A bag contains 4 red, 3 blue, and 5 green marbles.
One marble is drawn at random. What is the probability of drawing a blue marble?
✦ Solution
Total marbles = 4 + 3 + 5 = 12. Blue marbles = 3.
$$P(\text{blue}) = \frac{3}{12} = \frac{1}{4}$$
Key trap: Don't add only two colours as the total — always sum ALL categories.
Question 02Easy
Events $A$ and $B$ are such that $P(A) = 0.5$, $P(B) = 0.4$, and $P(A \cap B) = 0.2$.
Find $P(A \cup B)$.
📖 Recall
The Addition Rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
You subtract the intersection because it was counted twice — once in $P(A)$ and once in $P(B)$.
✦ Solution
$$P(A \cup B) = 0.5 + 0.4 - 0.2 = 0.7$$
Trap: Option A (0.9) is the answer if you forget to subtract the overlap. Always subtract $P(A \cap B)$.
Question 03Tricky ⚠
The probability that it rains on a given day is $0.3$. If it rains, the probability that Alice is late is $0.6$.
If it does NOT rain, the probability she is late is $0.1$.
What is the probability that Alice is late on a randomly chosen day?
📖 Tree Diagram Strategy
Rain → Late: $0.3 \times 0.6 = 0.18$
No Rain → Late: $0.7 \times 0.1 = 0.07$ Total = sum of all branches that lead to "Late"
✦ Solution — Total Probability
$$P(\text{Late}) = P(\text{Rain}) \cdot P(\text{Late}|\text{Rain}) + P(\text{No Rain}) \cdot P(\text{Late}|\text{No Rain})$$
$$= 0.3 \times 0.6 + 0.7 \times 0.1 = 0.18 + 0.07 = 0.25$$
This is the Law of Total Probability. Always identify your two (or more) exhaustive scenarios first.
Question 04Easy
A fair six-sided die is rolled twice. What is the probability of getting a 6 on both rolls?
✦ Solution
The two rolls are independent, so multiply:
$$P(6 \text{ and } 6) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$$
Trap: Adding $\frac{1}{6} + \frac{1}{6} = \frac{2}{6}$ is wrong — that would be for mutually exclusive events, not AND events.
Question 05Medium
In a class of 30 students, 18 study French, 14 study Spanish, and 8 study both.
A student is chosen at random. What is the probability they study neither?
✦ Solution — Venn Diagram
Students studying at least one = $|F \cup S| = 18 + 14 - 8 = 24$
Students studying neither = $30 - 24 = 6$
$$P(\text{neither}) = \frac{6}{30} = \frac{1}{5}$$
Step: Always find |A∪B| first, then subtract from total for "neither."
Question 06Tricky ⚠
Given $P(A) = 0.4$, $P(B) = 0.5$, and $A$ and $B$ are independent.
Find $P(A' \cap B)$ — the probability that $A$ does not occur but $B$ does.
✦ Solution
Since $A$ and $B$ are independent, $A'$ and $B$ are also independent.
$$P(A') = 1 - 0.4 = 0.6$$
$$P(A' \cap B) = P(A') \times P(B) = 0.6 \times 0.5 = 0.3$$
Key insight: Independence is preserved for complements too!
Question 07Medium
A card is drawn from a standard 52-card deck. Find $P(\text{King} \mid \text{Face card})$.
(Face cards = Jack, Queen, King — 4 of each suit.)
✦ Solution — Conditional Probability
Given it's a face card, the sample space shrinks to 12 face cards (4J + 4Q + 4K).
Kings among face cards = 4.
$$P(\text{King} \mid \text{Face}) = \frac{4}{12} = \frac{1}{3}$$
Trap: Option A uses the whole deck (52) — wrong! Conditional probability restricts the sample space to B.
Question 08Tricky ⚠
A bag has 5 red and 3 blue balls. Two balls are drawn without replacement.
What is the probability that both are red?
✦ Solution — Without Replacement
$$P(\text{1st red}) = \frac{5}{8}$$
After removing one red: 4 red, 7 total remain.
$$P(\text{2nd red} \mid \text{1st red}) = \frac{4}{7}$$
$$P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$$
Trap: Option B ($\frac{25}{64}$) is what you'd get with replacement. Without replacement, denominators change!
Question 09Medium
Are events $A$ and $B$ independent if $P(A) = 0.3$, $P(B) = 0.4$, and $P(A \cap B) = 0.12$?
What does this tell you?
✦ Solution
Test: $P(A) \times P(B) = 0.3 \times 0.4 = 0.12 = P(A \cap B)$ ✓ Definition of independence: $P(A \cap B) = P(A) \cdot P(B)$. Common mistake: Many students think "independent" means $P(A \cap B) = 0$. That's mutually exclusive, which is different!
Question 10Tricky ⚠
A test for a disease is 95% accurate. 2% of the population has the disease.
You test positive. What is the (approximate) probability you actually have the disease?
