Module 32 of 46Lesson 1 of 2 · HL · AA 4.13 — Bayes’ theorem

Course map · Topic 4: Statistics and Probability

4.13 Bayes’ theorem

HL only11 worked questions

Bayes’ theorem turns a conditional probability around: from P(positive test | disease), which a laboratory measures, to P(disease | positive test), which a patient cares about. In AA HL it appears with two or three possible causes, and the working is almost always by hand from a tree diagram.

✏️ Start with the questions. Try each one on paper, then open the solution and compare.

Stuck? Open the notes ↓

✏️ Practice questions

Try each question on paper first. Then open the solution and compare every step.

Question 1●○○ Warm-up✍️ By hand · Paper 1 · no calculator[4 marks]

P(A)=0.3

P(B\mid A)=0.6

and

P(B\mid A')=0.2

. Find (a)

P(B)

; (b)

P(A\mid B)

Show the full step-by-step solution

Step 1🧠 ThinkRead the tree

First branch:

A

A'

(

0.3

and

0.7

). Second branch:

B

or not. The probabilities

0.6

and

0.2

are on the

B

branches.

Step 2✍️ By hand(a) Add both paths that end in B

P(B)=0.3\times0.6+0.7\times0.2

=0.18+0.14=0.32

Step 3✍️ By hand(b) The path through A, out of all of B

P(A\mid B)=\dfrac{0.18}{0.32}

=\dfrac{9}{16}

Final answer (a)

0.32

(b)

\dfrac{9}{16}

Question 2●○○ Warm-up✍️ By hand · Paper 1 · no calculator[5 marks]

A factory has three machines. Machine A makes 50% of the items, B makes 30% and C makes 20%. The percentages of defective items are 2% for A, 3% for B and 5% for C. An item is chosen at random and is found to be defective. Find the probability that it was made by machine C.

Show the full step-by-step solution

Step 1🧠 ThinkThree causes

The defective item could come from A, B or C. Bottom: the total probability of “defective”. Top: the path through C.

Step 2✍️ By handTotal probability of a defective

P(D)=0.5\times0.02+0.3\times0.03+0.2\times0.05

=0.01+0.009+0.01=0.029

Step 3✍️ By handBayes

P(C\mid D)=\dfrac{0.2\times0.05}{0.029}=\dfrac{0.01}{0.029}

M1A1

=\dfrac{10}{29}

💡 Remember C makes only 20% of the items, but over a third of the defective ones.

Final answer

\dfrac{10}{29}

Question 3●○○ Warm-up✍️ By hand · Paper 1 · no calculator[4 marks]

Box A contains 3 red and 2 blue counters. Box B contains 1 red and 4 blue counters. A fair coin is tossed: heads → a counter is taken from box A, tails → from box B. The counter is red. Find the probability that it came from box A.

Show the full step-by-step solution

Step 1🧠 ThinkCauses: box A or box B, each $\dfrac{1}{2}$

P(red | A) =

\dfrac{3}{5}

, P(red | B) =

\dfrac{1}{5}

Step 2✍️ By handBayes

P(R)=\dfrac{1}{2}\times\dfrac{3}{5}+\dfrac{1}{2}\times\dfrac{1}{5}=\dfrac{4}{10}

M1A1

P(A\mid R)=\dfrac{\frac{3}{10}}{\frac{4}{10}}

=\dfrac{3}{4}

Final answer

\dfrac{3}{4}

Question 4●●○ Standard✍️ By hand · Paper 1 · no calculator[5 marks]

1% of a population has a certain disease. A test gives a positive result for 95% of people with the disease and for 5% of people without it. A person tests positive. Find the probability that they have the disease, as a fraction in its simplest form.

Show the full step-by-step solution

Step 1🧠 ThinkName the events

D

= has the disease,

+

= tests positive. Known:

P(D)=0.01

P(+\mid D)=0.95

P(+\mid D')=0.05

. Wanted:

P(D\mid +)

Step 2✍️ By handTotal probability of a positive test

P(+)=0.01\times0.95+0.99\times0.05

=0.0095+0.0495=0.059

Step 3✍️ By handBayes

P(D\mid +)=\dfrac{0.0095}{0.059}=\dfrac{95}{590}

M1A1

=\dfrac{19}{118}\ (\approx0.161)

💡 Why so small? In 10 000 people: 95 true positives but 495 false positives. Most positives come from the huge healthy group.

