Working with probability

Contents

10.1 Review of probability

10.2 Finding probabilities

10.3 Using binomial probability Chapter summary

Chapter review

Syllabus subject matter

Applied statistical analysis ■ Use of relative frequencies to estimate probabilities; the notion of probabilities of individual values for discrete variables and intervals for continuous variables

■ Identification of the binomial situation and use of tables or technology for binomial probabilities

Quantitative concepts and skills ■ Tree diagrams as a tool for defining sample spaces and estimating probabilities

■ The summation notation:

Syllabus Guide

Chapter 10

x_i

i=1 n

∑

Working with probability

Working with probability

10.1

Review of probability

We often use the past as a guide to predicting the future. When 10 000 people were surveyed, 1523 said that they were left-handed. This statistic can be written as

≈ 0.15 = 15%

suggesting that about 15 out of every 100 people can be expected to be left-handed.

The probability is a measure of how likely an event or outcome is to occur. Based on the above results, we would say that the probability that the next person surveyed is left-handed is about 0.15 or 15%. It doesn’t mean that there will be exactly 15 left-handed people in the next 100, but that in the long run the chance of picking a left-handed person will be 15%.

If a lot of people become sick in a particular workplace, how do doctors work out whether it is just coincidence or whether there is a problem? How do bookmakers set their odds? How do police forensic scientists work out whether a sample from a crime scene is significant? In all these cases, probability plays an important role.

1523 10 000

---Experimental probability

Experimental probability (empirical probability) uses data from experiments or statistics to work out chances.

Every item of data collected, or every attempt at an experiment, is called a trial. The result of each trial is called an outcome.

The outcome we are checking is called the favourable outcome or success. The number of times a particular outcome occurs is called its frequency.

Relative frequency

Empirical probabilities are calculated using the relative frequency, which is calculated using the formula

Relative frequency =frequency of favourable outcome number of trials

---!

In a survey of newspaper preference, 300 preferred the Courier-Mail, 260 the Australian, 180 the Australian Financial Review and 100 another paper. What is the probability of choosing someone at random who prefers the Courier-Mail?

Solution

Calculate the total number of trials. Number of trials = 300 + 260 + 180 + 100

Evaluate. = 840

Calculate the relative frequency of

the Courier-Mail. Relative frequency =

Evaluate. ≈ 0.36

State the result. The probability of choosing someone at random

who prefers the Courier-Mail is about 0.36 or 36%. 300

840

---Example

1

If the favourable outcome occurred every time, the relative frequency would be 1. If it never

occurred, the relative frequency would be 0. These are the extreme values because an outcome must occur either once or zero times in each trial.

We can use the relative frequency to predict the number of favourable outcomes in the future. In this investigation, you will need two normal 6-sided dice, and you should work in pairs.

1 Working in pairs, throw the dice and write down the smaller of the upper faces, giving a number from 1 to 6. Repeat this 50 times.

2 As a class, collate the frequencies for each pair in a large table with headings as follows.

3 Calculate the relative frequency of each number, using the totals for the whole class.

4 Choose one outcome, say a smaller number of 4.

a Work out the relative frequency of this outcome for each pair.

b Make a box-and-whisker plot of the relative frequencies for the outcome.

c What can you say about the frequency of this outcome for a limited number of trials? d What can you say about the frequency of this outcome for a large number of trials?

5 Make another box-and-whisker plot for a second outcome. Compare the two plots.

6 Write a report of your findings, including all information, plots and conclusions about the frequencies of outcomes.

Smaller number

Pair 1 2 3 4 5 6

1

2

3

.

.

.

50

Total

Investigation

Throwing dice

Expected frequency

Probabilities lie between 0 and 1. A probability of 1 means the outcome always occurs. A probability of 0 means it never occurs.

The expected frequency of an outcome is calculated by

Expected frequency = relative frequency × number of trials

!

Working with probability

When a normal die is rolled, we consider that it is equally likely to land with 1, 2, 3, 4, 5 or 6 uppermost. A die that behaves in this way is known as a fair die because no outcome is favoured more than another. From experience, we assume that for many trials the probability

of each outcome will become close to since there are 6 faces.

