Mathematics · Module 4

12 lessons · 48 problems

Probability & Statistics — full chapter

Learn each idea, drill it with Basic → Intermediate → Advanced problems, and finish with mixed practice. Every problem shows its thinking right after you answer.

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key formulas

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Lessons Formulas Traps Practice 48 Previous exams Checklist

After this chapter you can

Compute classical probabilities from a correctly counted sample space.
Choose permutations vs combinations by asking whether order matters.
Use complements for at-least-one phrasing (E1-48 zodiac pattern).
Apply the addition rule, subtracting overlaps except when mutually exclusive.
Condition with $P(A|B) = \frac{P(A \cap B)}{P(B)}$ after shrinking the space.
Test independence with the product rule; never multiply dependent draws.
Solve urn-count systems like E3-48 and weight cases with total probability.
Pick mean/median/mode to fit the data, ordering values for any median.
Report variance vs standard deviation with correct units.
Read frequency tables and charts literally; flag missing data (E2-47) instead of inventing it.

How to study this chapter

Learn lessons 1–4: sample spaces, counting, complements, addition.
Drill Basic practice, then read every trap you hit.
Learn lessons 5–7: conditioning, independence, urn systems.
Drill Intermediate practice, including the E1-48 and E3-48 patterns.
Learn lessons 8–12: centers, spread, tables, charts, mixed tactics.
Finish Advanced practice, then answer the 3 previous exam questions.

Lessons

Lesson 1 of 12

Sample Spaces and Classical Probability

Learn the idea

List all equally-likely outcomes first: that list is the sample space

\Omega

. Then

P(A) = \frac{|A|}{|\Omega|}

Classical probability needs EQUAL likelihood. A fair die has 6; two dice have

36

pairs — count, never guess.

Every probability lives in

[0, 1]

P(\varnothing) = 0

P(\Omega) = 1

. Outside that range, you slipped.

Key rules to memorize

$P(A) = \frac{|A|}{|\Omega|}$ over equally-likely outcomes
Die: $|\Omega| = 6$ ; two dice: $|\Omega| = 36$ ordered pairs
$0 \le P(A) \le 1$ ; $P(\varnothing) = 0$ , $P(\Omega) = 1$
Partition check: probabilities over all cases sum to $1$
Favourable means counted inside $\Omega$ , not wished for

Worked example

One fair die: $P(\text{even}) =$ ?

$\Omega = \{1,2,3,4,5,6\}$ , favourable $\{2,4,6\}$ .
$P = \frac{3}{6}$ .

$\frac{1}{2}$

Classic trap

Dividing by the wrong total, e.g.

P(\text{even}) = \frac{3}{5}

on a die: the total is

6

, so

\frac{3}{6} = \frac{1}{2}

Lesson 2 of 12

Permutations vs Combinations

Learn the idea

Ask one question before any factorial: does ORDER matter? Lineups and codes use permutations

A(n,k)

; teams and subsets use combinations

C(n,k)

Order-irrelevant selection overcounts by

k!

: dividing it out turns

A(n,k)

into

C(n,k)

Mirror shortcut:

C(n,k) = C(n,n-k)

. Choosing 2 to keep equals choosing 3 to leave.

Key rules to memorize

Order matters $\Rightarrow A(n,k) = n!/(n-k)!$
Order irrelevant $\Rightarrow C(n,k) = n!/(k!(n-k)!)$
$C(n,k) = C(n,n-k)$ ; $C(n,0) = C(n,n) = 1$
Team/committee wording $\Rightarrow$ combinations (never permute people into seats)
Arrangement/ranking/password wording $\Rightarrow$ permutations

Worked example

From 5 people choose a 2-person team. How many?

Order irrelevant: $C(5,2) = 5!/(2! \cdot 3!)$ .
Compute $\frac{20}{2}$ .