Final answer

\dfrac{19}{118}\approx0.161

Question 5●●○ Standard✍️ By hand · Paper 1 · no calculator[4 marks]

In Maun the probability of rain on a day in March is

0.2

. When it will rain, the forecast says “rain” with probability

0.9

; when it will not rain, the forecast says “rain” with probability

0.15

. The forecast says “rain”. Find the probability that it does rain.

Show the full step-by-step solution

Step 1✍️ By handBottom: all ways the forecast says rain

P(F)=0.2\times0.9+0.8\times0.15=0.18+0.12=0.3

M1A1

Step 2✍️ By handTop: rain and forecast rain

P(R\mid F)=\dfrac{0.18}{0.3}

=0.6

Final answer

0.6

Question 6●●○ Standard✍️ + 🖩 · Paper 2[5 marks]

For the test in Question 4 (prevalence 1%,

P(+\mid D)=0.95

P(+\mid D')=0.05

), a person who tests positive takes the test a second time, independently, and again tests positive. Find the probability that the person has the disease.

Show the full step-by-step solution

Step 1🧠 ThinkThe old answer becomes the new prior

After the first positive,

P(D)=\dfrac{19}{118}\approx0.161017

. Use this as the starting probability and apply Bayes again.

Step 2✍️ By handSet up

P(D\mid ++)=\dfrac{0.161017\times0.95}{0.161017\times0.95+0.838983\times0.05}

M1A1

Step 3🖩 On the GDCStore the prior, then compute

19/118›sto→💻 type :=›q›enter💻 Enter›q·0.95/(q·0.95+(1-q)·0.05)›ctrl+enter💻 Ctrl+Enter

press ctrl+enter for a decimal

Step 4✍️ By handAnswer

P(D\mid ++)=0.785

(exactly

\dfrac{361}{460}

)A1A1

Final answer

0.785

Question 7●●○ Standard✍️ By hand · Paper 1 · no calculator[4 marks]

P(A)=p

P(B\mid A)=0.8

and

P(B\mid A')=0.3

. Given that

P(A\mid B)=0.6

, find

p

Show the full step-by-step solution

Step 1🧠 ThinkWrite Bayes with p and solve

P(A')=1-p

. The equation is linear in

p

once we multiply out the fraction.

Step 2✍️ By handEquation

\dfrac{0.8p}{0.8p+0.3(1-p)}=0.6

M1A1

0.8p=0.48p+0.18-0.18p

0.5p=0.18\Rightarrow p=0.36

Final answer

p=0.36

Question 8●●● Exam-level✍️ + 🖩 · Paper 2[8 marks]

A supermarket in Gaborone buys eggs from three farms: 45% from farm A, 35% from farm B and 20% from farm C. The proportions of cracked eggs are 1%, 2% and 4% respectively.

(a) Find the probability that a randomly chosen egg is cracked.[2]
(b) Given that an egg is cracked, find the probability that it came from each farm.[3]
(c) Ten cracked eggs are chosen at random from a very large batch of cracked eggs. Find the probability that at least 4 of them came from farm C.[3]

Show the full step-by-step solution

Part (a)

Step 1🖩 On the GDCTotal probability, stored as t

0.45·0.01+0.35·0.02+0.2·0.04›sto→💻 type :=›t›enter💻 Enter

Part (b)

Step 1🖩 On the GDCEach branch divided by t

Step 2✍️ By handAnswers

P(A\mid C_r)=0.231,\ P(B\mid C_r)=0.359,\ P(C\mid C_r)=0.410

A1A1A1

Part (c)

Step 1🧠 ThinkBinomial with p from part (b)

X

= number from farm C among 10 cracked eggs,

X\sim B(10,\ 0.410256)

Step 2🖩 On the GDCBinomial Cdf from 4 to 10

Final answer (a)

0.0195

(b)

0.231,\ 0.359,\ 0.410

(c)

0.643

Question 9●●● Exam-level✍️ By hand · Paper 1 · no calculator[6 marks]

A bag contains two coins: one fair coin and one coin with heads on both sides. A coin is chosen at random and tossed

n

times. It shows heads every time.