Weather Bureau records show that the probability of rain in Tully in November is 35% each day. How many wet days would you expect in a school week in Tully in November?

Solution

State the number of trials. There are 5 days in the school week.

∴ Number of trials = 5

Write the rule. Expected frequency = relative frequency × no. of trials

Substitute relevant values. = 0.35 × 5

Evaluate. = 1.75

Round off. ≈ 2

State the result. About 2 wet days can be expected in a November

school week in Tully.

Example

2

For this investigation you will need a number of small rectangular cardboard boxes of different sizes, such as matchboxes, cigarette boxes and jewellery boxes.

1 Measure the faces of each box and calculate the area of each face. Number the faces from 1 to 6 and express the area of each face as a percentage of the total surface area of the box.

2 Predict the relative frequency of landing on each face when the box is flipped into the air.

3 Flip each box in the air at least 100 times and each time note the number of the side it lands on. Work out the relative frequency for each side. Present these results in a table.

Investigation

Flipping boxes

4 Compare the area of each face with the relative frequency of landing on that face.

5 Compare your results with those of others.

6 Make a conclusion about the relationship between the area of each face and the relative frequency of landing on that face.

7 Write a report of your findings, including all data, calculations and conclusions.

1 6

In cases like this, we assume the results without performing the experiments. Since the

probability is determined without experiment, we call this theoretical probability. In theoretical probability it is important to be able to list all the possible outcomes of an experiment. To calculate the probability of a particular outcome, we divide the number of favourable outcomes by the total of all possible outcomes, assuming that all outcomes are equally likely.

Theoretical probability

Theoretical probability is calculated using a list of equally likely possible outcomes. Each possible outcome is called a sample point or an element.

The list of possible outcomes is called the sample space. If an outcome can occur in more than one way, it is listed the corresponding number of times.

The number of elements in a sample space is written as n(sample space). An event is a subset of the sample space. It may have more than one element. The number of elements in an event is written as n(event).

The probability of an event is written as P(event) or P(E) and calculated using the formula

P(event) = n(event)

n(sample space)

---!

List the sample space for tossing two coins.

Solution

We could get a head or a tail on the first coin and a head or a tail on the second coin.

Sample space = {HH, HT, TH, TT}

Example

3

A bag contains 3 red and 5 green marbles. List the sample space for drawing out 1 marble.

Solution

There are 3 red marbles, so R is listed 3 times. Sample space = {R, R, R, G, G, G, G, G}

Write in a more convenient form. Sample space = {3R, 5G}

Example

4

Working with probability

Probability can be given as a fraction, decimal or percentage, and the probabilities of events can be compared to see which events are more likely.

Exercise 10.1

Review of probability

1 The watches worn by a group of students at a school were checked against the actual time. The approximate amounts they were out (in seconds) were as follows:

+10 0 0 +20 0 −10 −50 +40 +30 0 0 +20

0 0 0 0 −30 −10 0 0 +30 −20 +50 +50

−50 0 −30 0 +20 0 −10 −30 +10 0 0 −10

−20 +20 0 0 +10 0 −10 −20 0 0 +20 0

Random selection

The word ‘random’ is used to refer to the selection of a particular item or group from a list in such a way that each has an equal chance to be selected.

!

An event that is certain to happen has a probability of 1. An impossible event has a probability of 0.

A more likely event has a higher probability than a less likely event.

!

A cashbox contains 8 twenty-cent coins, 4 fifty-cent coins, 6 two-dollar coins, 10 one-dollar coins and 7 five-cent coins. A coin is taken out at random. a What is the probability that it is worth more than 1 cent?

b What is the probability that it is a ten-cent coin? c What is the probability that it is a fifty-cent coin? d Which is more likely: a ‘gold’ coin or a ‘silver’ coin?

Solution

a Certain event. P( 1c) = 1

b Impossible event. P(10c) = 0

c Write down the rule for probability. P(50c) =

Add up the number of coins. n(sample space) = 35

Substitute relevant values and evaluate. P(50c) =

Evaluate. ≈ 0.114

d Start by calculating P(silver) and P(gold).

