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01
Basic Probability
⚡ Quick Memory Key
PROB = WANT / TOTAL
P(A) = (favorable outcomes) ÷ (total outcomes)
Complement: P(A') = 1 − P(A) → "NOT A"
Impossible: P = 0 | Certain: P = 1
Always: 0 ≤ P(A) ≤ 1
Complement: P(A') = 1 − P(A) → "NOT A"
Impossible: P = 0 | Certain: P = 1
Always: 0 ≤ P(A) ≤ 1
Q 01
A bag contains 4 red, 3 blue, and 5 green marbles. One marble is picked at random. What is the probability it is not blue?
EASY
Solution: Total = 4+3+5 = 12. Blue = 3, so P(blue) = 3/12 = 1/4.
Using the complement rule: P(not blue) = 1 − 1/4 = 3/4.
⚠️ Trap: C and D look the same — but 9/12 simplifies to 3/4. Both are technically correct but C is fully simplified. The intended answer is 3/4.
Using the complement rule: P(not blue) = 1 − 1/4 = 3/4.
⚠️ Trap: C and D look the same — but 9/12 simplifies to 3/4. Both are technically correct but C is fully simplified. The intended answer is 3/4.
Q 02
A fair six-sided die is rolled. What is the probability of rolling a number greater than 4?
EASY
Solution: Numbers greater than 4 on a die: {5, 6} → 2 outcomes.
P = 2/6 = 1/3.
⚠️ Trap: "Greater than 4" does NOT include 4 itself! Students often count {4,5,6} = 3, giving 1/2. Read carefully: strictly greater.
P = 2/6 = 1/3.
⚠️ Trap: "Greater than 4" does NOT include 4 itself! Students often count {4,5,6} = 3, giving 1/2. Read carefully: strictly greater.
Q 03
The probability that it rains on any day is 0.35. What is the probability that it does not rain on a randomly chosen day?
EASY
Solution: Complement Rule: P(not rain) = 1 − 0.35 = 0.65.
⚠️ C is impossible — probability can never exceed 1!
⚠️ C is impossible — probability can never exceed 1!
02
Sets & Venn Diagrams
⚡ Quick Memory Key
AND = ∩ (overlap) | OR = ∪ (union)
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) ← Addition Rule
Mutually exclusive: A ∩ B = ∅ → P(A ∪ B) = P(A) + P(B)
Only A: in A but NOT in B = A \ B
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
Mutually exclusive: A ∩ B = ∅ → P(A ∪ B) = P(A) + P(B)
Only A: in A but NOT in B = A \ B
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
Q 04
In a class of 30 students: 18 play football, 12 play basketball, and 5 play both. How many students play neither?
MEDIUM
Solution: n(F ∪ B) = 18 + 12 − 5 = 25.
Neither = Total − Union = 30 − 25 = 5.
⚠️ Trap: Forgetting to subtract the overlap! 18+12 = 30, but those 5 who play both are counted twice.
Neither = Total − Union = 30 − 25 = 5.
⚠️ Trap: Forgetting to subtract the overlap! 18+12 = 30, but those 5 who play both are counted twice.
Q 05
P(A) = 0.4, P(B) = 0.5, P(A ∩ B) = 0.2. Find P(A ∪ B).
MEDIUM
Solution: Addition Rule: P(A ∪ B) = 0.4 + 0.5 − 0.2 = 0.7.
⚠️ A (0.9) forgets the subtraction. D (1.1) is impossible as it exceeds 1.
⚠️ A (0.9) forgets the subtraction. D (1.1) is impossible as it exceeds 1.
Q 06
Events A and B are mutually exclusive. P(A) = 0.3 and P(B) = 0.45. What is P(A ∪ B)?
EASY
Solution: Mutually exclusive means P(A ∩ B) = 0, so:
P(A ∪ B) = 0.3 + 0.45 = 0.75.
⚠️ A = 0.135 is the product (independent formula — wrong context!).
P(A ∪ B) = 0.3 + 0.45 = 0.75.
⚠️ A = 0.135 is the product (independent formula — wrong context!).
03
Permutations & Combinations
⚡ Quick Memory Key
ORDER MATTERS → P | ORDER DOESN'T → C
Permutation: ⁿPᵣ = n! ÷ (n−r)! "arrange r from n"
Combination: ⁿCᵣ = n! ÷ [r!(n−r)!] "choose r from n"
Trick: ⁿCᵣ = ⁿC(n−r) ← symmetry
Factorial: 0! = 1, 1! = 1, 5! = 120
Combination: ⁿCᵣ = n! ÷ [r!(n−r)!] "choose r from n"
Trick: ⁿCᵣ = ⁿC(n−r) ← symmetry
Factorial: 0! = 1, 1! = 1, 5! = 120
Permutation
ⁿPᵣ = n! / (n−r)!
