§ 1 Set Theory & Venn Diagrams
⚡ Quick Memory Points
A ∪ B= A OR B (Union — everything in either)A ∩ B= A AND B (Intersection — only the overlap)A'= NOT A (Complement — everything outside A)n(A ∪ B) = n(A) + n(B) − n(A ∩ B)← don't double-count!- Mutually exclusive:
A ∩ B = ∅, soP(A ∪ B) = P(A) + P(B)
Q1
Easy
In a class of 30 students, 18 study French (F), 14 study Spanish (S), and 8 study both. How many students study at least one of the two languages?
❌ Not quite!
Use the inclusion-exclusion formula:
$n(F \cup S) = n(F) + n(S) - n(F \cap S) = 18 + 14 - 8 = \mathbf{24}$
Trap: Simply adding 18 + 14 = 32 double-counts the 8 who study both.
$n(F \cup S) = n(F) + n(S) - n(F \cap S) = 18 + 14 - 8 = \mathbf{24}$
Trap: Simply adding 18 + 14 = 32 double-counts the 8 who study both.
Q2
Easy
If $n(\xi) = 50$, $n(A) = 20$, $n(B) = 25$, and $n(A \cap B) = 10$, find $n(A' \cap B')$ (students in neither A nor B).
❌ Not quite!
Step 1: $n(A \cup B) = 20 + 25 - 10 = 35$
Step 2: $n(A' \cap B') = n(\xi) - n(A \cup B) = 50 - 35 = \mathbf{15}$
Key idea: "Neither A nor B" = complement of $(A \cup B)$.
Step 2: $n(A' \cap B') = n(\xi) - n(A \cup B) = 50 - 35 = \mathbf{15}$
Key idea: "Neither A nor B" = complement of $(A \cup B)$.
Q3
Medium
Sets $A = \{1, 2, 3, 4, 5\}$ and $B = \{3, 4, 5, 6, 7\}$. What is $n(A \cup B) - n(A \cap B)$?
❌ Not quite!
$A \cap B = \{3,4,5\}$ so $n(A \cap B) = 3$
$A \cup B = \{1,2,3,4,5,6,7\}$ so $n(A \cup B) = 7$
$7 - 3 = \mathbf{4}$
$A \cup B = \{1,2,3,4,5,6,7\}$ so $n(A \cup B) = 7$
$7 - 3 = \mathbf{4}$
§ 2 Basic Probability
⚡ Quick Memory Points
P(A) = favourable / total— always between 0 and 1P(A) + P(A') = 1→ soP(A') = 1 − P(A)- Complementary shortcut: P(at least one) = 1 − P(none)
- Independent:
P(A∩B) = P(A) × P(B) - Mutually exclusive:
P(A∩B) = 0, soP(A∪B) = P(A)+P(B)
Q4
Easy
A bag contains 4 red, 6 blue, and 2 green balls. One ball is drawn at random. What is the probability it is NOT blue?
❌ Not quite!
Total balls = 4+6+2 = 12. Not blue = red + green = 4+2 = 6.
$P(\text{not blue}) = \dfrac{6}{12} = \mathbf{\dfrac{1}{2}}$
Shortcut: $P(\text{not blue}) = 1 - P(\text{blue}) = 1 - \dfrac{6}{12} = \dfrac{1}{2}$ ✓
$P(\text{not blue}) = \dfrac{6}{12} = \mathbf{\dfrac{1}{2}}$
Shortcut: $P(\text{not blue}) = 1 - P(\text{blue}) = 1 - \dfrac{6}{12} = \dfrac{1}{2}$ ✓
Q5
Easy
$P(A) = 0.4$ and $P(B) = 0.5$. If A and B are mutually exclusive, find $P(A \cup B)$.
❌ Not quite!
Mutually exclusive means $P(A \cap B) = 0$, so:
$P(A \cup B) = P(A) + P(B) = 0.4 + 0.5 = \mathbf{0.9}$
Trap: 0.20 = P(A)×P(B) — that's for independent events, not mutually exclusive!
$P(A \cup B) = P(A) + P(B) = 0.4 + 0.5 = \mathbf{0.9}$
Trap: 0.20 = P(A)×P(B) — that's for independent events, not mutually exclusive!
Q6
Medium
A fair die is rolled. What is the probability of getting a number that is even OR greater than 4?
Recall: A fair die has faces $\{1, 2, 3, 4, 5, 6\}$.
