DP Maths AI HL 4.5 Probability

DP Maths AI HL 4.5 Probability - Full notes

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DP Maths AI HL | 4.5 Probability

DP MATHEMATICS AI · TOPIC 4: STATISTICS AND PROBABILITY

4.5

Probability

Full notes for Applications and Interpretation HL

Contents

1	The language of probability
2	Theoretical probability
3	Experimental probability and relative frequency
4	Expected number of occurrences
5	Complementary events
6	Sample space diagrams
7	Two-way tables
8	Venn diagrams
9	The addition rule
10	Mutually exclusive events
11	Conditional probability
12	Independent events
13	Tree diagrams
14	Sampling with and without replacement
15	Reversing conditional probability with trees
16	Simulation and technology
17	Applications: testing, quality control and games
18	Mixed review: worked problems
19	Bayes' theorem (HL)
20	Probability and counting (HL)
21	Investigation: the birthday problem
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 practice exercise.

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
use the language of trials, outcomes, sample spaces and events
find theoretical and experimental probabilities and expected numbers
use sample space diagrams, two-way tables and Venn diagrams
apply the addition rule and identify mutually exclusive events
calculate conditional probabilities
test events for independence
use tree diagrams with and without replacement
solve probability problems in context

HL booklet: this booklet contains all the SL content plus Bayes' theorem and probability with counting.

Guiding questions

How can uncertainty be measured? How does new information change the probability of an event?

How this topic is examined

Probability is tested on both papers, usually with a two-way table, Venn diagram or tree diagram in context. Conditional probability ('given that...') appears in almost every examination.

Write events and probability statements clearly, show the calculation, and give answers as exact fractions or to 3 s.f.

Mathematics: Applications and Interpretation places strong emphasis on technology, modelling and interpretation. Most papers allow a GDC, so you are expected to use it efficiently, record the method you used, and interpret every answer in context.

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, including what you entered into a GDC
Show that	obtain the given result; every step must be shown and the result must not be assumed
Estimate	obtain an approximate value
Sketch	a diagram showing the general shape and key features, labelled
Interpret / Comment	explain the meaning of a result in context, or give a judgement
Justify / Explain	give valid reasons or evidence
Hence	use the previous result

How to use these notes

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or explanation of a key result
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
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), graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet

1 The language of probability

Trials, outcomes, events.

Think about it

A weather app says 'a 30% chance of rain'. What does this actually mean? How might the forecaster have arrived at 30%?

A trial is a single performance of an experiment. An outcome is a possible result; the sample space U lists all outcomes. An event A is a set of outcomes. n(A) denotes the number of outcomes in A. Probabilities lie between 0 (impossible) and 1 (certain).

2 Theoretical probability

Equally likely outcomes.

WORKED EXAMPLE 1 Cards (basic)

A card is drawn from a standard pack of 52. Find P(a red king) and P(a picture card).

Solution

2/52 = 1/26A1

12/52 = 3/13A1

Practice exercise 1

1. A letter is chosen from PROBABILITY. Find P(B).

2. A fair 12-sided die is rolled. Find P(a multiple of 3).

3 Experimental probability and relative frequency

Learning from data.

When outcomes are not equally likely, estimate P(A) by the relative frequency: the number of times A occurred divided by the number of trials. More trials give more reliable estimates (the law of large numbers).

Tosses of a drawing pin	Point up	Point down
Frequency (200 trials)	124	76

WORKED EXAMPLE 2 Relative frequency (standard)

Estimate P(point up) and comment on reliability.

Solution

124/200 = 0.62A1

200 trials gives a reasonable estimate; more trials would improve itR1

Practice exercise 2

1. A biased coin lands heads 340 times in 500. Estimate P(tails).

4 Expected number of occurrences

np.

WORKED EXAMPLE 3 Expected number (standard)

P(a seed germinates) = 0.85. How many of 600 seeds are expected to germinate?

Solution

600 × 0.85M1

= 510A1

Practice exercise 3

1. A fair die is rolled 180 times. How many sixes are expected?

2. Using the drawing pin data, how many 'point up' would you expect in 1500 tosses?

5 Complementary events

Not A.

A′ means 'A does not happen'. Complements are useful for 'at least one' problems: P(at least one) = 1 − P(none).

WORKED EXAMPLE 4 At least one (standard)

Three fair coins are tossed. Find P(at least one head).

Solution

P(no heads) = (1/2)3 = 1/8M1

= 1 − 1/8 = 7/8A1

Practice exercise 4

1. P(rain) = 0.35. Find P(no rain).

2. Find P(at least one 6) in four rolls of a die.

6 Sample space diagrams

Grids for two-stage experiments.