(a) Show that the probability that the double-headed coin was chosen is $\dfrac{2^{n}}{2^{n}+1}$ .[4]
(b) Find the least value of $n$ for which this probability is greater than $0.99$ .[2]

Show the full step-by-step solution

Part (a)

Step 1✍️ By handBayes with H = n heads in a row

P(H\mid \text{double})=1,\quad P(H\mid \text{fair})=\left(\dfrac{1}{2}\right)^{n}

P(\text{double}\mid H)=\dfrac{\frac{1}{2}\times1}{\frac{1}{2}\times1+\frac{1}{2}\left(\frac{1}{2}\right)^{n}}

Multiply top and bottom by

2\times2^{n}

=\dfrac{2^{n}}{2^{n}+1}

M1AG

Part (b)

Step 1✍️ By handInequality

\dfrac{2^{n}}{2^{n}+1}\gt 0.99\Rightarrow2^{n}\gt 0.99\cdot2^{n}+0.99\Rightarrow0.01\cdot2^{n}\gt 0.99

2^{n}\gt 99

;

2^{6}=64,\ 2^{7}=128\Rightarrow n=7

Final answer (a) shown (b)

n=7

Question 10●●● Exam-level✍️ + 🖩 · Paper 2[7 marks]

40% of the emails Boitumelo receives are spam. The word “winner” appears in 35% of spam emails and in 5% of other emails.

(a) Find the probability that an email contains “winner”.[2]
(b) An email contains “winner”. Find the probability that it is spam.[2]
(c) On one day 6 emails contain “winner”. A filter deletes them all. Find the probability that at least 5 of the deleted emails really were spam.[3]

Show the full step-by-step solution

Part (a)

Step 1✍️ By handTotal probability

P(W)=0.4\times0.35+0.6\times0.05=0.17

M1A1

Part (b)

Step 1✍️ By handBayes

P(S\mid W)=\dfrac{0.14}{0.17}=\dfrac{14}{17}\approx0.824

M1A1

Part (c)

Step 1🧠 ThinkEach deleted email is spam with probability $\dfrac{14}{17}$

Y\sim B\left(6,\ \dfrac{14}{17}\right)

; we want

P(Y\ge5)

Step 2🖩 On the GDCBinomial Cdf

binomCdf(6,14/17,5,6)›enter💻 Enter

📍 Where to find it on the TI-Nspire

binomCdf(is inmenu💻 click menu on keypad›5: Probability›5: Distributions›B: Binomial Cdf…

Final answer (a)

0.17

(b)

\dfrac{14}{17}\approx0.824

(c)

0.713

Question 11●●● Exam-level✍️ + 🖩 · Paper 3[15 marks]

Investigating a screening test. A disease affects a proportion

p

of a population. A screening test is positive for 98% of people with the disease and for 4% of people without it. Let

f(p)=P(D\mid +)

(a) Show that $P(+)=0.94p+0.04$ .[2]
(b) Hence show that $f(p)=\dfrac{0.98p}{0.94p+0.04}$ .[1]
(c) Find $f(0.01)$ and $f(0.05)$ .[2]
(d) Find the exact value of $p$ for which $f(p)=0.5$ .[3]
(e) Sketch $y=f(p)$ for $0\le p\le0.2$ and describe what happens to $f(p)$ as $p$ increases.[2]
(f) For $p=0.01$ , a person tests positive $k$ times in independent tests. Show that $P(D\mid k \text{ positives})=\dfrac{0.01\times0.98^{k}}{0.01\times0.98^{k}+0.99\times0.04^{k}}$ , and find the least $k$ for which this exceeds $0.99$ .[5]

Show the full step-by-step solution

Part (a)

Step 1✍️ By handTotal probability

P(+)=0.98p+0.04(1-p)

=0.94p+0.04

Part (b)

Step 1✍️ By handBayes

f(p)=\dfrac{P(D)P(+\mid D)}{P(+)}=\dfrac{0.98p}{0.94p+0.04}

Part (c)

Step 1🖩 On the GDCDefine f, then evaluate

f(p):=0.98·p/(0.94·p+0.04)›enter💻 Enter›f(0.01)›enter💻 Enter›f(0.05)›enter💻 Enter

