Syllabus topics: Probability — concept of uncertainty, axioms and rules of probability, conditional probability. Measures of central tendency — mean, median, mode. Measures of dispersion — range, interquartile range, variance, standard deviation. Introduction to correlation and covariance. Data representation — histograms, bar charts, scatter plots.
EXAMINED BUT NOT IN THE SYLLABUS
Plus Bayes' theorem, which is examined but missing from the syllabus list — see §1.6 and review finding D1.
THE BIG IDEA
Probability is a number between 0 and 1 that measures how likely something is. 0 means it will never happen, 1 means it certainly will, and 0.5 means it is as likely as not.
IN DEPTH
You are waiting for a bus. You cannot know whether it will arrive in the next five minutes — but you are not completely ignorant either. You have taken this bus a hundred times. It came within five minutes on about seventy of those. That "about 70 out of 100" is a probability, and it is enough to decide whether to wait or walk. Probability is the mathematics of acting sensibly when you cannot be certain.
FORMULA
For equally likely outcomes:
P(A) = number of favourable outcomes / total number of outcomes
Rolling a fair die and wanting an even number: favourable = {2, 4, 6} = 3, total = 6, so P(even) = 3/6 = 0.5.
Vocabulary you need:
| Term | Meaning |
|---|---|
| Experiment | A process with an uncertain outcome (rolling a die) |
| Sample space (S) | Every possible outcome: {1, 2, 3, 4, 5, 6} |
| Event | A subset of the sample space: "even" = {2, 4, 6} |
| Mutually exclusive | Two events that cannot both happen |
| Independent | One event happening does not change the other's probability |
| Exhaustive | Events that between them cover the whole sample space |
Everything else is derived from these three.
Complement rule
FORMULA
P(A′) = 1 − P(A)
If the chance of rain is 0.3, the chance of no rain is 0.7. Often the easiest route to an answer: "at least one" problems are usually solved as 1 − P(none).
Addition rule — for "A or B"
FORMULA
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Subtract the overlap, or you count it twice. When A and B are mutually exclusive, P(A ∩ B) = 0 and it simplifies to P(A) + P(B).
WORKED EXAMPLE
In a class of 100: 60 play cricket, 40 play football, 25 play both. P(cricket or football) = 0.60 + 0.40 − 0.25 = 0.75. Without subtracting the 25, you would get 1.00 — claiming every student plays a sport, which is false.
Multiplication rule — for "A and B"
P(A ∩ B) = P(A) × P(B|A) and if A and B are independent, P(A ∩ B) = P(A) × P(B)
Example. Two cards drawn without replacement from a pack. P(both aces) = (4/52) × (3/51) = 12/2652 ≈ 0.0045. The second fraction is 3/51 because one ace is already gone — the events are not independent.
With replacement they would be independent: (4/52) × (4/52) ≈ 0.0059.
THE BIG IDEA
Conditional probability is the probability of A given that you already know B happened. New information changes the odds.
IN DEPTH
A friend rolls a die behind a screen and tells you the result is even. Before that, P(6) was 1/6. Now you know the outcome is one of {2, 4, 6}, so P(6) has risen to 1/3. Nothing about the die changed — only what you know did.
FORMULA
P(A|B) = P(A ∩ B) / P(B), provided P(B) > 0
In plain English:
Reading it as "restrict the world to B, then ask what fraction of that world is also A" makes it obvious.
Worked example. Of 100 students, 40 study Python, 30 study R, 20 study both. A student studies R. What is the probability they also study Python?
P(Python | R) = P(both) / P(R) = 0.20 / 0.30 = 0.667
Compare with the unconditional P(Python) = 0.40. Knowing they study R raised it from 0.40 to 0.67.
A and B are independent when knowing one tells you nothing about the other:
FORMULA
P(A|B) = P(A), equivalently P(A ∩ B) = P(A) × P(B)
In the example above, 0.667 ≠ 0.40, so the two subjects are not independent — students who take R are more likely to take Python.
Do not confuse independent with mutually exclusive. Mutually exclusive events cannot both occur, so knowing one happened tells you the other did not — which is a very strong dependence. Two mutually exclusive events with non-zero probability are never independent.
THE BIG IDEA
One number that stands in for a whole set of numbers — where the data "centres".
IN DEPTH
Someone asks how your class did in the statistics test. You could read out all sixty marks. Instead you say "about 72". That single number is a measure of central tendency, and it communicates more in one word than the full list does in sixty.
FORMULA
x̄ = Σxᵢ / n
Add everything, divide by how many.