+	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

WORKED EXAMPLE 5 Two dice (standard)

Two dice are rolled and the scores added. Find P(sum = 7) and P(sum ≥ 10).

Solution

6 outcomes give 7: 1/6A1

6 outcomes give 10, 11 or 12: 1/6A1

Practice exercise 5

1. Find P(the sum is prime).

2. Find P(both dice show the same number).

7 Two-way tables

Two categorical variables.

	Walks	Bus	Car	Total
Year 11	32	45	23	100
Year 12	28	38	34	100
Total	60	83	57	200

WORKED EXAMPLE 6 Two-way table (standard)

A student is chosen at random. Find P(takes the bus), P(Year 12 and walks), and P(Year 11 or car).

Solution

83/200 = 0.415A1

28/200 = 0.14A1

(100 + 34)/200 = 0.67A1

Practice exercise 6

1. Find P(car | Year 12).

2. Find P(Year 11 | walks).

8 Venn diagrams

Overlapping sets.

A Venn diagram shows events as circles inside the rectangle U. A ∩ B ('A and B') is the overlap; A ∪ B ('A or B or both') is everything in either circle; A′ is everything outside A.

Figure 1. 50 students: F plays football, T plays tennis.

WORKED EXAMPLE 7 Venn diagram (standard)

Find P(F), P(F ∩ T), P(F ∪ T) and P(neither).

Solution

P(F) = 25/50 = 0.5; P(F ∩ T) = 0.14A1 A1

P(F ∪ T) = 37/50 = 0.74; P(neither) = 0.26A1 A1

Practice exercise 7

1. In a class of 30, 17 study French, 12 study Spanish and 6 study both. Draw a Venn diagram and find P(neither).

9 The addition rule

Avoid counting the overlap twice.

WORKED EXAMPLE 8 Addition rule (standard)

P(A) = 0.45, P(B) = 0.3, P(A ∩ B) = 0.12. Find P(A ∪ B).

Solution

0.45 + 0.3 − 0.12M1

= 0.63A1

Practice exercise 8

1. P(A ∪ B) = 0.8, P(A) = 0.5, P(B) = 0.6. Find P(A ∩ B).

10 Mutually exclusive events

Cannot happen together.

A and B are mutually exclusive if P(A ∩ B) = 0. Then P(A ∪ B) = P(A) + P(B). Example: rolling a 2 and rolling a 5 on one die.

Practice exercise 9

1. Are 'rolling an even number' and 'rolling a 3' mutually exclusive?

2. Are 'drawing a king' and 'drawing a heart' mutually exclusive?

11 Conditional probability

Given that...

P(A | B) is the probability of A given that B has occurred. Knowing B restricts the sample space to B. From a table or Venn diagram, count within the given group.

WORKED EXAMPLE 9 From the Venn diagram (standard)

Find P(T | F) and P(F | T).

Solution

P(T | F) = 7/25 = 0.28A1

P(F | T) = 7/19 = 0.368A1

Common mistake

Confusing P(A | B) with P(B | A). The event after the bar is the one known to have happened.

Practice exercise 10

1. P(A ∩ B) = 0.18, P(B) = 0.45. Find P(A | B).

2. Two dice: given that the sum is 9, find P(at least one 6).

12 Independent events

One does not affect the other.

WORKED EXAMPLE 10 Testing independence (exam level)

Are F and T independent?

Solution

P(F)P(T) = 0.5 × 0.38 = 0.19M1

P(F ∩ T) = 0.14 ≠ 0.19A1

Not independentR1

Practice exercise 11

1. P(A) = 0.3, P(B) = 0.5 and A, B are independent. Find P(A ∪ B).

2. Explain why two mutually exclusive events with non-zero probabilities cannot be independent.

13 Tree diagrams

Multiply along, add between.

Tree diagrams show sequences of events. Probabilities on branches from one point add to 1. Multiply along a path for the probability of a combined outcome; add the probabilities of the paths that satisfy an event.

Figure 2. L = late for the bus; W = walks to school.

WORKED EXAMPLE 11 Using a tree (standard)

Using the tree, find P(W).

Solution

0.3 × 0.4 + 0.7 × 0.75M1

= 0.12 + 0.525 = 0.645A1

Practice exercise 12

1. Find P(L ∩ W′).

2. A team wins 60% of home games and 35% of away games. Half of games are at home. Find P(win).

14 Sampling with and without replacement

Does the second stage change?

With replacement, the probabilities stay the same for each draw (independent). Without replacement, the second-stage probabilities change because the population is smaller.