📍 Where to find it on the TI-Nspire

⌨️ store a value: type := (colon, then equals) on the keyboard

Part (d)

Step 1✍️ By handSolve by hand

0.98p=0.5(0.94p+0.04)

0.51p=0.02

p=\dfrac{2}{51}\ (\approx0.0392)

Part (e)

Step 1🖩 On the GDCGraph f1(x) = $\dfrac{0.98x}{0.94x+0.04}$

Increasing, steep at first, then levelling off

Step 2✍️ By handDescribe

f

increases: the rarer the disease, the less a positive result means; it rises steeply for small

p

and then levels off towards 1A1A1

Part (f)

Step 1✍️ By handk independent positives

P(k+\mid D)=0.98^{k},\ P(k+\mid D')=0.04^{k}

P(D\mid k+)=\dfrac{0.01\times0.98^{k}}{0.01\times0.98^{k}+0.99\times0.04^{k}}

Step 2🖩 On the GDCTry k = 1, 2, 3

g(k):=0.01·0.98^k/(0.01·0.98^k+0.99·0.04^k)›enter💻 Enter

📍 Where to find it on the TI-Nspire

⌨️ store a value: type := (colon, then equals) on the keyboard

Step 3✍️ By handConclude

g(2)=0.858\lt 0.99,\ g(3)=0.993\gt 0.99

M1A1

least

k=3

Final answer (c)

0.198,\ 0.563

(d)

p=\dfrac{2}{51}

(e) increasing, levelling off (f)

k=3

🆘 Stuck on a question? Open the notes, GDC toolkit and common mistakes

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

🎯 What you must know

Idea: $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$ . Bayes just works out the top and the bottom from a tree drawn in the “other” order.
Two causes: $P(A\mid B)=\dfrac{P(A)P(B\mid A)}{P(A)P(B\mid A)+P(A')P(B\mid A')}$ .
Three or more causes $A_1, A_2, A_3$ (they must cover everything and not overlap): $P(A_i\mid B)=\dfrac{P(A_i)P(B\mid A_i)}{\sum_j P(A_j)P(B\mid A_j)}$ .
Tree method: (1) draw the tree with the causes first; (2) the bottom = add all branches that end in $B$ ; (3) the top = the one branch through the cause you want.
The bottom line is the law of total probability: $P(B)=\sum P(A_j)P(B\mid A_j)$ .
Rare conditions: even a good test gives many false positives, so $P(\text{disease}\mid +)$ can be surprisingly small. A second positive test updates the probability again (the old answer becomes the new prior).

Bayes’ theorem (two events)

P(B\mid A)=\dfrac{P(B)P(A\mid B)}{P(B)P(A\mid B)+P(B')P(A\mid B')}

📘 In the formula booklet

Bayes’ theorem (three events)

P(B_i\mid A)=\dfrac{P(B_i)P(A\mid B_i)}{P(B_1)P(A\mid B_1)+P(B_2)P(A\mid B_2)+P(B_3)P(A\mid B_3)}

📘 In the formula booklet

🖩 Your GDC toolkit for this subtopic

Follow the steps on your own TI-Nspire CX II. The screens show what you should see.

Store the bottom line (total probability), then divide

0.45·0.01+0.35·0.02+0.2·0.04›sto→💻 type :=›t›enter💻 Enter

Type 0.45·0.01+0.35·0.02+0.2·0.04.
Press sto→ (ctrl then var).
Type t.
Press enter.

Worked example: Question 11 — Define f, then evaluate

Exact fraction ↔ decimal

ctrl+enter💻 Ctrl+Enter

Press ctrl then enter to get the answer as a decimal.

press ctrl+enter for a decimal

Worked example: Question 6 — Store the prior, then compute

⚠️ Where students lose marks

Mixing up $P(B\mid A)$ and $P(A\mid B)$ . Read the sentence: the part after “given” is what you already know.
Putting only one branch in the denominator. The bottom must include every path to the observed event.
Using percentages and decimals together carelessly: 2% → 0.02, not 0.2.
Rounding the denominator early on Paper 1 — keep exact fractions.

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL HL · AA 4.13 — Bayes’ theorem

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL

4.13 Bayes’ theorem

✏️ Practice questions

🎯 What you must know

🖩 Your GDC toolkit for this subtopic

⚠️ Where students lose marks