DP Maths AA HL - Statistics and Probability - Inferential Statistics

DP Maths AA HL - Statistics and Probability - Inferential Statistics - Full notes

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DP Maths AA HL | Inferential Statistics

Inferential Statistics

This topic contains 2 notes:

4.4 Correlation and Regression
4.13 Bayes' Theorem

DP MATHEMATICS AA · TOPIC 4: STATISTICS AND PROBABILITY

4.4

Correlation and Regression

Full notes for HL · Syllabus SL 4.4, 4.10

Contents

1	Bivariate data and scatter diagrams
2	Describing correlation
3	Pearson's product-moment correlation coefficient
4	Correlation and causation
5	The line of best fit by eye
6	The least-squares regression line
7	Interpolation and extrapolation
8	Regression of x on y
9	Scatter diagrams in practice
10	Calculating r with technology
11	The regression line of y on x
12	The regression line of x on y
13	Interpolation, extrapolation and reliability
14	Correlation and causation
15	Residuals and the quality of fit
16	Non-linear relationships
17	Mixed review: worked problems
18	Additional exam-style practice
19	Spearman's rank correlation (extension)
20	Technology checklist for bivariate data
21	Exam-style questions with markscheme
22	Common misconceptions and links
23	Check your understanding
24	Summary and topic map
25	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
draw and interpret scatter diagrams
describe correlation by direction, strength and form
calculate and interpret Pearson's correlation coefficient r using technology
distinguish correlation from causation
draw a line of best fit through the mean point
find and use the least-squares regression line of y on x
decide when predictions are reliable (interpolation versus extrapolation)
use the regression line of x on y to predict x from y

This booklet covers SL 4.4, 4.10, which is common to SL and HL.

Guiding questions

How can we measure the strength of a relationship between two variables? When can a regression line be trusted to make predictions?

How this topic is examined

Correlation and regression appear almost every session on Paper 2: find r and interpret it, find the regression equation, interpret its gradient in context, make a prediction and comment on its reliability.

Interpret coefficients in context: 'for each additional hour studied, the score increases by about 2.8 marks'.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 Bivariate data and scatter diagrams

Two variables for each individual.

Think about it

Do students who revise for longer score higher? How could you display data to see whether there is a relationship?

Bivariate data consist of pairs of values (x, y) for each individual. The independent (explanatory) variable x is plotted horizontally and the dependent (response) variable y vertically. A scatter diagram shows the pattern of the relationship.

Figure 1. Revision time against test score for eight students, with the regression line through the mean point (square).

Hours, x	2	3	5	6	8	9	11	12
Score, y	52	55	61	64	70	71	78	80

2 Describing correlation

Direction, strength and form.

Direction: positive (y tends to increase as x increases), negative (y tends to decrease), or none. Strength: strong (points close to a line), moderate or weak. Form: linear or non-linear. Also comment on any outliers.

Figure 2. Types of correlation.

3 Pearson's product-moment correlation coefficient

A number from −1 to 1.

Pearson's correlation coefficient r measures the strength and direction of a linear relationship. It always satisfies −1 ≤ r ≤ 1. r = 1 is perfect positive linear correlation, r = −1 perfect negative, and r = 0 no linear correlation. In this course r is found with technology.

\|r\|	Strength
0 to 0.25	very weak
0.25 to 0.5	weak
0.5 to 0.75	moderate
0.75 to 1	strong

WORKED EXAMPLE 1 Calculating r (standard)

Find r for the revision data and interpret it.

Solution

GDC linear regression: r = 0.998A1

Very strong positive linear correlation: students who revise longer tend to score higherR1

Common mistake

Saying r = 0 means there is no relationship. It means no linear relationship; the data may follow a curve.

Practice exercise 1

1. Find r for x = 1, 2, 3, 4, 5 and y = 10, 8, 7, 4, 2, and describe the correlation.

2. A scatter diagram shows a clear U-shaped curve. Explain why r may be close to 0.

4 Correlation and causation

Association is not proof of cause.

A strong correlation does not show that one variable causes the other. Both may be driven by a third (confounding or lurking) variable: ice cream sales and drowning incidents are correlated because both increase in hot weather. The correlation may also be coincidence, especially with small samples.

5 The line of best fit by eye

Through the mean point.

A line of best fit drawn by eye should follow the trend of the points, with roughly equal numbers above and below, and should pass through the mean point (x̄, ȳ).

6 The least-squares regression line

Minimizing squared vertical distances.