Combination
ⁿCᵣ = n! / r!(n−r)!
Factorial
n! = n×(n−1)×…×1
Special
0! = 1 ⁿC₀ = 1
Q 07
How many different ways can 5 students be arranged in a line?
EASY
Solution: All 5 arranged = 5! = 5×4×3×2×1 = 120.
Order matters here (a line has first, second… positions), so use factorial.
⚠️ A = 5² = 25 (wrong — that's not how permutations work).
Order matters here (a line has first, second… positions), so use factorial.
⚠️ A = 5² = 25 (wrong — that's not how permutations work).
Q 08
A team of 3 people is chosen from a group of 8. How many ways can this be done? (Order does NOT matter.)
EASY
Solution: ⁸C₃ = 8! / (3! × 5!) = (8×7×6) / (3×2×1) = 336/6 = 56.
⚠️ A = 336 = ⁸P₃ — that's the permutation (order matters). Since it's a team, order doesn't matter → use combination.
⚠️ A = 336 = ⁸P₃ — that's the permutation (order matters). Since it's a team, order doesn't matter → use combination.
Q 09
The value of ⁷C₅ equals:
MEDIUM
Solution: Use symmetry: ⁷C₅ = ⁷C₂ = (7×6)/(2×1) = 42/2 = 21.
⚠️ ⁷C₅ and ⁷C₂ are equal — this symmetry trick saves time! C = ⁷C₃ = 35 (off by one error in index). D = 7! ÷ 2! (over-calculated).
⚠️ ⁷C₅ and ⁷C₂ are equal — this symmetry trick saves time! C = ⁷C₃ = 35 (off by one error in index). D = 7! ÷ 2! (over-calculated).
Q 10
A president, vice-president, and secretary are chosen from 10 candidates. How many arrangements are possible?
MEDIUM
Solution: Roles are different → order matters → Permutation:
¹⁰P₃ = 10! / 7! = 10×9×8 = 720.
⚠️ A = 10C3 = 120 (wrong — these are distinct roles, so order matters!).
¹⁰P₃ = 10! / 7! = 10×9×8 = 720.
⚠️ A = 10C3 = 120 (wrong — these are distinct roles, so order matters!).
Q 11
A pizza shop offers 10 toppings. A customer picks exactly 4. How many different pizzas are possible?
EASY
Solution: Order of toppings on pizza doesn't matter → combination:
¹⁰C₄ = (10×9×8×7)/(4×3×2×1) = 5040/24 = 210.
⚠️ A = ¹⁰P₄ = 5040 — permutation (wrong, toppings have no order).
¹⁰C₄ = (10×9×8×7)/(4×3×2×1) = 5040/24 = 210.
⚠️ A = ¹⁰P₄ = 5040 — permutation (wrong, toppings have no order).
Q 12
How many 4-digit codes can be made from digits 1–9 if no digit repeats?
HARD
Solution: A code has order (1234 ≠ 4321) → Permutation:
⁹P₄ = 9×8×7×6 = 3024.
⚠️ D = 9⁴ = 6561 assumes repetition is allowed. B = ⁹C₄ = 126 ignores order.
⁹P₄ = 9×8×7×6 = 3024.
⚠️ D = 9⁴ = 6561 assumes repetition is allowed. B = ⁹C₄ = 126 ignores order.
04
Conditional & Independent Probability
⚡ Quick Memory Key
CONDITIONAL = "given that" = P(A|B)
P(A|B) = P(A ∩ B) ÷ P(B) "restrict universe to B"
Independent: P(A|B) = P(A) → B tells nothing about A
Independent test: P(A ∩ B) = P(A) × P(B)
Dependent: multiply conditionally → P(A) × P(B|A)
Independent: P(A|B) = P(A) → B tells nothing about A
Independent test: P(A ∩ B) = P(A) × P(B)
Dependent: multiply conditionally → P(A) × P(B|A)
Q 13
P(A) = 0.5, P(B) = 0.4, P(A ∩ B) = 0.2. Find P(A|B).
MEDIUM
Solution: P(A|B) = P(A ∩ B) / P(B) = 0.2 / 0.4 = 0.5.
Notice: P(A|B) = P(A) = 0.5 → A and B are actually independent!
⚠️ D = 0.5 × 0.4 = 0.2 × 0.4 — students confuse the formula.
Notice: P(A|B) = P(A) = 0.5 → A and B are actually independent!
⚠️ D = 0.5 × 0.4 = 0.2 × 0.4 — students confuse the formula.
Q 14
A box has 6 red and 4 blue balls. Two balls are drawn without replacement. What is the probability both are red?
MEDIUM
Solution: Without replacement → dependent events:
P = 6/10 × 5/9 = 30/90 = 1/3.
