7 students. How many ways can this be done?
💡 Explanation
Choosing a team = order doesn't matter → Combination.
₇C₃ = 7! / (3! × 4!) = (7×6×5)/(3×2×1) = 210/6 =
35.
Option B (210) is ₇P₃ — the permutation (order matters). Same numbers, different concept!
Q 12
★ Hard · Permutations
How many different 3-digit codes can be made from the digits {1, 2, 3, 4, 5} if no digit repeats?
💡 Explanation
A "code" has order (123 ≠ 321) → Permutation.
₅P₃ = 5!/(5−3)! = 5!/2! = 5×4×3 =
60.
Option B (125) = 5³ — this counts codes WITH repetition allowed. The restriction "no repeats" reduces the count!
Q 13
⚠ Tricky · nCr Symmetry
Which of the following equals \(\dbinom{12}{9}\)?
Use the symmetry property of combinations.
💡 Explanation
Key property: ₙCᵣ = ₙC(ₙ₋ᵣ).
So ₁₂C₉ = ₁₂C(12−9) = ₁₂C₃ =
220.
This makes sense: choosing 9 to include is the same as choosing 3 to exclude. Option C is invalid (r can't exceed n).
Q 14
★ Hard · Combo + Prob
A committee of 2 is chosen at random from 4 men and 3 women.
What is the probability that both members are women?
C
\(\dfrac{3}{21} = \dfrac{1}{7}\)
💡 Explanation
Total ways to choose 2 from 7: ₇C₂ = 21.
Ways to choose 2 women from 3: ₃C₂ = 3.
P(both women) = 3/21 =
1/7.
Note: options B and C are both equal — C just shows the working. In IB, showing the fraction before simplifying is good practice!
Chapter 5 · Tree Diagrams & Sample Space
🌳
⚡ Quick Memory Points
Along branches: multiply probabilities
Between branches: add probabilities
All branch ends must sum to 1 (sanity check!)
Sample space: list ALL possible outcomes — use grid or tree
Q 15
⚠ Tricky · Tree Diagram
A coin is flipped twice. What is the probability of getting exactly one Head?
💡 Explanation
Sample space: {HH, HT, TH, TT}. Exactly one H: {HT, TH} = 2 outcomes.
P = 2/4 =
1/2.
Using tree: P(HT) = 1/2 × 1/2 = 1/4. P(TH) = 1/4. Add them: 1/4 + 1/4 = 1/2 ✓
Q 16
★ Hard · Two-stage Tree
Box A has 2 red and 1 blue. Box B has 1 red and 2 blue. A box is chosen at random, then a ball is drawn.
What is the probability of drawing a red ball?
💡 Explanation
P(A)=1/2, P(B)=1/2. P(red|A)=2/3. P(red|B)=1/3.
P(red) = P(A)×P(red|A) + P(B)×P(red|B)
= 1/2×2/3 + 1/2×1/3 = 2/6 + 1/6 = 3/6 =
1/2.
This is the law of total probability. Always multiply along branches, add between routes!
Chapter 6 · Expected Value & Distributions
📊
⚡ Quick Memory Points
E(X) = Σ x·P(x) ← sum of (value × probability)
Fair game: E(X) = 0 → expected win = expected loss
All P(x) must sum to 1 — use this to find missing values!
E(aX+b) = aE(X)+b ← linear transformation rule
Q 17
⚠ Tricky · Missing Probability
A random variable X has the distribution:
\(P(X=1)=0.2,\quad P(X=2)=0.5,\quad P(X=3)=k\)
Find \(k\).
💡 Explanation
All probabilities must sum to 1: 0.2 + 0.5 + k = 1.
0.7 + k = 1 → k =
0.3.
This "sum to 1" rule is a key IB exam skill and a quick sanity check for any probability table!
Q 18
★ Hard · Expected Value
A game costs $3 to play. You roll a die: score 6 → win $10; any other → win nothing.
What is the expected profit per game?
A
\(\$\tfrac{10}{6} \approx \$1.67\)
B
\(-\$\tfrac{4}{6} \approx -\$0.67\)
💡 Explanation
Profit if win: 10−3 = $7. Probability: 1/6.
Profit if lose: 0−3 = −$3. Probability: 5/6.
E(profit) = 7×(1/6) + (−3)×(5/6) = 7/6 − 15/6 = −8/6 = −
4/6 ≈ −$0.67.
Negative expected value = house always wins. This game is not worth playing!
Q 19
★ Hard · Independent Events Test
\(P(A)=0.4\), \(P(B)=0.3\), \(P(A \cup B)=0.58\).
Are A and B independent?
A
Yes, because \(P(A)+P(B)=0.7 \neq 1\)
B
No, because \(P(A \cup B) \neq 1\)
C
Yes, because the events don't share outcomes
D
Yes, because \(P(A \cap B)=P(A) \cdot P(B)\)
💡 Explanation
Find P(A∩B): P(A∪B) = P(A)+P(B)−P(A∩B) → 0.58 = 0.4+0.3−P(A∩B) → P(A∩B) = 0.12.
Test independence: P(A)×P(B) = 0.4×0.3 = 0.12 = P(A∩B) ✓
So yes, A and B are
independent, confirmed by the multiplication rule!
Q 20
★ Hard · Combo in Probability
A hand of 5 cards is dealt from a standard deck of 52. How many possible hands contain exactly 3 aces?
Hint: Choose 3 aces from 4 available, then choose the remaining 2 cards from the non-ace cards.
💡 Explanation
Ways to choose 3 aces from 4: ₄C₃ = 4.
Ways to choose remaining 2 cards from 48 non-aces: ₄₈C₂ = (48×47)/2 = 1128.
Total = 4 × 1128 =
4512.
This "two-step combination" technique is essential for IB — always split into independent choices and multiply!