The regression line of y on x, y = ax + b, is the line that minimizes the sum of the squared vertical distances (residuals) from the points to the line. It always passes through (x̄, ȳ). It is found using technology.

Interpretation: the gradient a is the estimated change in y for each unit increase in x; the intercept b is the estimated value of y when x = 0 (which may not be meaningful in context).

WORKED EXAMPLE 2 Regression line (standard)

Find the regression line of y on x for the revision data and interpret the gradient.

Solution

GDC: y = 2.80x + 46.7A1 A1

For each additional hour of revision, the score increases by about 2.8 marks on averageR1

Practice exercise 2

1. Show that the regression line passes through the mean point (7, 66.4) for the revision data.

2. A regression line for the price p ($) of a used car against age t (years) is p = −1850t + 21 400. Interpret both coefficients.

7 Interpolation and extrapolation

Predict inside the data range.

Interpolation predicts within the range of the observed x values and is reasonably reliable when the correlation is strong. Extrapolation predicts outside the range and is unreliable, because the relationship may not continue. Predictions are also unreliable when r is weak, or when the sample is small.

WORKED EXAMPLE 3 Making predictions (exam level)

Use the revision regression line to predict the score for (a) 7 hours; (b) 25 hours. Comment on reliability.

Solution

(a) 2.80(7) + 46.7 = 66.4: reliable (interpolation, r strong)A1 R1

(b) 2.80(25) + 46.7 = 117: impossible if scores are out of 100; extrapolation is unreliableA1 R1

8 Regression of x on y

A different line for predicting x.

To predict x from a given y, use the regression line of x on y, x = cy + d, which minimizes horizontal distances. It is a different line from y on x (unless r = ±1), and both pass through the mean point. Do not rearrange the y-on-x equation to predict x.

WORKED EXAMPLE 4 Predicting x (standard)

Using technology, find the regression line of x on y for the revision data and estimate the revision time needed for a score of 75.

Solution

GDC: x = 0.355y − 16.6A1

x = 0.355(75) − 16.6 = 10.1 hoursA1

Practice exercise 3

1. Explain why the y-on-x line should not be rearranged to predict x.

2. For data with r = 0.95, would the two regression lines be close together? Explain.

9 Scatter diagrams in practice

Plotting and describing.

Plot the explanatory variable on the x-axis. Describe the relationship by direction (positive or negative), strength (strong, moderate, weak), form (linear or not), and note any outliers or clusters.

Height (cm)	150	155	160	165	170	175	180	185
Mass (kg)	48	53	55	61	63	69	72	78

Figure 3. Height against mass for eight students, with the regression line.

WORKED EXAMPLE 5 Describing (standard)

Describe the correlation between height and mass.

Solution

Strong positive linear correlation; no outliersR1

10 Calculating r with technology

Interpret with care.

Enter the data in two lists and run linear regression to obtain r. Interpret using direction and strength. r measures linear association only; a strong non-linear relationship can have r near 0.

WORKED EXAMPLE 6 Pearson's r (standard)

Find r for the height and mass data.

Solution

GDC: r = 0.995A1

Very strong positive linear correlationR1

Practice exercise 4

1. Data: x = 1, 2, 3, 4, 5, 6; y = 1, 4, 9, 16, 25, 36. Find r and comment.

2. Explain why r = 0 does not mean there is no relationship.

11 The regression line of y on x

Least squares.

The regression line minimizes the sum of squared vertical distances from the points. It passes through (x̄, ȳ). Use it to predict y from x within the range of the data.

WORKED EXAMPLE 7 Regression (standard)

Find the regression line of mass on height and predict the mass of a student 168 cm tall.

Solution

GDC: m = 0.831h − 76.8A1 A1

h = 168: m = 62.8 kgA1

Practice exercise 5

1. Interpret the gradient of the regression line in context.

2. Verify that the line passes through the mean point.

12 The regression line of x on y

Predicting x from y.

To predict x from a given y, use the regression line of x on y, which minimizes horizontal squared distances. It differs from the y-on-x line unless |r| = 1. Both lines pass through the mean point.

WORKED EXAMPLE 8 x on y (standard)

Estimate the height of a student with mass 70 kg.

Solution

GDC: h = 1.19m + 93.1A1

m = 70: h = 177 cmA1

Practice exercise 6

1. Why should you not rearrange the y-on-x line to predict x?

13 Interpolation, extrapolation and reliability

When to trust a prediction.

