⚡ Quick Memory Points
Set Key Words
∪ = OR = Union
∩ = AND = Intersection
A′ = NOT A = Complement
n(A∪B) = n(A)+n(B)−n(A∩B)
∅ = Empty Set
A⊆B → A is subset of B
In a class of 30 students, 18 play football and 14 play basketball. 8 students play both sports.
How many students play at least one sport?
Formula to remember
$n(F \cup B) = n(F) + n(B) - n(F \cap B)$
Why subtract? Because students in both are counted twice!
Given $U = \{1,2,3,4,5,6,7,8,9,10\}$, $A = \{2,4,6,8\}$, $B = \{1,2,3,4,5\}$.
Find $n(A' \cap B)$ — the number of elements in B but NOT in A.
Step by step
$A' = $ everything NOT in $A$ $= \{1,3,5,7,9,10\}$
$A' \cap B = $ elements in both $A'$ and $B$
In a survey of 50 people: 30 like tea, 25 like coffee, and 10 like neither.
How many people like both tea and coffee?
Trap Alert ⚠️
First find those who like at least one:
$n(T \cup C) = 50 - 10 = 40$
Then use: $n(T \cup C) = n(T) + n(C) - n(T \cap C)$
$40 = 30 + 25 - n(T \cap C)$ → solve for $n(T \cap C)$
⚡ Quick Memory Points
AND vs OR
AND → Multiply (×)
OR → Add (+)
Independent choices → Multiply all together
Mutually exclusive → Add
A restaurant offers 4 starters, 6 main courses, and 3 desserts.
How many different 3-course meals are possible?
Key Idea — AND rule
Each choice is independent → multiply:
$4 \times 6 \times 3 = ?$
How many 3-digit numbers can be formed using digits $\{1, 2, 3, 4, 5\}$ if repetition is NOT allowed?
Think it through
1st digit: 5 choices → 2nd digit: 4 choices (one used) → 3rd digit: 3 choices
Each position has one fewer choice because you can't repeat!
⚡ Quick Memory Points
Permutation = ORDER matters (P = Position)
P = Position = ORDER matters
$^nP_r = \frac{n!}{(n-r)!}$
$n! = n \times (n-1) \times \cdots \times 1$
$0! = 1$ (always!)
ABC ≠ BAC → different arrangements
In how many ways can 5 students be arranged in a straight line?
Formula
All $n$ objects arranged: $n!$ ways
$5! = 5 \times 4 \times 3 \times 2 \times 1$
From 8 runners, in how many ways can gold, silver, and bronze medals be awarded?
Why Permutation?
Gold ≠ Silver ≠ Bronze →
order matters (rank matters)
$^8P_3 = \dfrac{8!}{(8-3)!} = \dfrac{8!}{5!} = 8 \times 7 \times 6$
How many 4-letter arrangements can be made from the word MATH where M must be first?
Strategy — Fix then count
M is fixed in position 1 → arrange remaining 3 letters (A, T, H) in 3 positions:
$1 \times 3! = 1 \times 6$
⚡ Quick Memory Points
Combination = CHOOSE (C = Choose, no order)
C = Choose = ORDER doesn't matter
$\binom{n}{r} = \frac{n!}{r!(n-r)!}$
$\binom{n}{r} = \binom{n}{n-r}$ (symmetry!)
$\binom{n}{0} = \binom{n}{n} = 1$
ABC = BAC = CAB → same group
A committee of 3 people is to be chosen from a group of 7.
How many different committees are possible?
Why Combination?
A committee {Alice, Bob, Carol} = {Carol, Bob, Alice} →
order doesn't matter!
$\binom{7}{3} = \dfrac{7!}{3! \cdot 4!} = \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1}$
From 10 students, a president and vice-president are chosen. How many ways is this possible?
(Compare this to choosing a 2-person committee!)
The Classic P vs C Trap
President + VP → roles are
different → ORDER matters → Use $^{10}P_2$
Committee of 2 → same role → order doesn't matter → Use $\binom{10}{2}$
$^{10}P_2 = 10 \times 9 = ?$ vs $\binom{10}{2} = 45$
A hand of 5 cards is dealt from a standard deck of 52 cards.
How many different 5-card hands are possible?