$10$

Classic trap

Permuting a team:

C(5,2) = 10

, not

A(5,2) = 20

. Order never matters among teammates.

Lesson 3 of 12

Complements and At-Least-One

Learn the idea

P(A^{c}) = 1 - P(A)

. When the direct count explodes, count the opposite and subtract from

1

The phrase AT LEAST ONE is the complement signal:

P(\text{at least one}) = 1 - P(\text{none})

Exam proof (E1-48): three zodiac animals — all-different is easy, so

1 - 12 \cdot 11 \cdot \frac{10}{12^{3}} = \frac{17}{72}

Key rules to memorize

$P(A^{c}) = 1 - P(A)$
$P(\text{at least one}) = 1 - P(\text{none})$
E1-48: $P(\text{all differ}) = 12 \cdot 11 \cdot \frac{10}{12^{3}} = \frac{55}{72}$
E1-48 answer: $1 - \frac{55}{72} = \frac{17}{72}$ (option A)
Use complements before inclusion-exclusion whenever you see $\ge 1$

Worked example

Zodiac: $P(\text{at least two of three share an animal}) =$ ?

$P(\text{all differ}) = 12 \cdot 11 \cdot \frac{10}{12^{3}} = \frac{55}{72}$ .
Complement: $1 - \frac{55}{72}$ .

$\frac{17}{72}$

Classic trap

Computing at-least-one directly case by case (1 match + 2 matches + 3 matches). One subtraction replaces all three cases.

Lesson 4 of 12

Addition Rule and Mutual Exclusivity

Learn the idea

General rule:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

. The overlap was counted twice — subtract it once.

Mutually exclusive means

P(A \cap B) = 0

: then simply add. Die faces are the classic case.

Overlap traps (heart-king, red-face) are where the

- P(A \cap B)

term earns its mark.

Key rules to memorize

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Mutually exclusive: $P(A \cap B) = 0 \Rightarrow$ add directly
Die: $P(1 \text{ or } 6) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}$
Cards: $P(\text{heart or king}) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$
Never add overlapping events without subtracting the intersection

Worked example

$P(A) = 0.4$ , $P(B) = 0.5$ , $P(A \cap B) = 0.2$ : $P(A \cup B) =$ ?

Add then subtract overlap: $0.4 + 0.5 - 0.2$ .
Simplify.

$0.7$

Classic trap

Adding

\frac{13}{52} + \frac{4}{52} = \frac{17}{52}

for heart-or-king: the king of hearts sits in both — subtract it once for

\frac{16}{52} = \frac{4}{13}

Lesson 5 of 12

Conditional Probability

Learn the idea

P(A|B) = \frac{P(A \cap B)}{P(B)}

: restrict the world to

B

, then measure

A

inside it.

Shrink the sample space first. A die given EVEN leaves

\{2,4,6\}

— 3 outcomes, not 6.

Direction matters:

P(A|B) \ne P(B|A)

. Divide by the probability of the GIVEN condition.

Key rules to memorize

$P(A|B) = \frac{P(A \cap B)}{P(B)}$
Restrict $\Omega$ to $B$ first, count $A$ inside
Die given even, $P(> 3) = \frac{2}{3}$ ( $\{4,6\}$ of $\{2,4,6\}$ )
$P(A|B)$ divides by $P(B)$ ; $P(B|A)$ divides by $P(A)$
Without replacement, the condition is the shrunken bag

Worked example

Fair die, given the result is even: $P(> 3) =$ ?

Restricted space: $\{2,4,6\}$ .
Favourable inside it: $\{4,6\}$ , so $\frac{2}{3}$ .

$\frac{2}{3}$

Classic trap

Keeping 6 outcomes after being told EVEN: the space shrank to 3 —

P(>3|\text{even}) = \frac{2}{3}

, not

\frac{2}{6}

Lesson 6 of 12

Independence and the Product Rule

Learn the idea

Test, never assume: independent

\iff P(A \cap B) = P(A) \cdot P(B)

. Seeming unrelated is not a proof.