Predictions within the data range (interpolation) are reasonable when |r| is large. Predictions outside it (extrapolation) are unreliable because the relationship may change. Small samples and outliers also reduce reliability.

WORKED EXAMPLE 9 Reliability (standard)

Comment on using the height-mass line to predict the mass of a child of height 110 cm.

Solution

110 cm is outside the data range (150–185 cm): extrapolation, unreliableR1

Practice exercise 7

1. A regression gives a predicted value of −5 for a quantity that cannot be negative. What does this show?

14 Correlation and causation

Confounding variables.

Correlation between two variables does not prove that one causes the other. A third (confounding) variable may affect both, or the association may be coincidence. Controlled experiments are needed to establish causation.

WORKED EXAMPLE 10 Confounding (standard)

Towns with more doctors have more deaths. Explain.

Solution

Larger populations have both more doctors and more deaths: population is confoundingR1

Practice exercise 8

1. Suggest a confounding variable for the correlation between ice cream sales and sunburn cases.

15 Residuals and the quality of fit

Observed minus predicted.

A residual is the observed y minus the predicted y. A good linear model has residuals scattered randomly around zero with no pattern. A curved pattern in the residuals suggests a non-linear model is better.

WORKED EXAMPLE 11 Residual (standard)

Find the residual for the student of height 175 cm and mass 69 kg.

Solution

Predicted 0.831(175) − 76.8 = 68.6M1

Residual 69 − 68.6 = 0.4 kgA1

Practice exercise 9

1. What does a large residual for one point suggest?

16 Non-linear relationships

When a line is not enough.

Some relationships are curved: exponential growth, inverse variation, power laws. Plotting the data, checking residuals, or comparing models with technology (exponential, power or quadratic regression) helps choose a better model.

Figure 4. Exponential data would curve away from a straight line.

WORKED EXAMPLE 12 Choosing a model (standard)

Data double every 2 units of x. Suggest a model.

Solution

Exponential y = abx with b = √2R1

17 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 13 Mixed problem (exam level)

For 10 cities, the mean temperature x (°C) and gas use y (units) give x̄ = 12, ȳ = 340, regression y = −14.5x + b, r = −0.92. (a) Find b. (b) Predict y for x = 8. (c) Comment.

Solution

(a) 340 = −14.5(12) + b: b = 514M1 A1

(b) 398 unitsA1

Practice exercise 10

1. The regression line is y = 2.4x + 7 and x̄ = 15. Find ȳ.

18 Additional exam-style practice

Paper 2.

WORKED EXAMPLE 14 Paper 2 question (exam level)

Hours of revision x and test score y for 7 students: x = 2, 4, 5, 6, 8, 9, 11; y = 45, 52, 58, 61, 70, 72, 81. (a) Find r. (b) Find the regression line. (c) Predict the score for 7 hours.

Solution

(a) r = 0.998A1

(b) y = 4.00x + 37.0A2

Practice exercise 11

1. Explain why predicting the score for 20 hours is unreliable.

19 Spearman's rank correlation (extension)

Monotonic relationships.

Spearman's rank correlation coefficient rs is Pearson's r applied to the ranks. It measures whether the relationship is monotonic (consistently increasing or decreasing) and is less affected by outliers.

WORKED EXAMPLE 15 Ranks (standard)

Two judges rank 5 entries: A: 1, 2, 3, 4, 5; B: 2, 1, 3, 5, 4. Find rs.

Solution

Pearson's r on ranksM1

rs = 0.8A1

20 Technology checklist for bivariate data

Lists and outputs.

1. Enter x and y in two lists. 2. Plot a scatter diagram. 3. Run LinReg(ax + b) with diagnostics on to see r. 4. Store the regression equation and graph it with the data. 5. Use the equation for predictions within the data range.

WORKED EXAMPLE 16 Mean point (standard)

Use 2-Var Stats to find (x̄, ȳ) for the revision data.

Solution

(6.43, 62.7)A1

GDC tip

Turn on 'Diagnostics' (or 'Stat Diagnostics') so that r and r2 are displayed with the regression output.

21 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 2 style) (about 15 minutes)

The table shows the average daily temperature T (°C) and the number of hot drinks H sold by a café on 8 days: T = 4, 7, 9, 12, 15, 18, 21, 24; H = 92, 85, 81, 70, 64, 55, 48, 40.