WORKED EXAMPLE 12 Without replacement (standard)

Two sweets are taken from a bag of 5 red and 7 green without replacement. Find P(both red) and P(one of each).

Solution

(5/12)(4/11) = 5/33A1

(5/12)(7/11) + (7/12)(5/11) = 35/66M1 A1

Practice exercise 13

1. Repeat the first part with replacement.

2. Three cards are drawn without replacement from a pack. Find P(all hearts).

15 Reversing conditional probability with trees

Given the outcome, find the cause.

WORKED EXAMPLE 13 Given W, find L (exam level)

Using the tree, find P(L | W).

Solution

P(L ∩ W) = 0.12; P(W) = 0.645A1

P(L | W) = 0.12/0.645 = 0.186M1 A1

Practice exercise 14

1. A factory: machine A makes 60% of items (3% defective), B makes 40% (5% defective). An item is defective. Find P(made by B).

16 Simulation and technology

Estimating by repetition.

Random number generators simulate experiments that are hard to analyse exactly. For example, randInt(1, 6) simulates a die. Repeating a simulation many times gives a relative frequency estimate of a probability.

GDC tip

Use randInt(1, 6, 100) to generate 100 die rolls in a list, then count the sixes. Use randBin for binomial experiments.

17 Applications: testing, quality control and games

Probability in decisions.

WORKED EXAMPLE 14 Screening test (exam level)

2% of a population has a condition. A test detects it 95% of the time and gives a false positive 6% of the time. Find P(positive) and P(condition | positive).

Solution

P(+) = 0.02(0.95) + 0.98(0.06) = 0.0778M1 A1

P(C | +) = 0.019/0.0778 = 0.244M1 A1

Most positive results come from healthy people because the condition is rare.

Practice exercise 15

1. A game costs $2. You roll a die: a 6 wins $10. Find the expected gain.

2. A batch is accepted if two randomly chosen items (without replacement) are both good. A batch of 20 has 3 defective. Find P(accepted).

18 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 15 Mixed problem (exam level)

P(A) = 0.4, P(B | A) = 0.25, P(B | A′) = 0.5. Find P(B) and P(A | B).

Solution

P(B) = 0.1 + 0.3 = 0.4M1 A1

P(A | B) = 0.1/0.4 = 0.25A1

Practice exercise 16

1. P(A) = 0.6, P(B) = 0.5, P(A ∪ B) = 0.8. Find P(A | B) and decide whether A and B are independent.

19 Bayes' theorem (HL)

Reversing conditional probabilities formally.

The denominator is P(B) by the law of total probability. Bayes' theorem explains why rare conditions give many false positives, and how evidence updates a prior probability to a posterior.

WORKED EXAMPLE 16 Three causes (exam level)

Suppliers X, Y, Z provide 50%, 30%, 20% of parts with fault rates 1%, 2%, 4%. A part is faulty. Find P(from Z).

Solution

P(F) = 0.005 + 0.006 + 0.008 = 0.019M1 A1

P(Z | F) = 0.008/0.019 = 0.421A1

Practice exercise 17

1. A test with sensitivity 0.99 and false positive rate 0.01 is used where prevalence is 0.1%. Find P(condition | positive).

20 Probability and counting (HL)

Using nCr.

When outcomes are equally likely, count with combinations: P = (number of favourable selections)/(number of possible selections).

WORKED EXAMPLE 17 Lottery (standard)

In a lottery, 6 numbers are chosen from 49. Find P(winning with one ticket) and P(exactly 3 correct).

Solution

1/49C6 = 7.15 × 10−8A1

6C3 × 43C3/49C6 = 0.0177M1 A1

Practice exercise 18

1. A committee of 3 is chosen from 5 men and 4 women. Find P(at least one woman).

21 Investigation: the birthday problem

A Paper 3 style exploration.

1. Find the probability that two people have different birthdays (365 days, equally likely). 2. Show that for n people, P(all different) = (365/365)(364/365)...((365 − n + 1)/365). 3. Use technology to find the smallest n for which P(at least two share a birthday) > 0.5. 4. Find this probability for n = 30 and n = 50. 5. Discuss why the assumptions (no leap years, equally likely birthdays) might not hold.

Answers. 1. 364/365. 3. n = 23 (P = 0.507). 4. 0.706; 0.970. 5. Birth rates vary seasonally; 29 February exists; twins.

22 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 10 minutes)

In a group of 80 people, 45 own a car (C), 30 own a bicycle (B) and 15 own both.