Independent events multiply across trials: two coins give

P(HH) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

Mutually exclusive (with positive probability) means DEPENDENT: if

A

happened,

B

cannot.

Key rules to memorize

Independent $\iff P(A \cap B) = P(A) \cdot P(B)$
Across independent trials, multiply: $P(6,6) = \frac{1}{36}$
Never multiply for dependent draws (no replacement) — condition instead
Mutually exclusive + $P > 0$ $\Rightarrow$ dependent, not independent
$P(A) = 0.3$ , $P(B) = 0.4$ independent $\Rightarrow P(A \cap B) = 0.12$

Worked example

$P(A) = 0.3$ , $P(B) = 0.4$ , independent: $P(A \cap B) =$ ?

Independence licenses the product.
Compute $0.3 \times 0.4$ .

$0.12$

Classic trap

Multiplying dependent draws: bag

3R2B

P(RR)

without replacement is

\frac{3}{5} \times \frac{2}{4} = \frac{3}{10}

, not

\frac{9}{25}

Lesson 7 of 12

Urn Systems and Total Probability

Learn the idea

Exam proof (E3-48): 12 balls,

P(R) = \frac{1}{3}

R = 4

;

Y+G = 5

B+Y = 5

, total

12

forces

G = 3

Y = 2

B = 3

Turn each probability into a COUNT first (multiply by the total), then solve the little system.

Split-sample extension:

P(A) = \sum P(A|B_{i})P(B_{i})

— weight each case by how likely the case is.

Key rules to memorize

Counts first: $P \times$ total $=$ count
E3-48: $R = 4$ , $G = 3$ , $Y = 2$ , $B = 3$
E3-48 answers: $P(Y) = \frac{1}{6}$ , $P(B) = \frac{1}{4}$ , $P(G) = \frac{1}{4}$ (option A)
Total probability: $P(A) = \sum P(A|B_{i})P(B_{i})$
Bayes flips the condition: $P(B|A) = \frac{P(A|B)P(B)}{P(A)}$

Worked example

12 balls: $P(R) = \frac{1}{3}$ , $P(Y \text{ or } G) = \frac{5}{12}$ , $P(B \text{ or } Y) = \frac{5}{12}$ : find $P(Y)$ .

Counts: $R = 4$ ; $Y+G = 5$ ; $B+Y = 5$ ; total $12$ gives $G = 3$ , $Y = 2$ , $B = 3$ .
So $P(Y) = \frac{2}{12}$ .

$\frac{1}{6}$

Classic trap

Reading

P(\text{yellow or green}) = \frac{5}{12}

P(Y) = \frac{5}{12}

: it is a SUM — split it with the total-count system.

Lesson 8 of 12

Mean, Median and Mode

Learn the idea

Mean uses EVERY value (

\bar{x} = \frac{\sum x}{n}

) — efficient but dragged by outliers. Median uses ORDER only — robust.

Sort first, always: odd

n

takes the middle value; even

n

averages the middle two.

Mode is the most frequent value. A dataset can have no mode, one mode, or several.

Key rules to memorize

Mean $\bar{x} = \frac{\sum x}{n}$
Median (odd $n$ ): the $\frac{n+1}{2}$ -th ordered value
Median (even $n$ ): average of the two middle ordered values
Mode: most frequent (may not exist, or tie)
Example: $2,4,4,9 \Rightarrow$ mean $4.75$ , median $4$ , mode $4$

Worked example

Mean and median of $2, 4, 4, 9$ .

Mean: $\frac{2+4+4+9}{4} = \frac{19}{4} = 4.75$ .
Ordered already; median $= \frac{4+4}{2}$ .

Mean $4.75$ ; median $4$

Classic trap

Median of unsorted data, or averaging ALL values for even

n

: only the middle two get averaged.