(a) Find Pearson's correlation coefficient and describe the correlation. [3]

(b) Find the regression line of H on T. [2]

(d) Estimate H when T = 16 and comment on the reliability. [2]

(e) Explain why the model should not be used for T = 35. [1]

Markscheme

Part	Answer	Marks
(a)	r = −0.998	A2
	very strong negative linear correlation	R1
(b)	H = −2.64T + 103	A2
(c)	about 2.6 fewer hot drinks are sold per extra °C	R1
(d)	H ≈ 61	A1
	reliable: interpolation with strong correlation	R1
(e)	extrapolation beyond the data range	R1

Examiner's comments. Answers must be given in context.

22 Common misconceptions and links

Students often think…	Actually…
correlation implies causation	a third variable may cause both
r = 0 means no relationship	no linear relationship
rearrange y on x to find x	use the x-on-y line
extrapolation is reliable when r is strong	it is always risky

How 4.4 connects to the rest of the course

Topic	Connection
4.1 Statistics	means and spread
2.1 Lines	gradient and intercept
2.8 Exponential models	non-linear relationships
4.9 Normal distribution	residual assumptions

23 Check your understanding

Section A: multiple choice

1. r = −0.92 indicates

A. strong negative linear correlation B. weak positive C. no correlation D. causation

2. The regression line passes through

A. the origin B. the first point C. the mean point D. the median point

3. Predicting outside the data range is

A. interpolation B. extrapolation C. regression D. correlation

4. r always lies between

A. 0 and 1 B. −∞ and ∞ C. −10 and 10 D. −1 and 1

Section B: structured

5. Explain what is meant by a confounding variable, with an example. [2]

6. A regression line is y = 0.8x + 12 with r = 0.3. Comment on using it to predict y. [2]

Answers

1. A 2. C 3. B 4. D

5. A third variable affecting both: e.g. temperature drives both ice cream sales and swimming accidents.

6. Weak correlation, so predictions are unreliable.

24 Summary and topic map

Idea	Remember
Scatter diagrams	direction, strength, form, outliers
Pearson's r	−1 ≤ r ≤ 1; linear association
Causation	correlation does not imply cause
Regression y on x	least squares; through (x̄, ȳ)
Interpretation	gradient = change in y per unit x
Predictions	interpolation reliable; extrapolation not
x on y	use to predict x

Before the exam, make sure you…

•interpret in context
•use the correct regression line
•comment on reliability
•mention linearity when using r

Correlation and Regression on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 5. Topic map for Correlation and Regression.

25 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. r = −0.990: very strong negative linear correlation

2. r only measures linear association; the U shape is non-linear

Practice exercise 2

1. 2.804(7) + 46.74 = 66.4, the mean of y

2. the value falls by about $1850 per year; a new car would cost about $21 400

Practice exercise 3

1. it minimizes vertical, not horizontal, distances

2. yes: the closer |r| is to 1, the closer the lines

Practice exercise 4

1. r = 0.979: strongly positive, but the relationship is quadratic

2. the relationship could be non-linear

Practice exercise 5

1. mass increases by about 0.83 kg per cm of height

2. x̄ = 167.5, ȳ = 62.375: 0.831(167.5) − 76.8 = 62.4 (rounding)

Practice exercise 6

1. it minimizes errors in y, not in x

Practice exercise 7

1. the model is being used outside its valid range

Practice exercise 8

1. temperature or sunshine

Practice exercise 9

1. a possible outlier or an unusual individual

Practice exercise 10

1. 43

Practice exercise 11

1. extrapolation; scores cannot exceed the maximum mark

DP MATHEMATICS AA · TOPIC 4: STATISTICS AND PROBABILITY

4.13

Bayes' Theorem

Full notes for HL · Syllabus AHL 4.13

Contents

1	Revisiting conditional probability
2	The law of total probability
3	Bayes' theorem for two events
4	Medical testing and false positives
5	Bayes' theorem with three or more events
6	Repeated tests and updating probabilities
7	Partitions of the sample space
8	The law of total probability in practice
9	Bayes' theorem with two causes
10	Bayes' theorem with three causes
11	Medical tests: sensitivity and specificity
12	Two-way tables and Bayes' theorem
13	Repeated tests
14	Bayes' theorem in quality control and forensics
15	Bayes' theorem with counting
16	Mixed review: worked problems
17	Additional exam-style practice
18	Common errors with conditional probability
19	How the prior affects the posterior
20	Further exam-style practice
21	Bayes' theorem summary and strategy
22	Exam-style questions with markscheme
23	Common misconceptions and links
24	Check your understanding
25	Summary and topic map
26	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
recall the definition of conditional probability
use the law of total probability to find P(B)
derive and apply Bayes' theorem for two events
interpret false positives and the effect of a low base rate
apply Bayes' theorem with three mutually exclusive events
update probabilities after repeated independent evidence

This booklet covers Additional Higher Level (AHL) content (AHL 4.13). It is studied by HL students only.