(Use Bayes' idea: compare "true positive" vs "false positive".)
📖 Imagine 10,000 people
Disease: 200 people → Test +: ~190 (true positive)
No disease: 9,800 people → Test + (false positive): ~490
Total positives ≈ 680
✦ Solution — Bayes' Theorem Intuition
$$P(\text{Disease} \mid +) = \frac{190}{190 + 490} \approx \frac{190}{680} \approx 28\%$$
This is the famous "base rate fallacy"! Even a 95%-accurate test gives many false positives when the disease is rare. The low prevalence (2%) overwhelms the accuracy. Always consider the base rate!
A
Algebra & Expressions
Unit 2 · Questions 11–14 — Expanding, Factoring, Quadratics
⚡ Quick Memory Points
Difference of squares: $a^2-b^2=(a+b)(a-b)$Perfect square: $(a+b)^2=a^2+2ab+b^2$Quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$Discriminant: $\Delta = b^2-4ac$$\Delta>0$: 2 roots · $\Delta=0$: 1 root · $\Delta<0$: no real roots
Question 11Easy
Expand and simplify: $(x + 3)(x - 5)$
✦ Solution — FOIL
$$(x+3)(x-5) = x^2 - 5x + 3x - 15 = x^2 - 2x - 15$$
Trap: Middle term: $-5x + 3x = -2x$, not $+2x$. Sign errors in FOIL are the #1 mistake!
✦ Solution
$$\cos C = \frac{7^2 + 9^2 - 5^2}{2(7)(9)} = \frac{49 + 81 - 25}{126} = \frac{105}{126} = \frac{5}{6}$$
$$C = \cos^{-1}\!\left(\frac{5}{6}\right) \approx 33.6° \approx 33°$$
Trap: Make sure to use the sides adjacent to the angle ($a$ and $b$) in the denominator, and the opposite side ($c$) is subtracted in the numerator.
S
Statistics & Data
Unit 4 · Questions 17–18 — Mean, Median, Standard Deviation
⚡ Quick Memory Points
Mean = sum ÷ countMedian = middle value (sort first!)Mode = most frequentRange = max − minOutlier pulls mean, NOT median → median more robustIQR = Q3 − Q1 (resistant to outliers)
Question 17Medium
Data set: $\{3, 7, 7, 9, 10, 12, 100\}$.
Which measure of central tendency is least affected by the outlier (100)?
✦ Solution
Mean = $(3+7+7+9+10+12+100)/7 = 148/7 \approx 21.1$ — pulled up hugely by 100.
Median = middle value = 9 (sorted: 3,7,7,9,10,12,100) — barely changes!
Mode = 7, Range = 97 (also very sensitive). Rule: Outliers → use Median & IQR. No outliers → Mean & Std Dev are fine.
Question 18Tricky ⚠
A dataset has mean $\bar{x} = 50$ and standard deviation $\sigma = 10$.
Each value is multiplied by 2, then 5 is added.
What are the new mean and standard deviation?
✦ Solution — Linear Transformation
New transformation: $y = 2x + 5$
New mean: $\bar{y} = 2(50) + 5 = 105$
New SD: $\sigma_y = |2| \times \sigma_x = 2 \times 10 = 20$ Golden rule: Adding a constant shifts the mean but does NOT change spread. Multiplying changes BOTH mean AND spread.
F
Functions & Graphs
Unit 5 · Questions 19–20 — Domain, Range, Linear & Quadratic Graphs
⚡ Quick Memory Points
Domain = allowed x values · Range = output y values$f(x) = ax^2+bx+c$: vertex at $x = -\frac{b}{2a}$a > 0: opens up (min) · a < 0: opens down (max)y-intercept: set $x=0$ · x-intercept: set $y=0$Vertical line test: one output per x = function
Question 19Medium
For $f(x) = \sqrt{2x - 6}$, what is the domain?
✦ Solution — Square Root Domain
For a square root to be defined (real), the inside must be $\geq 0$:
$$2x - 6 \geq 0 \implies x \geq 3$$
Domain: $x \in [3, \infty)$ Trap: Option B says $x > 3$ (strict inequality) — wrong! At $x = 3$, $f(3) = \sqrt{0} = 0$, which is perfectly valid.
Question 20Tricky ⚠
The quadratic $f(x) = -2x^2 + 8x - 3$ has a maximum value. What is it?
📖 Vertex of a Parabola
The maximum/minimum occurs at $x = -\dfrac{b}{2a}$, then substitute back.
✦ Solution
Since $a = -2 < 0$, parabola opens downward → has a maximum.
Vertex x-coordinate: $x = -\dfrac{8}{2(-2)} = -\dfrac{8}{-4} = 2$
Maximum value: $f(2) = -2(4) + 8(2) - 3 = -8 + 16 - 3 = \mathbf{5}$ Trap: Don't forget to substitute back! Many students find $x = 2$ but forget to compute $f(2)$.