Lesson 9 of 12

Variance and Standard Deviation

Learn the idea

Variance averages SQUARED deviations:

\sigma^{2} = \frac{\sum(x-\bar{x})^{2}}{n}

. Squaring punishes far values.

Standard deviation

\sigma = \sqrt{\text{variance}}

restores the data's own units — report

\sigma

when units are asked.

\sigma = 0

exactly when all values are identical; otherwise bigger

\sigma

means wider spread.

Key rules to memorize

$\sigma^{2} = \frac{\sum(x-\bar{x})^{2}}{n}$
$\sigma = \sqrt{\sigma^{2}}$ (same units as the data; variance is squared units)
$\sigma = 0 \iff$ all values identical
Example: $1,2,3 \Rightarrow$ mean $2$ , variance $\frac{2}{3}$ , $\sigma = \sqrt{2/3}$
Range $= \max - \min$ is the quick spread check

Worked example

Data $1, 2, 3$ : population variance?

Mean $= 2$ ; squared deviations $1, 0, 1$ .
Average: $\frac{1+0+1}{3}$ .

$\frac{2}{3}$

Classic trap

Giving variance with data units: variance is SQUARED units — take the root for

\sigma

before attaching

cm

kg

, etc.

Lesson 10 of 12

Frequency Tables and Grouped Means

Learn the idea

Grouped mean weights by frequency:

\bar{x} = \frac{\sum f_{i}x_{i}}{\sum f_{i}}

. Divide by total FREQUENCY, not by group count.

Expand the table on scrap:

0,0,1,1,1,2,\ldots

— then median and mode are read-offs.

Range from a table is still max-value minus min-value, frequencies aside.

Key rules to memorize

$\bar{x} = \frac{\sum f_{i}x_{i}}{\sum f_{i}}$
Denominator is total frequency $\sum f_{i}$
Example: $0(2), 1(3), 2(5) \Rightarrow \bar{x} = \frac{13}{10} = 1.3$
Median position: odd $n \to \frac{n+1}{2}$ -th; even $n \to$ middle-two average
Mode of a table: the value with the largest $f$

Worked example

Scores $0$ (freq 2), $1$ (freq 3), $2$ (freq 5): mean?

Total $= 0 \cdot 2 + 1 \cdot 3 + 2 \cdot 5 = 13$ ; students $= 10$ .
Divide: $\frac{13}{10}$ .

$1.3$

Classic trap

Dividing by 3 groups instead of 10 students:

\frac{13}{3}

is wrong —

\frac{13}{10} = 1.3

Lesson 11 of 12

Reading Charts and the E2-47 Warning

Learn the idea

Read literally first: are entries COUNTS or PERCENTAGES? Check axes, units, and the total before any computation.

Pie: sector

\times

total; bars: read heights, then sum for shares. Convert only at the end.

Exam warning (E2-47): the running-apps survey gave NO preference numbers, so no probability exists — mark unsolvable, never invent data.

Key rules to memorize

Counts vs percentages: confirm before computing
Pie $25\%$ of $1200 = 300$
Bars $5, 8, 12 \Rightarrow$ total $25$ , shares $\frac{5}{25}, \frac{8}{25}, \frac{12}{25}$
E2-47: no distribution stated $\Rightarrow$ answer $-1$ (disputed source)
Inventing numbers is always wrong — flag missing data instead

Worked example

Survey names no app-preference counts. $P(\text{two picked differ}) =$ ?

No distribution $\Rightarrow$ no sample space.
Cannot be computed from the stem.

Unsolvable as written (source issue)

Classic trap

Computing

\frac{23}{80}

\frac{57}{80}

for E2-47 from thin air: with no distribution given, the only honest answer is unsolvable.

Lesson 12 of 12

Mixed Exam Tactics and Bounds Checks

Learn the idea

Close every question with a bounds check: probabilities in

[0,1]

, case-probabilities summing to

1

\sigma \ge 0

Skewed data with outliers wants the MEDIAN as center; symmetric clean data is fine with the mean.