Guiding questions

How should new evidence change our beliefs? Why can a positive result from an accurate test still mean you are probably healthy?

How this topic is examined

Bayes' theorem appears on Papers 1 and 2, usually through a tree diagram: find P(B) by total probability, then 'given B, find the probability of A'. Context questions (medical tests, quality control, weather) are common, sometimes with three branches.

Draw the tree, label events clearly, and write the conditional probability you want in notation before calculating.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 Revisiting conditional probability

Two directions.

Think about it

A test is 95% accurate for people with a rare disease. You test positive. Is the probability that you have the disease 95%?

A tree diagram usually gives probabilities in one direction, such as P(positive | disease). Bayes' theorem reverses them to find P(disease | positive).

2 The law of total probability

Add over all routes.

If A and A′ split the sample space, then B can happen 'via A' or 'via A′':

More generally, if A1, A2, ..., An are mutually exclusive and cover all possibilities, P(B) = Σ P(B | Ai)P(Ai). On a tree diagram, this is adding the probabilities of all paths that end in B.

3 Bayes' theorem for two events

Reversing a conditional.

Why it works

By the definition of conditional probability, P(A | B) = P(A ∩ B)/P(B). The numerator is P(B | A)P(A) (multiplying along a path), and the denominator is P(B) by the law of total probability. Substituting gives the formula.

In words: the probability of A given B equals the probability of the path through A to B, divided by the total probability of reaching B by any path.

4 Medical testing and false positives

The base rate matters.

For a diagnostic test: the sensitivity is P(positive | disease) and the false positive rate is P(positive | no disease). The prevalence (base rate) is P(disease). When a condition is rare, most positive results can come from the much larger healthy group, so P(disease | positive) can be surprisingly small.

Figure 1. A test with sensitivity 95% and false positive rate 8% for a condition with prevalence 1%.

WORKED EXAMPLE 1 The base-rate effect (standard)

Using the tree, find the probability that a person who tests positive has the condition.

Solution

P(+) = 0.01(0.95) + 0.99(0.08) = 0.0095 + 0.0792 = 0.0887M1 A1

P(D | +) = 0.0095/0.0887M1

= 0.107: only about 11%A1

Although the test is quite accurate, most positive results are false positives because the condition is rare.

WORKED EXAMPLE 2 Negative result (standard)

Find P(D′ | −).

Solution

P(−) = 1 − 0.0887 = 0.9113A1

P(D′ ∩ −) = 0.99(0.92) = 0.9108A1

P(D′ | −) = 0.99945A1

Practice exercise 1

1. A test has sensitivity 0.9 and false positive rate 0.05. Prevalence is 2%. Find P(disease | positive).

2. Repeat with prevalence 20%. Comment on the change.

5 Bayes' theorem with three or more events

Several possible causes.

WORKED EXAMPLE 3 Three machines (exam level)

Machines X, Y and Z make 50%, 30% and 20% of a factory's components, with defect rates 2%, 3% and 5%. A component is defective. Find the probability it came from each machine.

Solution

P(D) = 0.010 + 0.009 + 0.010 = 0.029M1 A1

P(X | D) = 0.010/0.029 = 0.345; P(Y | D) = 0.009/0.029 = 0.310M1 A1

P(Z | D) = 0.345; the probabilities sum to 1A1

Practice exercise 2

1. Weather: P(sunny) = 0.5, P(cloudy) = 0.3, P(rain) = 0.2. P(late | sunny) = 0.05, P(late | cloudy) = 0.1, P(late | rain) = 0.3. Given that a bus is late, find P(rain).

2. Three boxes contain 2 red/3 blue, 4 red/1 blue and 3 red/3 blue. A box is chosen at random and a red ball drawn. Find P(box 2 | red).

6 Repeated tests and updating probabilities

Today's posterior is tomorrow's prior.

The probability before evidence is the prior; after the evidence it is the posterior. If a second independent test is performed, the posterior from the first test becomes the prior for the second. Repeated positive results increase confidence sharply.