(a) Draw a Venn diagram. [2]

(b) Find P(C ∪ B). [2]

(d) Determine whether C and B are independent. [2]

Markscheme

Part	Answer	Marks
(a)	30, 15, 15, 20	A2
(b)	60/80 = 0.75	M1 A1
(c)	15/45 = 1/3	M1 A1
(d)	P(C)P(B) = 0.5625 × 0.375 = 0.211 ≠ 0.1875	M1
	not independent	R1

Examiner's comments. Show the comparison numerically.

Question 2 (Paper 2 style) (about 12 minutes)

A bag contains 4 blue and 6 yellow counters. Two are taken without replacement.

(a) Draw a tree diagram. [2]

(b) Find P(both yellow). [2]

(d) Given that the second is blue, find P(the first is yellow). [2]

Markscheme

Part	Answer	Marks
(a)	correct branches 4/10, 6/10; 3/9, 6/9; 4/9, 5/9	A2
(b)	(6/10)(5/9) = 1/3	M1 A1
(c)	(4/10)(3/9) + (6/10)(4/9) = 2/5	M1 A1
(d)	(24/90)/(36/90)	M1
	2/3	A1

Examiner's comments. Part (d) reverses the conditioning.

Question 3 (Paper 2 style) (about 8 minutes)

The two-way table in this booklet (transport by year group) is used.

(a) Find P(bus | Year 11). [2]

(b) Find the expected number of bus users in a school of 1200 with the same proportions. [2]

Markscheme

Part	Answer	Marks
(a)	45/100 = 0.45	A2
(b)	1200 × 0.415	M1
	498	A1

Examiner's comments. Use the overall proportion for (b).

23 Common misconceptions and links

Students often think…	Actually…
mutually exclusive = independent	different ideas
P(A \| B) = P(B \| A)	not in general
add along branches	multiply along, add between
relative frequency is exact	it is an estimate

How 4.5 connects to the rest of the course

Topic	Connection
4.2 Statistics	relative frequency
4.7 Distributions	random variables
4.8 Binomial distribution	repeated trials
1.9 Binomial theorem	counting

24 Check your understanding

Section A: multiple choice

1. P(A′) =

A. 1 − P(A) B. P(A) C. 0 D. 1

2. Independent events satisfy

A. P(A ∩ B) = 0 B. P(A) = P(B) C. P(A ∩ B) = P(A)P(B) D. P(A ∪ B) = 1

3. Expected number =

A. P(A)/n B. nP(A) C. n + P(A) D. n

4. Two dice: P(sum 2) =

A. 1/6 B. 1/12 C. 2/36 D. 1/36

5. Without replacement, the trials are

A. independent B. dependent C. mutually exclusive D. impossible

Section B: structured

6. P(A) = 0.7, P(B) = 0.4, independent. Find P(A ∪ B). [2]

7. Find P(two cards drawn without replacement are both aces). [2]

Answers

1. A 2. C 3. B 4. D 5. B

6. 0.82

7. 1/221

25 Summary and topic map

Idea	Remember
Theoretical	n(A)/n(U)
Experimental	relative frequency
Expected number	np
Addition rule	P(A) + P(B) − P(A ∩ B)
Conditional	P(A ∩ B)/P(B)
Independence	P(A ∩ B) = P(A)P(B)
Trees	multiply along, add between

Before the exam, make sure you…

•define events
•use the right conditioning
•test independence numerically
•adjust without replacement

Probability 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 Probability.

26 Answers to practice exercises

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

Practice exercise 1

1. 2/11

2. 1/3

Practice exercise 2

1. 0.32

Practice exercise 3

1. 30

2. 930

Practice exercise 4

1. 0.65

2. 1 − (5/6)4 = 0.518

Practice exercise 5

1. 15/36 = 5/12

2. 1/6

Practice exercise 6

1. 34/100 = 0.34

2. 32/60 = 0.533

Practice exercise 7

1. 7/30

Practice exercise 8

1. 0.3

Practice exercise 9

1. yes

2. no: the king of hearts

Practice exercise 10

1. 0.4

2. 2/4 = 1/2

Practice exercise 11

1. 0.65

2. P(A ∩ B) = 0 but P(A)P(B) > 0

Practice exercise 12

1. 0.18

2. 0.475

Practice exercise 13

1. 25/144

2. (13/52)(12/51)(11/50) = 0.0129

Practice exercise 14

1. 0.02/(0.018 + 0.02) = 0.526

Practice exercise 15

1. 10/6 − 2 = −$0.33

2. (17/20)(16/19) = 0.716

Practice exercise 16

1. P(A ∩ B) = 0.3; P(A | B) = 0.6 = P(A): independent

Practice exercise 17

1. 0.00099/(0.00099 + 0.00999) = 0.0902

Practice exercise 18

1. 1 − 10/84 = 0.881

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