WORKED EXAMPLE
Marks: 5, 8, 10, 12, 15
Sort the data; take the middle value. With an even count, average the two middle values.
Same data: 5, 8, 10, 12, 15 → median = 10
With an even count: 5, 8, 10, 12, 15, 18 → (10 + 12)/2 = 11
The most frequent value. There may be none, one, or several.
3, 7, 7, 9, 12 → mode = 7
KEY INSIGHT
Add one outlier and watch what happens:
| Data | Mean | Median |
|---|---|---|
| 5, 8, 10, 12, 15 | 10.0 | 10 |
| 5, 8, 10, 12, 500 | 107.0 | 10 |
The mean moved from 10 to 107 — it now describes nobody in the dataset. The median did not move at all.
That is why incomes and house prices are always reported as medians. A handful of billionaires would drag the mean income far above what anyone actually earns. When a question asks "which measure is appropriate and why", this is the answer: use the median when the data is skewed or contains outliers.
| Use | When |
|---|---|
| Mean | Data is roughly symmetric with no extreme outliers |
| Median | Data is skewed, or has outliers (income, house prices) |
| Mode | Data is categorical (most common blood group, shoe size) |
Empirical relationship for a moderately skewed distribution:
NOTE
Mode ≈ 3 × Median − 2 × Mean
THE BIG IDEA
The centre is only half the story. Dispersion says how spread out the values are around it.
IN DEPTH
Two students both average 60 across five tests.
Same mean. Completely different students. If you had to pick one for a team, you would want to know the spread, not just the average. The mean without a measure of spread is close to useless.
FORMULA
Range = maximum − minimum
Ravi: 62 − 58 = 4. Priya: 100 − 20 = 80.
Simple, but it uses only two values — one outlier defines it entirely.
FORMULA
IQR = Q₃ − Q₁
Q₁ is the 25th percentile, Q₃ the 75th. The IQR is the spread of the middle 50%, so outliers cannot inflate it.
Outlier rule: anything below Q₁ − 1.5×IQR or above Q₃ + 1.5×IQR.
The sequence matters — never skip a step:
FORMULA
Population variance: σ² = Σ(xᵢ − μ)² / N Sample variance: s² = Σ(xᵢ − x̄)² / (n − 1)
In plain English:
WORKED EXAMPLE
Data: 5, 8, 10, 12, 15 (treat as a sample)
| xᵢ | xᵢ − x̄ | (xᵢ − x̄)² |
|---|---|---|
| 5 | −5 | 25 |
| 8 | −2 | 4 |
| 10 | 0 | 0 |
| 12 | 2 | 4 |
| 15 | 5 | 25 |
| Σ | 0 | 58 |
Notice the differences sum to zero. They always do — that is precisely why we square them. It is also a free check on your arithmetic.
KEY INSIGHT
You measured the deviations from the sample's own mean, not the true population mean. The sample mean sits, by construction, right in the middle of your sample — so the deviations you observe are slightly smaller than the true ones. Dividing by n−1 instead of n inflates the answer just enough to correct for that bias.
Rule for the exam: dividing a sample by n instead of n−1 is the most frequent arithmetic error in this course. If the question says "sample", use n−1.
FORMULA
CV = (s / x̄) × 100%
A unit-free measure, so it compares spread across different scales. Comparing the variability of heights (cm) with weights (kg) requires CV; standard deviation alone cannot do it.
Full treatment in Unit 4; this is the introduction the syllabus places here.
Covariance measures whether two variables move together:
FORMULA
Cov(X,Y) = Σ(xᵢ − x̄)(yᵢ − ȳ) / (n − 1)
Positive means they rise together; negative means one rises as the other falls. But its magnitude is meaningless — it depends on the units. Measure height in centimetres instead of metres and the covariance multiplies by 100.
Correlation fixes that by standardising:
FORMULA
r = Cov(X,Y) / (sₓ × s_y)
r is unit-free and always between −1 and +1, which is why it is the number you report.
NOTE
Unit 1 lists only "conditional probability". Bayes' theorem appears in the prescribed activities ("quiz on probability, conditional probability, Bayes") and in lab experiment 2 — but in no unit. Study it. See review finding D1.
THE BIG IDEA
Bayes' theorem lets you reverse a conditional probability. If you know the chance of a positive test given the disease, Bayes gives you the chance of the disease given a positive test — which is what you actually want to know.
This one trips up medical doctors, let alone students. Take it slowly.
IN DEPTH
A disease affects 1 in 100 people. A test is very good: it catches 99% of people who have it, and correctly clears 95% of those who do not. You test positive.