WORKED EXAMPLE 4 Two positive tests (exam level)

Using the earlier test (sensitivity 0.95, false positive rate 0.08, prevalence 0.01), a person tests positive twice (independent tests). Find P(D | two positives).

Solution

New prior 0.107 (from one positive)M1

P(D | ++) = 0.107(0.95)/(0.107(0.95) + 0.893(0.08))M1

= 0.1017/(0.1017 + 0.0714) = 0.587A1

Practice exercise 3

1. Find P(D | three positives) for the same test.

7 Partitions of the sample space

Mutually exclusive and exhaustive.

Events A1, A2, ..., An form a partition if exactly one of them occurs: they are mutually exclusive and their probabilities add to 1. Bayes' theorem and the law of total probability both use a partition.

WORKED EXAMPLE 5 Partition (basic)

A factory uses machines X, Y, Z for 45%, 35% and 20% of production. Explain why these events form a partition.

Solution

Each item is made by exactly one machine; 0.45 + 0.35 + 0.2 = 1R1

Practice exercise 4

1. Do the events 'even' and 'greater than 3' partition the outcomes of a die?

8 The law of total probability in practice

Add the paths.

P(B) = ΣP(B | Ai)P(Ai). On a tree, this is the sum of the probabilities of all paths ending in B.

WORKED EXAMPLE 6 Machines (standard)

Defect rates are 2%, 3% and 6% for X, Y, Z above. Find P(defective).

Solution

0.45(0.02) + 0.35(0.03) + 0.2(0.06)M1

= 0.009 + 0.0105 + 0.012 = 0.0315A1

Practice exercise 5

1. P(rain) = 0.3. P(late | rain) = 0.4, P(late | no rain) = 0.1. Find P(late).

9 Bayes' theorem with two causes

Reversing the condition.

P(A | B) = P(B | A)P(A)/P(B). The numerator is the path through A to B; the denominator is the total probability of B.

WORKED EXAMPLE 7 Late for school (standard)

Using the rain example, given that a student is late, find P(it rained).

Solution

0.12/0.19M1

= 0.632A1

Practice exercise 6

1. Given the student is not late, find P(it rained).

10 Bayes' theorem with three causes

Several possible sources.

P(Ai | B) = P(B | Ai)P(Ai)/ΣP(B | Aj)P(Aj). The posterior probabilities for all causes sum to 1.

WORKED EXAMPLE 8 Which machine? (exam level)

Using the factory example, a defective item is found. Find the probability it came from each machine.

Solution

P(X | D) = 0.009/0.0315 = 0.286A1

P(Y | D) = 0.333; P(Z | D) = 0.381A1 A1

Z is most likely despite making only 20%R1

Practice exercise 7

1. Find P(X | not defective).

11 Medical tests: sensitivity and specificity

Interpreting test results.

Sensitivity = P(positive | disease); specificity = P(negative | no disease). The positive predictive value P(disease | positive) depends strongly on the prevalence.

Prevalence	P(disease \| positive), sensitivity 0.95, specificity 0.95
0.1%	0.0187
1%	0.161
10%	0.679
30%	0.891

WORKED EXAMPLE 9 Rare condition (exam level)

A test has sensitivity 0.95 and specificity 0.95. Prevalence is 1%. Find P(disease | positive).

Solution

P(+) = 0.0095 + 0.0495 = 0.059M1 A1

0.0095/0.059 = 0.161A1

Practice exercise 8

1. For the same test, find P(no disease | negative) with 1% prevalence.

12 Two-way tables and Bayes' theorem

Natural frequencies.

Imagine a large population (for example 10 000 people) and fill a two-way table of expected counts. Conditional probabilities can then be read directly, which makes Bayes' theorem intuitive.

	Positive	Negative	Total
Disease	95	5	100
No disease	495	9405	9900
Total	590	9410	10 000

WORKED EXAMPLE 10 Natural frequencies (standard)

Use the table to find P(disease | positive).

Solution

95/590M1

= 0.161A1

Practice exercise 9

1. Use the table to find P(positive).

13 Repeated tests

Updating with new evidence.

After a positive test, the posterior probability becomes the prior for a second, independent test. Two positive results increase the probability of disease substantially.

WORKED EXAMPLE 11 Second test (exam level)

After one positive result (posterior 0.161), the test is repeated and again positive. Find P(disease).