‘Silver’ includes the 20c, 50c and 5c coins. P(silver) = ≈ 0.543

‘Gold’ includes the $1 and $2 coins. P(gold) = ≈ 0.457

Compare P(gold) and P(silver). P(silver) P(gold)

State the result. A ‘silver’ coin is more likely.

n(50c) n(sample space)

---4 35

---19 35

---16 35

---Example

5

Additional Exercise

10 .1

a What was the number of trials?

b What was the frequency of the correct time (0)? c What was the relative frequency of the correct time?

d What was the relative frequency of being 20 seconds fast (+20)?

e If you checked another group of 20 students at the school, how many would you expect to have the correct time?

2 A packaging machine fills 100 g packets of potato chips. As part of a quality assurance program, the contents of 40 packets are weighed. The results are shown in the table.

Estimate the probability that the next packet filled by the machine will: a contain less than 95 g of chips b contain at least 110 g of chips c be underweight ( 100 g).

3 The reasons that Queensland drivers had their licences cancelled or were disqualified from holding a driver’s licence during a particular year are shown in the table.

Use this information to estimate the probability that the next cancellation or disqualification results from:

a speeding b the accumulation of demerit points

c an alcohol-related cause.

4 Ten identical pieces of cardboard are tossed into a wastepaper basket and the number of cards that make it into the bin is counted. The results are:

0 1 1 1 1 2 3 3 3 3 1 1 1 1 2 2 2 2 2 2 1 1 1 1 1 4 4 5 5 2 2 2 6 5 1 2 2 5 5 2 2 2 2 2 2 2 2 4 6 6 2 2 3 3 3 3 2 2 2 2 2 0 0 0 3 3 4 4 4 4 0 2 2 3 3 3 3 3 3 0 0 2 2 6 7 2 3 3 3 3 3 3 3 4 4 4 4 6 6 9 a How many trials were conducted in this experiment?

b Construct a frequency distribution table for this data.

c If the 10 cards are again thrown into the same wastepaper basket in the same way, what chance is there that exactly 2 will land in the basket?

Number of packets 3 8 21 7 1

Weight of contents (g) 90–94 95–99 100–104 105–109 110–114

Reason for cancellation Number of drivers

Alcohol related: • blood alcohol > 0.05%

• refusal to provide breath sample • other

3 357 417 525

Dangerous driving 1 589

Speeding 3 792

Driving under disqualification 1 516

Driving without a licence 1 413

Demerit points suspension 3 159

Other 2 846

Working with probability

5 People walking down George Street were asked how they had travelled to the city.

a Find the relative frequency for each method of transport.

b What is the probability that the next person spoken to would have travelled by car?

c If you asked another 20 people, how many would you expect to have travelled by train?

d For 30 people, how many could you expect to travel by bus?

6 In cases of the measles, complications are relatively common. From 4000 cases in children in a recent outbreak, 54 developed complications of some sort, resulting in death for 3 children. In 5 cases where children survived, they suffered serious brain damage, and in 2 other cases the children became deaf.

Complications from the measles vaccination also occur. From 50 000 vaccinations, 80 children had some complications. In 75 of these, it was a fever that subsided after a few days. In 1 case, the child suffered some loss of hearing, and the other 4 cases developed mild cases of measles that did not result in any permanent loss of hearing.

a What is the relative frequency of complications from measles? b What is the probability of complications from the vaccine? c What is the probability of serious complications from measles? d What is the probability of serious complications from the vaccine?

e If 1000 children were vaccinated, how many would you expect to suffer complications?

7 A single card is drawn from a standard pack of 52 playing cards. a How many elements are in the sample space?

b Are all outcomes equally likely?

c List the outcomes for the event ‘A queen is selected’. d What is the probability that a queen is selected? e What is the probability that a heart is selected?

f What is the probability that the queen of hearts is selected?

8 A point is selected at random inside a circle. Find the probability that the point is closer to the centre of the circle than to its circumference.