❌ Not quite!
Even = $\{2,4,6\}$, Greater than 4 = $\{5,6\}$
$A \cup B = \{2,4,5,6\}$ → 4 outcomes
$P = \dfrac{4}{6} = \mathbf{\dfrac{2}{3}}$
Trap: $n(\text{even}) + n(>4) = 3+2=5$, but $\{6\}$ is counted twice! Subtract the overlap $\{6\}$: $3+2-1=4$.
$A \cup B = \{2,4,5,6\}$ → 4 outcomes
$P = \dfrac{4}{6} = \mathbf{\dfrac{2}{3}}$
Trap: $n(\text{even}) + n(>4) = 3+2=5$, but $\{6\}$ is counted twice! Subtract the overlap $\{6\}$: $3+2-1=4$.
§ 3 Conditional Probability
⚡ Quick Memory Points
P(A|B) = P(A∩B) / P(B)— "given B happened, what's P(A)?"- If independent:
P(A|B) = P(A)— knowing B doesn't change A - Tree diagram tip: multiply along branches, add between branches
- Conditional = restrict sample space to the given condition
Q7
Medium
$P(A) = 0.6$, $P(B) = 0.5$, $P(A \cap B) = 0.3$. Find $P(A \mid B)$.
❌ Not quite!
$P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.3}{0.5} = \mathbf{0.6}$
Notice $P(A|B) = P(A) = 0.6$ → this means A and B are actually independent!
Check: $P(A) \times P(B) = 0.6 \times 0.5 = 0.3 = P(A\cap B)$ ✓
Notice $P(A|B) = P(A) = 0.6$ → this means A and B are actually independent!
Check: $P(A) \times P(B) = 0.6 \times 0.5 = 0.3 = P(A\cap B)$ ✓
Q8
Medium
A box has 5 red and 3 blue balls. Two balls are drawn without replacement. Find the probability both are red.
❌ Not quite!
Without replacement — the denominator changes!
$P(\text{1st red}) = \dfrac{5}{8}$, then $P(\text{2nd red} \mid \text{1st red}) = \dfrac{4}{7}$
$P(\text{both red}) = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \mathbf{\dfrac{5}{14}}$
Trap: $\dfrac{25}{64} = \left(\dfrac{5}{8}\right)^2$ — that's WITH replacement!
$P(\text{1st red}) = \dfrac{5}{8}$, then $P(\text{2nd red} \mid \text{1st red}) = \dfrac{4}{7}$
$P(\text{both red}) = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \mathbf{\dfrac{5}{14}}$
Trap: $\dfrac{25}{64} = \left(\dfrac{5}{8}\right)^2$ — that's WITH replacement!
Q9
Hard
From a group of 10 students, 6 passed Math (M) and 7 passed English (E). 4 passed both. A student is selected at random and passed English. What is the probability they also passed Math?
❌ Not quite!
We need $P(M \mid E)$:
$P(M \mid E) = \dfrac{P(M \cap E)}{P(E)} = \dfrac{4/10}{7/10} = \dfrac{4}{7}$
Shortcut: "Restrict to the 7 English passers. Of those, 4 also passed Math."
$P = \dfrac{4}{7}$ ✓
$P(M \mid E) = \dfrac{P(M \cap E)}{P(E)} = \dfrac{4/10}{7/10} = \dfrac{4}{7}$
Shortcut: "Restrict to the 7 English passers. Of those, 4 also passed Math."
$P = \dfrac{4}{7}$ ✓
§ 4 Permutations
⚡ Quick Memory Points
- Order MATTERS → use Permutation
nPr = n! / (n−r)! - All arrangements of n things:
n!(n factorial) - With repeated letters: divide by each repeated letter's factorial
- Memory: Permutation = Position matters (passwords, races, rankings)
0! = 1— memorise this!
Q10
Easy
How many ways can 5 different books be arranged on a shelf?
❌ Not quite!
Order matters (different arrangements of a shelf):
$5! = 5 \times 4 \times 3 \times 2 \times 1 = \mathbf{120}$
Visualise: 5 slots. 1st slot: 5 choices, 2nd: 4, 3rd: 3, 4th: 2, 5th: 1 → multiply.
$5! = 5 \times 4 \times 3 \times 2 \times 1 = \mathbf{120}$
Visualise: 5 slots. 1st slot: 5 choices, 2nd: 4, 3rd: 3, 4th: 2, 5th: 1 → multiply.