Solution

0.161(0.95)/(0.161(0.95) + 0.839(0.05))M1

= 0.153/(0.153 + 0.042) = 0.785A1

Practice exercise 10

1. Find P(disease) after three positive results.

14 Bayes' theorem in quality control and forensics

Applications.

Bayes' theorem is used to identify the likely source of defects, to interpret DNA or fingerprint evidence, in spam filtering (P(spam | word appears)), and in diagnosis. In each case, the prior probability matters as much as the evidence.

WORKED EXAMPLE 12 Spam filter (exam level)

40% of emails are spam. The word 'prize' appears in 25% of spam and 2% of other emails. Find P(spam | 'prize').

Solution

0.4(0.25) = 0.1; 0.6(0.02) = 0.012A1

0.1/0.112 = 0.893M1 A1

Practice exercise 11

1. Find P(spam | 'prize' does not appear).

15 Bayes' theorem with counting

Equally likely selections.

When probabilities come from selections (balls from bags, cards), find the conditional probabilities with counting, then apply Bayes' theorem.

WORKED EXAMPLE 13 Bags (exam level)

Bag A has 3 red, 2 blue; bag B has 1 red, 4 blue. A bag is chosen at random and two balls drawn without replacement; both are red. Find P(bag A).

Solution

P(RR | A) = (3/5)(2/4) = 0.3; P(RR | B) = 0A1

P(A | RR) = 0.15/0.15 = 1M1 A1

Bag B cannot give two redsR1

Practice exercise 12

1. In the same setting, find P(bag A | one red and one blue).

16 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 14 Mixed problem (exam level)

70% of students take the bus, 30% walk. 10% of bus users and 25% of walkers are late. (a) Find P(late). (b) Find P(walk | late). (c) Find P(bus | not late).

Solution

(a) 0.07 + 0.075 = 0.145A1

(b) 0.075/0.145 = 0.517A1

Practice exercise 13

1. Show that P(bus | late) + P(walk | late) = 1.

17 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 15 Paper 1 question (exam level)

P(A) = 0.6, P(B | A) = 0.5, P(B | A′) = 0.25. Find P(B), P(A | B) and P(A′ | B′).

Solution

P(B) = 0.3 + 0.1 = 0.4A1

P(A | B) = 0.3/0.4 = 0.75A1

P(A′ | B′) = 0.3/0.6 = 0.5M1 A1

Practice exercise 14

1. Show that A and B in this question are not independent.

18 Common errors with conditional probability

The prosecutor's fallacy.

The prosecutor's fallacy confuses P(evidence | innocent) with P(innocent | evidence). A small chance that an innocent person matches some evidence does not mean that a matching person is unlikely to be innocent: with a large population, many innocent people could match.

WORKED EXAMPLE 16 Fallacy (exam level)

A DNA profile matches 1 in 100 000 people. In a city of 2 million, one suspect matches. Assuming the culprit is in the city and there is no other evidence, estimate P(suspect is guilty).

Solution

About 20 innocent people would also matchM1

P(guilty | match) ≈ 1/21 = 0.048A1

Not 0.99999R1

19 How the prior affects the posterior

Sensitivity analysis.

The posterior probability P(A | B) depends on the prior P(A). When the evidence is strong (high sensitivity, low false-positive rate), the posterior is high even for moderate priors; when the prior is tiny, even strong evidence may leave the posterior small. Plotting the posterior against the prior shows this clearly.

Figure 2. The posterior rises steeply with the prior.

WORKED EXAMPLE 17 Comparing tests (exam level)

With prevalence 1%, compare P(disease | positive) for specificity 0.95 and 0.99 (sensitivity 0.95).

Solution

0.95: 0.161A1

0.99: 0.0095/(0.0095 + 0.0099) = 0.490A1

Improving specificity reduces false positives dramaticallyR1

Practice exercise 15

1. Find the prevalence at which P(disease | positive) = 0.5 for sensitivity and specificity 0.95.

20 Further exam-style practice

Paper 2.

WORKED EXAMPLE 18 Paper 2 question (exam level)

A biased coin (P(H) = 0.8) and a fair coin are in a bag. One is chosen at random and tossed three times, giving three heads. Find P(biased coin).

Solution

P(HHH | biased) = 0.512; P(HHH | fair) = 0.125A1

P(biased | HHH) = 0.256/(0.256 + 0.0625)M1

= 0.804A1

Practice exercise 16

1. In the same setting, the result is HHT. Find P(biased coin).

21 Bayes' theorem summary and strategy

A reliable method.