Two dice sums to memorize:

P(\text{sum } 7) = \frac{6}{36} = \frac{1}{6}

P(\text{sum } 9) = \frac{4}{36} = \frac{1}{9}

Key rules to memorize

Bounds: $0 \le P \le 1$ ; partition sums to $1$
E3-48 check: $\frac{1}{3}+\frac{1}{6}+\frac{1}{4}+\frac{1}{4} = 1$ $\checkmark$
Outlier present $\Rightarrow$ median; symmetric $\Rightarrow$ mean is fine
$P(\text{sum } 7) = \frac{1}{6}$ ; $P(\text{sum } 9) = \frac{1}{9}$
Syllabus time targets: probability $75s$ , statistics $70s$ per question

Worked example

Two dice: $P(\text{sum} = 9) =$ ?

Pairs: $(3,6),(4,5),(5,4),(6,3)$ — 4 of 36.
Simplify $\frac{4}{36}$ .

$\frac{1}{9}$

Classic trap

Reporting

1.2

-0.3

as a probability and moving on: bounds-check every final number.

Master formula sheet

Classical

$P(A) = \frac{|A|}{|\Omega|}$ (equally-likely $\Omega$ )
$0 \le P(A) \le 1$ ; $P(\varnothing) = 0$ , $P(\Omega) = 1$
$P(A^{c}) = 1 - P(A)$ ; $P(\text{at least one}) = 1 - P(\text{none})$

Counting

$A(n,k) = n!/(n-k)!$ — order matters
$C(n,k) = n!/(k!(n-k)!)$ — order irrelevant
$C(n,k) = C(n,n-k)$ ; $C(n,0) = C(n,n) = 1$

Combine & condition

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Mutually exclusive: $P(A \cap B) = 0 \Rightarrow P(A \cup B) = P(A) + P(B)$
$P(A|B) = \frac{P(A \cap B)}{P(B)}$
Independent: $P(A \cap B) = P(A) \cdot P(B)$
Total: $P(A) = \sum P(A|B_{i})P(B_{i})$ ; Bayes: $P(B|A) = \frac{P(A|B)P(B)}{P(A)}$

Statistics

Mean $\bar{x} = \frac{\sum x}{n}$ ; grouped $\bar{x} = \frac{\sum f_{i}x_{i}}{\sum f_{i}}$
Median: odd $n \to \frac{n+1}{2}$ -th ordered; even $n \to$ middle-two average
Mode: most frequent value
Variance $\sigma^{2} = \frac{\sum(x-\bar{x})^{2}}{n}$ ; $\sigma = \sqrt{\sigma^{2}}$
Range $= \max - \min$ ; $\sigma = 0 \iff$ all values identical

Traps & recognition

12 classic traps

Wrong total in $P(A)$ : die even is $\frac{3}{6} = \frac{1}{2}$ , not $\frac{3}{5}$ .
Permuting a team: $C(5,2) = 10$ , not $A(5,2) = 20$ .
Case-bashing at-least-one instead of one complement subtraction (E1-48).
Forgetting $-P(A \cap B)$ : heart-or-king is $\frac{16}{52} = \frac{4}{13}$ , not $\frac{17}{52}$ .
Conditioning without shrinking: $P(>3|\text{even}) = \frac{2}{3}$ , not $\frac{2}{6}$ .
Multiplying dependent draws: no-replacement $P(RR) = \frac{3}{5} \times \frac{2}{4}$ .
Mutually exclusive $\ne$ independent (positive probabilities are dependent).
Reading $P(Y \text{ or } G) = \frac{5}{12}$ as $P(Y)$ : split the system (E3-48).
Unsorted median, or averaging all values for even $n$ .
Variance in data units: root it for $\sigma$ ; divide grouped means by $\sum f$ .
Inventing E2-47 numbers: no distribution means unsolvable — flag it.
Probabilities outside $[0,1]$ : bounds-check every answer.