Q11
Medium
In how many ways can a President, Vice-President, and Secretary be chosen from a group of 8 people? (One person cannot hold two roles.)
❌ Not quite!
Roles are different → order matters → use $^nP_r$:
$^8P_3 = \dfrac{8!}{(8-3)!} = \dfrac{8!}{5!} = 8 \times 7 \times 6 = \mathbf{336}$
Trap: $\binom{8}{3} = 56$ — that's choosing 3 people with no assigned roles (order doesn't matter).
$^8P_3 = \dfrac{8!}{(8-3)!} = \dfrac{8!}{5!} = 8 \times 7 \times 6 = \mathbf{336}$
Trap: $\binom{8}{3} = 56$ — that's choosing 3 people with no assigned roles (order doesn't matter).
Q12
Hard
How many distinct arrangements are there of the letters in the word BANANA?
❌ Not quite!
BANANA has 6 letters: B×1, A×3, N×2
Divide by repeated letters' factorials:
$\dfrac{6!}{3! \times 2!} = \dfrac{720}{6 \times 2} = \dfrac{720}{12} = \mathbf{60}$
Rule: When letters repeat, identical swaps don't make new arrangements.
Divide by repeated letters' factorials:
$\dfrac{6!}{3! \times 2!} = \dfrac{720}{6 \times 2} = \dfrac{720}{12} = \mathbf{60}$
Rule: When letters repeat, identical swaps don't make new arrangements.
§ 5 Combinations
⚡ Quick Memory Points
- Order does NOT matter → use Combination
nCr = n! / (r!(n−r)!) - Memory: Combination = Choosing a team/committee (no ranks)
nCr = nC(n−r)— symmetry shortcut: $\binom{10}{8} = \binom{10}{2} = 45$nC0 = nCn = 1,nC1 = n- When to use nPr vs nCr: ask "does the order change the meaning?"
Q13
Easy
A committee of 3 students is to be chosen from a group of 7. In how many ways can this be done?
❌ Not quite!
A committee has no ranked roles → order doesn't matter → $\binom{7}{3}$:
$\binom{7}{3} = \dfrac{7!}{3! \cdot 4!} = \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} = \dfrac{210}{6} = \mathbf{35}$
Trap: $^7P_3 = 210$ — that counts order (for ranked roles, not a plain committee).
$\binom{7}{3} = \dfrac{7!}{3! \cdot 4!} = \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} = \dfrac{210}{6} = \mathbf{35}$
Trap: $^7P_3 = 210$ — that counts order (for ranked roles, not a plain committee).
Q14
Medium
A team of 4 is chosen from 5 boys and 4 girls. How many teams have exactly 2 boys and 2 girls?
❌ Not quite!
Choose 2 boys from 5 AND 2 girls from 4 — multiply the choices:
$\binom{5}{2} \times \binom{4}{2} = 10 \times 6 = \mathbf{60}$
Rule: When you need this group AND that group → multiply (AND = ×).
$\binom{5}{2} \times \binom{4}{2} = 10 \times 6 = \mathbf{60}$
Rule: When you need this group AND that group → multiply (AND = ×).
Q15
Hard
How many ways can you choose at least 2 items from a set of 4 distinct items?
❌ Not quite!
"At least 2" = choose 2, 3, or 4:
$\binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 6 + 4 + 1 = \mathbf{11}$
Alternative: Total subsets = $2^4 = 16$. Subtract choose-0 and choose-1: $16 - 1 - 4 = 11$ ✓
$\binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 6 + 4 + 1 = \mathbf{11}$
Alternative: Total subsets = $2^4 = 16$. Subtract choose-0 and choose-1: $16 - 1 - 4 = 11$ ✓
§ 6 Combined & Tricky Problems
⚡ Common Traps & Fixes
- "At least one" → use complement:
1 − P(none)(much faster!) - With vs without replacement: denominator changes each draw
- AND → multiply probabilities; OR → add (and subtract overlap)
- Independent ≠ mutually exclusive (they're opposite concepts!)
- Tree diagram: always check branches sum to 1 at each node
Q16
Medium
A coin is flipped 3 times. What is the probability of getting at least one head?
❌ Not quite!