1. Define events and identify the partition. 2. Draw a tree (or two-way table). 3. Find P(B) by total probability. 4. Divide the relevant path by P(B). 5. Interpret in context, considering the prior.

WORKED EXAMPLE 19 Strategy (standard)

List the probabilities needed to find P(A | B) from a tree.

Solution

P(A), P(B | A), P(A′), P(B | A′)A1

22 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 2 style) (about 15 minutes)

In a city, 30% of commuters cycle and the rest drive. 10% of cyclists and 25% of drivers are late for work on a given day.

(a) Find the probability that a randomly chosen commuter is late. [3]

(b) Given that a commuter is late, find the probability that they cycle. [3]

Markscheme

Part	Answer	Marks
(a)	0.3(0.1) + 0.7(0.25)	M1 A1
	0.205	A1
(b)	0.03/0.205	M1 A1
	0.146	A1
(c)	0.7(0.75)/0.795	M1 A1
	0.660	A1

Examiner's comments. A clearly labelled tree diagram earned the method marks even with arithmetic slips.

23 Common misconceptions and links

Students often think…	Actually…
P(A \| B) = P(B \| A)	Bayes' theorem relates them via P(A) and P(B)
an accurate test gives accurate positives	the base rate matters
P(B) = P(B \| A)	use total probability

How 4.13 connects to the rest of the course

Topic	Connection
4.5 Probability	conditional probability
4.7 Discrete random variables	distributions
4.1 Statistics	relative frequency

24 Check your understanding

Section A: multiple choice

1. P(B) =

A. P(B | A)P(A) + P(B | A′)P(A′) B. P(B | A) C. P(A)P(B) D. P(A | B)

2. The prior probability is

A. after the evidence B. always 0.5 C. the probability before the evidence D. the false positive rate

3. A rare condition with an accurate test means P(condition | positive) is

A. always above 0.9 B. possibly small C. equal to the sensitivity D. 1

Section B: structured

4. P(A) = 0.4, P(B | A) = 0.7, P(B | A′) = 0.2. Find P(A | B). [3]

5. Explain in words why a positive test for a rare disease is often a false positive. [2]

Answers

1. A 2. C 3. B

4. 0.28/(0.28 + 0.12) = 0.7.

5. The healthy group is so large that even a small false positive rate produces more positives than the small diseased group.

25 Summary and topic map

Idea	Remember
Conditional	P(A \| B) = P(A ∩ B)/P(B)
Total probability	P(B) = ΣP(B \| Ai)P(Ai)
Bayes	P(A \| B) = P(B \| A)P(A)/P(B)
Base rate	low prevalence reduces P(condition \| positive)
Updating	posterior becomes the new prior

Before the exam, make sure you…

•draw the tree
•write the target probability in notation
•find P(B) first
•interpret in context

Bayes' Theorem on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 3. Topic map for Bayes' Theorem.

26 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. 0.018/(0.018 + 0.049) = 0.269

2. 0.18/(0.18 + 0.04) = 0.818: a higher base rate makes a positive result far more reliable

Practice exercise 2

1. 0.06/0.115 = 0.522

2. (1/3)(4/5) ÷ [(1/3)(2/5 + 4/5 + 1/2)] = 0.8/1.7 = 0.471

Practice exercise 3

1. 0.587(0.95)/(0.587(0.95) + 0.413(0.08)) = 0.944

Practice exercise 4

1. no: they overlap (4, 6) and miss 1, 3

Practice exercise 5

1. 0.12 + 0.07 = 0.19

Practice exercise 6

1. 0.18/0.81 = 0.222

Practice exercise 7

1. 0.441/0.9685 = 0.455

Practice exercise 8

1. 0.9405/0.941 = 0.99947

Practice exercise 9

1. 0.059

Practice exercise 10

1. 0.785(0.95)/(0.785(0.95) + 0.215(0.05)) = 0.986

Practice exercise 11

1. 0.3/0.888 = 0.338

Practice exercise 12

1. P(RB | A) = 0.6, P(RB | B) = 0.4: 0.3/0.5 = 0.6

Practice exercise 13

1. 0.483 + 0.517 = 1

Practice exercise 14

1. P(A | B) = 0.75 ≠ P(A) = 0.6

Practice exercise 15

1. 0.95p = 0.05(1 − p): p = 0.05

Practice exercise 16

1. 0.5(0.128)/(0.5(0.128) + 0.5(0.125)) = 0.506

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