⚠️ B = (6/10)² = 36/100 = 9/25 assumes replacement (independent) — wrong! After drawing 1 red, only 5 red remain and total = 9.
P = 6/10 × 5/9 = 30/90 = 1/3.
⚠️ B = (6/10)² = 36/100 = 9/25 assumes replacement (independent) — wrong! After drawing 1 red, only 5 red remain and total = 9.
Q 15
Are events A and B independent? Given: P(A) = 0.6, P(B) = 0.5, P(A ∩ B) = 0.3.
MEDIUM
Solution: Test: P(A)×P(B) = 0.6×0.5 = 0.3 = P(A ∩ B) ✓
Since they match → independent.
⚠️ A is a common misconception: mutually exclusive means P(A∩B)=0, NOT independent. These are different concepts!
Since they match → independent.
⚠️ A is a common misconception: mutually exclusive means P(A∩B)=0, NOT independent. These are different concepts!
Q 16
A student passes Maths with probability 0.7 and Science with probability 0.8. If the events are independent, what is the probability of passing at least one?
HARD
Solution: "At least one" → use complement: P(neither) = P(fail M) × P(fail S)
= 0.3 × 0.2 = 0.06
P(at least one) = 1 − 0.06 = 0.94.
⚠️ A = 0.7×0.8 = 0.56 is P(both pass). B = 0.7+0.8 exceeds 1 — impossible!
= 0.3 × 0.2 = 0.06
P(at least one) = 1 − 0.06 = 0.94.
⚠️ A = 0.7×0.8 = 0.56 is P(both pass). B = 0.7+0.8 exceeds 1 — impossible!
05
Tree Diagrams & Combined Problems
⚡ Quick Memory Key
ALONG branches: MULTIPLY | BETWEEN branches: ADD
Tree diagram: multiply along each path (AND)
Multiple paths: add the path probabilities (OR)
Check: all final branch probs must sum to 1
With/without replacement: changes 2nd branch values!
Multiple paths: add the path probabilities (OR)
Check: all final branch probs must sum to 1
With/without replacement: changes 2nd branch values!
Q 17
A coin is flipped twice. What is the probability of getting exactly one Head?
EASY
Solution: 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 and P(TH) = 1/4 → add = 1/2.
⚠️ B = 3/4 is P(at least one Head) — a different question!
P = 2/4 = 1/2.
Using tree: P(HT) = 1/2×1/2 = 1/4 and P(TH) = 1/4 → add = 1/2.
⚠️ B = 3/4 is P(at least one Head) — a different question!
Q 18
A bag has 3 white (W) and 2 black (B) balls. Two are drawn without replacement. What is P(W then B)?
MEDIUM
Solution: P(W₁) = 3/5. After drawing W, 4 balls remain (2W, 2B).
P(B₂|W₁) = 2/4 = 1/2.
P(W then B) = 3/5 × 1/2 = 3/10.
⚠️ A = (3/5)×(2/5) — uses replacement (wrong denominator 5 instead of 4 on 2nd draw).
P(B₂|W₁) = 2/4 = 1/2.
P(W then B) = 3/5 × 1/2 = 3/10.
⚠️ A = (3/5)×(2/5) — uses replacement (wrong denominator 5 instead of 4 on 2nd draw).
Q 19
From a standard deck of 52 cards, 2 cards are drawn without replacement. What is the probability that both are Aces?
HARD
Solution: P(A₁) = 4/52 = 1/13. After 1 ace drawn: 3 aces, 51 cards.
P(A₂|A₁) = 3/51 = 1/17.
P(both aces) = 1/13 × 1/17 = 1/221.
⚠️ B = (1/13)² = 1/169 assumes replacement. Always reduce denominator by 1 when drawing without replacement.
P(A₂|A₁) = 3/51 = 1/17.
P(both aces) = 1/13 × 1/17 = 1/221.
⚠️ B = (1/13)² = 1/169 assumes replacement. Always reduce denominator by 1 when drawing without replacement.
Q 20
A committee of 4 is chosen from 5 men and 3 women. What is the probability the committee has exactly 2 women?
HARD
Solution:
Total ways = ⁸C₄ = 70.
Favourable (2W from 3, 2M from 5) = ³C₂ × ⁵C₂ = 3 × 10 = 30.
P = 30/70 = 3/7.
⚠️ This combines combinatorics WITH probability — the most tested IB question type! Identify: (choose the right group) / (choose from all).
Total ways = ⁸C₄ = 70.
Favourable (2W from 3, 2M from 5) = ³C₂ × ⁵C₂ = 3 × 10 = 30.
P = 30/70 = 3/7.
⚠️ This combines combinatorics WITH probability — the most tested IB question type! Identify: (choose the right group) / (choose from all).
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