9 A book has its pages numbered from i to xii (preliminary pages) and then from 1 to 142 (text pages). A page is selected at random.

a What is the probability that it is a preliminary page? b Which is more likely: a preliminary page or a text page?

c What is the probability that a page with three digits in the number is selected? d What is the probability that the page number is greater than 200?

e What is the probability that the page number is greater than 100? Main method of

transport

Number of people

Taxi 18

Train 101

Bus 78

Car 86

Bicycle 6

Other 11

10 A spinner made in the shape of a regular pentagon is spun. The equal-sized sectors are coloured blue, red, green, pink and orange. a List the sample space for this experiment.

b What is the probability that the pointer stops on green? c What is the probability that the pointer stops on a

primary colour (red or blue)?

11 Three horses A, B and C are in a race. A is twice as likely to win as B and B is twice as likely to win as C. Calculate their respective probabilities of winning (i.e. P(A), P(B) and P(C)).

12 A roulette wheel in a casino has numbers 0 through 36. Bets are placed on a table marked as shown at right. Eight different types of bets are possible. They are indicated in the diagram by the letters A to H, which indicate the positions at which chips would be placed to indicate the bets. The bets are: A a single number—‘straight up’

B two numbers—‘split’ (either of the numbers the chip touches wins)

C three numbers— ‘street’ (any of the three numbers in the row wins)

D four numbers— ‘corner’ (any of the four numbers the chip touches wins) E six numbers — ‘six line’ (any of the numbers in the two rows wins) F first, second or third dozen — ‘dozen’

G left, middle or right twelve-number column — ‘column’ H odd/even, red/black, first/second half — ‘even chances’.

Note: In some casinos, players cannot bet on 0 and the house takes all bets when the ball

lands on 0. If betting on 0 is allowed, a player can win when the ball lands on 0 only if the bet is on this position, as 0 is neither odd nor even, red nor black, and so on.

For a chip placed at each of the eight different positions described, calculate the probability that when the ball comes to rest, the player will win. (Assume that bets can be placed on 0.)

10.2

Finding probabilities

A sample space that can be thought of as a combination of events in stages is often most easily organised as a branching diagram. The simplest example of such a sample space is successive

random selection of items.

34

36

30

32

19

16

13

9 _{12 18 21 27}

5 _{14 23 35}

31

33

29

25

22

26

24

20

17

13

7

4

2 8 ₁₁

15

6

0

1–18 EVEN ODD 19–36

10 28

1st DOZEN 2nd DOZEN 3rd DOZEN

A B D

C

E F

G

H

A tree diagram of a probability situation shows successive stages of a compound sample space as branching. At each stage, multiple forks are shown to indicate the possible outcomes. The final ‘twigs’ show the final outcomes of the sample space. In most cases it is convenient to label each outcome branch according to the number of that outcome and the possible number of outcomes. This simplifies the tree diagram so that it has fewer branches. If there are 16 possible outcomes and 5 of them are the same, this combined branch is labelled as 5 .

16

---!

Blue Red

Orange

Working with probability

Consider the case where a bag containing 3 green and 2 red marbles has two marbles taken out. There are 3 outcomes that are green and 2 that are red. This can be shown as:

Whichever way it is shown, it means there are 5 branches of which 3 are green and 2 are red.

Branches cannot be combined in the next example. G G G R R G R G R 3 5 2 5

or or just

3 5

2 5

A bag contains 3 green and 2 red marbles. Two marbles are randomly withdrawn. What is the probability that they are of different colours?

Solution

We can think of the selection as being first one marble, then the second marble without the first being replaced. The first branch has 5 possibilities, but each of the second branches has only 4 possibilities because 1 marble has been removed.

Show this as a simplified tree diagram. There are 5 branches for the first marble and

each has 4 branches for the second marble. n(sample space) = 4 × 5 = 20

There are 3 × 2GRbranches and 2 × 3RGbranches.

n(different colours) = 12

Calculate the probability. P(different colours) =

Evaluate. = , 0.6 or 60%

Write the answer in a sentence. The probability of different colours is 0.6.