Key reasoning
A hand of cards has no order → combination
$\binom{52}{5} = \dfrac{52!}{5! \times 47!} = \dfrac{52 \times 51 \times 50 \times 49 \times 48}{5!}$
⚡ Quick Memory Points
Probability Essentials
$P(A) = \frac{\text{favourable}}{\text{total}}$
$P(A') = 1 - P(A)$
$0 \leq P(A) \leq 1$ always
$P(\text{impossible}) = 0$
$P(\text{certain}) = 1$
Complement = easier path!
A fair die is rolled. What is the probability of NOT rolling a 6?
Use the complement — always faster!
$P(\text{not 6}) = 1 - P(6) = 1 - \dfrac{1}{6}$
A bag contains 4 red, 3 blue, and 5 green balls. One ball is drawn at random.
What is the probability it is red or green?
OR rule for mutually exclusive events
Red and Green cannot happen at same time → add
$P(R \cup G) = P(R) + P(G)$ (since $P(R \cap G) = 0$)
⚡ Quick Memory Points
Independence vs Dependence
Independent: $P(A \cap B) = P(A) \cdot P(B)$
Conditional: $P(A|B) = \frac{P(A \cap B)}{P(B)}$
With replacement → Independent
Without replacement → Dependent
A|B = "A given B already happened"
A bag has 5 red and 3 blue balls. Two balls are drawn without replacement.
What is the probability that both are red?
Without replacement = dependent events
$P(\text{both red}) = P(1\text{st red}) \times P(2\text{nd red} \mid 1\text{st red})$
$= \dfrac{5}{8} \times \dfrac{4}{7}$
After drawing 1 red, only 4 red remain out of 7 total!
B
$\dfrac{20}{56} = \dfrac{5}{14}$
Two fair coins are tossed. Given that at least one head appeared, what is the probability that both are heads?
Conditional probability formula
Sample space: $\{HH, HT, TH, TT\}$
Event B = "at least one head" $= \{HH, HT, TH\}$ → $P(B) = \frac{3}{4}$
Event A = "both heads" $= \{HH\}$ → $P(A \cap B) = \frac{1}{4}$
$P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{1/4}{3/4}$
Events $A$ and $B$ are independent. $P(A) = 0.4$ and $P(B) = 0.3$.
Find $P(A \cup B)$.
Don't forget to subtract the overlap!
Independent → $P(A \cap B) = P(A) \cdot P(B) = 0.4 \times 0.3 = 0.12$
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Common mistake: students forget to subtract $P(A \cap B)$!
A committee of 4 is chosen from 6 men and 4 women.
What is the probability the committee has exactly 2 women?
Strategy — Favourable ÷ Total
Total ways to choose 4 from 10: $\binom{10}{4}$
Favourable: choose 2 women from 4 AND 2 men from 6:
$\binom{4}{2} \times \binom{6}{2}$
$P = \dfrac{\binom{4}{2} \times \binom{6}{2}}{\binom{10}{4}}$
How many ways can 6 people sit in a row if two specific friends must sit together?
Glue Trick — treat the pair as one unit!
Step 1: Glue the 2 friends → now you have 5 "units" to arrange: $5!$ ways
Step 2: The 2 friends can swap within their pair: $2!$ ways
Total $= 5! \times 2! = 120 \times 2$
$P(A) = 0.5$, $P(B) = 0.4$, $P(A \cup B) = 0.7$.
Are events $A$ and $B$ independent?
Independence Test
Step 1: Find $P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.5 + 0.4 - 0.7 = 0.2$
Step 2: Check if $P(A) \times P(B) = P(A \cap B)$
$0.5 \times 0.4 = 0.2$ ← does this equal $P(A \cap B)$?
A
Yes, independent because $P(A)\cdot P(B) = P(A\cap B)$
B
No, not independent because $P(A \cup B) \neq 1$
C
No, not independent because $P(A \cap B) = 0$
A password consists of 3 letters followed by 2 digits (0–9).
Letters are from A–Z (26 letters), no repetition allowed for letters; digits can repeat.
How many passwords are possible?
Break it into parts — AND rule
Letters (order matters, no repeat): $^{26}P_3 = 26 \times 25 \times 24$
Digits (order matters, repeat allowed): $10 \times 10$
Total $= ^{26}P_3 \times 10^2$
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