Use the complement shortcut:
$P(\text{at least one H}) = 1 - P(\text{no heads}) = 1 - \left(\dfrac{1}{2}\right)^3 = 1 - \dfrac{1}{8} = \mathbf{\dfrac{7}{8}}$
Why faster? There's only 1 way to get all tails (TTT), vs 7 ways to get at least one head.
$P(\text{at least one H}) = 1 - P(\text{no heads}) = 1 - \left(\dfrac{1}{2}\right)^3 = 1 - \dfrac{1}{8} = \mathbf{\dfrac{7}{8}}$
Why faster? There's only 1 way to get all tails (TTT), vs 7 ways to get at least one head.
Q17
Medium
Two independent events: $P(A) = \dfrac{1}{3}$ and $P(B) = \dfrac{1}{4}$. Find $P(A \cup B)$.
❌ Not quite!
Independent → $P(A \cap B) = P(A) \cdot P(B) = \dfrac{1}{3} \times \dfrac{1}{4} = \dfrac{1}{12}$
Then: $P(A \cup B) = \dfrac{1}{3} + \dfrac{1}{4} - \dfrac{1}{12} = \dfrac{4}{12} + \dfrac{3}{12} - \dfrac{1}{12} = \mathbf{\dfrac{6}{12} = \dfrac{1}{2}}$
Trap: $\dfrac{7}{12}$ = forgetting to subtract the intersection!
Then: $P(A \cup B) = \dfrac{1}{3} + \dfrac{1}{4} - \dfrac{1}{12} = \dfrac{4}{12} + \dfrac{3}{12} - \dfrac{1}{12} = \mathbf{\dfrac{6}{12} = \dfrac{1}{2}}$
Trap: $\dfrac{7}{12}$ = forgetting to subtract the intersection!
Q18
Hard
From a deck of 52 cards, two cards are drawn without replacement. What is the probability that both are aces?
A standard deck has 4 aces out of 52 cards.
❌ Not quite!
Without replacement:
$P = \dfrac{4}{52} \times \dfrac{3}{51} = \dfrac{12}{2652} = \mathbf{\dfrac{1}{221}}$
Trap: $\dfrac{1}{169} = \dfrac{1}{13} \times \dfrac{1}{13}$ — that's WITH replacement (independent draws).
$P = \dfrac{4}{52} \times \dfrac{3}{51} = \dfrac{12}{2652} = \mathbf{\dfrac{1}{221}}$
Trap: $\dfrac{1}{169} = \dfrac{1}{13} \times \dfrac{1}{13}$ — that's WITH replacement (independent draws).
Q19
Hard
A student takes a multiple-choice test with 5 questions, each having 4 options (exactly one correct). If the student guesses randomly, what is the probability of getting exactly 2 correct?
Use the binomial formula: $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$
❌ Not quite!
$n=5$, $k=2$, $p=\dfrac{1}{4}$, $1-p=\dfrac{3}{4}$
$P(X=2) = \binom{5}{2}\left(\dfrac{1}{4}\right)^2\left(\dfrac{3}{4}\right)^3 = 10 \times \dfrac{1}{16} \times \dfrac{27}{64}$
$= \dfrac{270}{1024} = \mathbf{\dfrac{135}{512}}$
Steps: 1) Pick which 2 correct ($\binom{5}{2}=10$), 2) Multiply probabilities.
$P(X=2) = \binom{5}{2}\left(\dfrac{1}{4}\right)^2\left(\dfrac{3}{4}\right)^3 = 10 \times \dfrac{1}{16} \times \dfrac{27}{64}$
$= \dfrac{270}{1024} = \mathbf{\dfrac{135}{512}}$
Steps: 1) Pick which 2 correct ($\binom{5}{2}=10$), 2) Multiply probabilities.
Q20
Hard
How many different 4-digit passwords can be formed using digits $\{1, 2, 3, 4, 5, 6\}$ if no digit may repeat and the password must be even?
❌ Not quite!
Even password → last digit must be even: $\{2, 4, 6\}$ → 3 choices.
Remaining 3 positions filled from 5 remaining digits:
$3 \times 5 \times 4 \times 3 = 3 \times 60 = \mathbf{180}$
Strategy: Always fill the restricted position first, then fill the rest.
Remaining 3 positions filled from 5 remaining digits:
$3 \times 5 \times 4 \times 3 = 3 \times 60 = \mathbf{180}$
Strategy: Always fill the restricted position first, then fill the rest.
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