How worried should you be? Most people say "99% — the test is 99% accurate."
The real answer is about 17%.
FORMULA
P(A|B) = P(B|A) × P(A) / P(B)
where P(B) = P(B|A)×P(A) + P(B|A′)×P(A′) — the law of total probability
In plain English:
Worked solution.
P(− | not D) = 0.95, so P(+ | not D) = 0.05 (false positive rate)
Step 2: find P(+) using total probability
= 0.0099 + 0.0495 = 0.0594
Step 3: apply Bayes
KEY INSIGHT
Think of 10,000 people:
| Have the disease | Do not | Total | |
|---|---|---|---|
| Test positive | 99 | 495 | 594 |
| Test negative | 1 | 9,405 | 9,406 |
| Total | 100 | 9,900 | 10,000 |
Of the 594 who test positive, only 99 are genuinely ill — 99/594 = 17%.
The disease is rare. So the 5% of false positives, drawn from the enormous healthy majority, vastly outnumber the true positives drawn from the tiny sick minority. The test is excellent; the base rate is doing the damage.
Confusing P(D|+) with P(+|D) is called the base rate fallacy, and it is the single most examined idea in this part of the course. Draw the 10,000-person table every time — it makes the answer obvious and it earns marks.
| Chart | Use for | Key point |
|---|---|---|
| Bar chart | Categorical data | Bars have gaps; order is arbitrary |
| Histogram | Continuous data | Bars touch; classes are adjacent intervals |
| Scatter plot | Two numeric variables | Points only — never join them with lines |
| Pie chart | Parts of a whole | Only when the parts sum to 100% |
| Box plot | Spread and outliers | Shows min, Q₁, median, Q₃, max |
Bar chart vs histogram is a guaranteed two-mark question. Bar chart: categories, gaps between bars. Histogram: continuous intervals, no gaps.
FORMULA
Pearson's skewness = 3(mean − median) / standard deviation
PROBLEM 1
A bag contains 5 red, 3 blue and 2 green balls. One ball is drawn at random. Find (a) P(red), (b) P(not red), (c) P(red or blue).
Solution.
Total balls = 5 + 3 + 2 = 10.
(a) P(red) = 5/10 = 0.5
(b) P(not red) = 1 − 0.5 = 0.5 (complement rule)
(c) Red and blue are mutually exclusive — one ball cannot be both — so the overlap term is zero: P(red or blue) = 5/10 + 3/10 = 8/10 = 0.8
PROBLEM 2
The marks of 8 students are: 45, 52, 48, 60, 55, 49, 58, 51. Find the mean, median, range, sample variance and sample standard deviation.
Solution.
Step 1 — Mean: sum = 45+52+48+60+55+49+58+51 = 418; n = 8; x̄ = 418/8 = 52.25
Step 2 — Median: sorted → 45, 48, 49, 51, 52, 55, 58, 60. n is even, so average the 4th and 5th: (51 + 52)/2 = 51.5
Step 3 — Range: 60 − 45 = 15
Step 4 — Variance:
| xᵢ | xᵢ − 52.25 | (xᵢ − 52.25)² |
|---|---|---|
| 45 | −7.25 | 52.5625 |
| 52 | −0.25 | 0.0625 |
| 48 | −4.25 | 18.0625 |
| 60 | 7.75 | 60.0625 |
| 55 | 2.75 | 7.5625 |
| 49 | −3.25 | 10.5625 |
| 58 | 5.75 | 33.0625 |
| 51 | −1.25 | 1.5625 |
| Σ | 0.00 ✓ | 183.50 |
s² = 183.50 / (8 − 1) = 183.50 / 7 = 26.21
(The deviations summing to zero confirms the mean is right.)
PROBLEM 3
In a factory, machine A makes 60% of the items and machine B makes 40%. Two percent of A's output is defective, and 5% of B's. An item is picked at random and found defective. What is the probability it came from machine B?
Solution.
P(D|A) = 0.02, P(D|B) = 0.05
Step 2 — total probability of a defective item
= 0.012 + 0.020 = 0.032
Step 3 — apply Bayes
So there is a 62.5% chance it came from machine B — even though B makes only 40% of the output, because its defect rate is more than twice A's.
Check: P(A|D) = 0.012/0.032 = 0.375, and 0.625 + 0.375 = 1 ✓
Two marks
Five marks
Compute the mean, median, mode, variance and standard deviation for a given dataset.
Explain covariance and correlation, and why correlation is preferred.
Ten marks
COMMON ERRORS