Fast problem recognition

Favourable over total
Verify equal likelihood, count $|\Omega|$ , then $\frac{|A|}{|\Omega|}$ . Bounds-check $[0,1]$ .
Choose / team / committee
Order irrelevant $\to C(n,k)$ ; arrangements / ranks / codes $\to A(n,k)$ .
At least one
Complement at once: $1 - P(\text{none})$ . E1-48 zodiac is the model.
Or with overlap
Add then subtract the intersection; add directly only if mutually exclusive.
Given / knowing that
Shrink to the given set, then $\frac{P(A \cap B)}{P(B)}$ .
Two trials / with replacement
Test independence; if independent, multiply.
Balls with or-probabilities
Convert to counts ( $P \times$ total), solve the system — E3-48 pattern.
Mean / median / mode
Sort first; even $n$ averages the middle two; outliers $\to$ median.

Practice — 48 problems

0/0 correct

PP-C-01basic · prob-sample-space
One fair die: $P(\text{even}) =$
PP-C-02basic · prob-sample-space
Two fair coins: $P(\text{at least one head}) =$
PP-C-03basic · prob-sample-space
One card from 52: $P(\text{ace}) =$
PP-C-04basic · prob-sample-space
$P(\varnothing) =$
PP-C-05basic · prob-counting
$C(5,2) =$
PP-C-06basic · prob-counting
$A(5,2) =$
PP-C-07basic · prob-counting
$C(6,4) =$
PP-C-08basic · prob-counting
Choose a 3-person team from 8 (order irrelevant): how many?
PP-C-09basic · prob-complement
Zodiac: $P(\text{at least two of three share an animal}) =$
PP-C-10basic · prob-complement
$P(\text{none}) = \frac{1}{4}$ : $P(\text{at least one}) =$
PP-C-11basic · prob-complement
Die rolled 3 times: $P(\text{at least one } 6) =$
PP-C-12basic · prob-complement
$P(A) = 0.3$ : $P(A^{c}) =$
PP-C-13basic · prob-addition
Die: $P(1 \text{ or } 6) =$
PP-C-14basic · prob-addition
$P(A) = 0.4$ , $P(B) = 0.5$ , $P(A \cap B) = 0.2$ : $P(A \cup B) =$
PP-C-15basic · prob-addition
Mutually exclusive $A, B$ : $P(A) = 0.5$ , $P(B) = 0.3$ : $P(A \cup B) =$
PP-C-16basic · prob-addition
Card: $P(\text{heart or king}) =$
PP-C-17intermediate · prob-conditional
Die, given even: $P(> 3) =$
PP-C-18intermediate · prob-conditional
$P(A \cap B) = 0.2$ , $P(B) = 0.5$ : $P(A|B) =$
PP-C-19intermediate · prob-conditional
Bag $3R2B$ , first drawn red (kept out): $P(\text{second red}) =$
PP-C-20intermediate · prob-conditional
$P(A \cap B) = 0.12$ , $P(B) = 0.3$ : $P(A|B) =$
PP-C-21intermediate · prob-independence
Two fair coins: $P(HH) =$
PP-C-22intermediate · prob-independence
Independent $A, B$ : $P(A) = 0.3$ , $P(B) = 0.4$ : $P(A \cap B) =$
PP-C-23intermediate · prob-independence
Mutually exclusive $A, B$ with positive probabilities are
PP-C-24intermediate · prob-independence
Die twice: $P(\text{both } 6) =$
PP-C-25intermediate · prob-urn-systems
E3-48 balls: $P(\text{yellow}) =$
PP-C-26intermediate · prob-urn-systems
E3-48 balls: $P(\text{black}) =$
PP-C-27intermediate · prob-urn-systems
E3-48 balls: $P(\text{green}) =$
PP-C-28intermediate · prob-urn-systems
Urn I $(1R2B)$ , urn II $(2R1B)$ , urn picked fairly: $P(R) =$
PP-C-29intermediate · stat-centers
Mean of $2, 4, 4, 9 =$
PP-C-30intermediate · stat-centers
Median of $2, 4, 4, 9 =$
PP-C-31intermediate · stat-centers
Mode of $1, 2, 2, 3, 3, 3 =$
PP-C-32intermediate · stat-centers
Median of $5, 1, 4 =$
PP-C-33advanced · stat-spread
Population variance of $1, 2, 3 =$
PP-C-34advanced · stat-spread
Std dev of $1, 2, 3 =$
PP-C-35advanced · stat-spread
$\sigma = 0$ means
PP-C-36advanced · stat-spread
Range of $2, 4, 4, 9 =$
PP-C-37advanced · stat-tables
Scores $0$ (2), $1$ (3), $2$ (5): mean =
PP-C-38advanced · stat-tables
Same table: total frequency =
PP-C-39advanced · stat-tables
Same table: range =
PP-C-40advanced · stat-tables
Same table: median =
PP-C-41advanced · stat-charts
Running-apps survey (E2-47, no counts given): $P(\text{two picked differ}) =$
PP-C-42advanced · stat-charts
Pie sector $25\%$ of $1200$ students =
PP-C-43advanced · stat-charts
Bars $5, 8, 12$ : total =
PP-C-44advanced · stat-charts
Heavily skewed data with outliers: best center is the
PP-C-45advanced · prob-stat-mixed
Which cannot be a probability?
PP-C-46advanced · prob-stat-mixed
E3-48 colour probabilities sum to
PP-C-47advanced · prob-stat-mixed
Two dice: $P(\text{sum} = 9) =$
PP-C-48advanced · prob-stat-mixed
Two dice: $P(\text{sum} = 7) =$

Previous CSCA exam questions

Real questions from the Dec 2025, Jan 2026 and Apr 2026 papers that test this chapter — solved with the thinking shown. This chapter has 3 probability questions (E1-48, E2-47, E3-48) and 0 statistics questions so far, which is why the lessons teach the full syllabus, not just past papers. Answers solved by the app (the source papers publish no official key); items with source errors are marked and still teach the correct math.

3 exam questions · 0/0 correct

Dec 2025 · Q48probability
Zodiac: P(at least two of three share an animal) =
Jan 2026 · Q47probabilitysource issue
Running apps survey: P(two picked students prefer different apps) =
No correct option in the source paper
Source issue: the question states no app-preference numbers — unsolvable as written.
Correct math
No preference distribution is given, so the probability cannot be computed.
Full solution — 3 steps
1. The question states no app-preference numbers.
2. Without a distribution no probability can be computed.
3. Source issue — unsolvable as written.
Apr 2026 · Q48probability
12 balls: P(red)=1/3, P(yellow or green)=5/12, P(black or yellow)=5/12: P(yellow, black, green) =

Exam checklist

0/10 checked

Probability

Statistics

One-minute memory sheet

Order? Teams $C$ , lineups $A$ .
At least one $= 1 - P(\text{none})$ : zodiac $\frac{17}{72}$ .
Or means minus overlap: heart-king $\frac{4}{13}$ .
Given shrinks the space: even $\to \frac{2}{3}$ .
Independent multiplies; exclusive adds; never both.
Balls: counts first — $R4$ , $Y2$ , $B3$ , $G3$ .
Sort before median; even $n$ averages two.
Variance squared, $\sigma$ plain; $\sum f$ divides.

Related guides

30-day study plan → fit this chapter into a week-by-week system
75-second pacing rule → hold 75s/question once the content clicks
Top 10 mistakes → the error patterns that cost the most points

Last reviewed Sep 2026 · GetCSCA editors

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