G =GG

R =GR

G =RG

R =RR G R 2 4 1 4 3 4 2 4 3 5 2 5 12 20 ---3 5

---Example

6

A coin is tossed 3 times. Show the sample space as a tree diagram and calculate the probability of getting exactly 2 heads, in any order.

Solution

It is possible to get a head or a tail for each toss, so there are

2 branches at each stage. Draw the tree diagram.

H = HHH

T = HHT

H = HTH

T = HTT

H = THH

T = THT

H = TTH

T = TTT

1st toss 2nd toss 3rd toss

H T H T H T

Example

7

A sample space that can be thought of as a combination of two different sets of outcomes taken together is often best organised as a grid.

Count the sample space. n(sample space) = 8

Count the elements in the event ‘exactly 2 heads’. n(exactly 2 heads) = 3

Calculate the probability. P(exactly 2 heads) =

Evaluate. = 0.375

3 8

---A table or grid shows a sample space that can be considered as a two-dimensional combination of outcomes. The headings on the first row show one set of outcomes, and the first column shows the other set of outcomes. Each cell of the table shows a combination of outcomes from each part of the space.

!

Two tetrahedral dice with faces numbered 1 to 4 are thrown. The sum of the faces that the dice rest on is then worked out. What is the probability of a sum of 6?

Solution

Draw the table.

Fill out the cells of the grid.

Count the sample space. n(sample space) = 16

Count the elements in the event ‘sum = 6’. n(sum = 6) = 3

Calculate the probability. P(sum = 6) =

Evaluate. = 0.1875

1 2 3 4

1 1 + 1 = 2 1 + 2 = 3 1 + 3 = 4 1 + 4 = 5

2 2 + 1 = 3 2 + 2 = 4 2 + 3 = 5 2 + 4 = 6

3 3 + 1 = 4 3 + 2 = 5 3 + 3 = 6 3 + 4 = 7

4 4 + 1 = 5 4 + 2 = 6 4 + 3 = 7 4 + 4 = 8

First die

Second die

3 16

---Example

8

4 3

Complementary events

Two events are complementary if, together, they make up the entire sample space for an experiment and they have no common elements.

The complement of event E is symbolised E′.

Because an event and its complement make up the entire sample space, we can say

P(E) + P(E′) = 1

Working with probability

The idea of tree diagrams can be extended to include cases where the probabilities on each branch are not the same.

A card is drawn from a normal deck of playing cards. Find the probability of: a choosing a heart

b choosing a card other than a heart.

Solution

a Let H = choosing a heart.

Calculate the probability. P(H) =

Evaluate. =

Write the result. The probability of choosing a heart is .

b Let H′= choosing a suit other than a heart.

H and H′ are complementary. P(H) + P(H′) = 1

Substitute for P(H) and rearrange. P(H′) = 1 –

Evaluate. =

Write the result. The probability of choosing a card other

than a heart is . 13 52 ---1 4 ---1 4 ---1 4 ---3 4 ---3 4

---Example

9

Two balls are randomly selected from a bag containing 6 green and 4 red balls. If the first ball is replaced before drawing the second ball, find the probability that:

a both balls are red

b the first ball is green and the second is red.

Solution

We are only interested in two options at each draw (green or red). This means that the tree diagram can be simplified.

Draw the tree diagram.

Assign probabilities to the branches.

There is replacement, so the probabilities for the first and second draws will be the same.

a There are 10 × 10 branches altogether. There are 4 × 4 branches that are RR.

n(sample space)= 10 × 10 = 100

n(RR) = 4 × 4 = 16

Work out the probability. P(RR) = = =

b There are 10 × 10 branches altogether. There are 6 × 4 branches that are GR.

n(sample space)= 10 × 10 = 100

n(GR) = 6 × 4 = 24

Work out the probability. P(GR) = = =

G=GG

R=GR

G=RG

R=RR

G R 6 10 4 10

1st draw 2nd draw

6 10 6 10 4 10 4 10

4×4 10×10 --- 16

100 --- 4

25

---6×4 10×10 --- 24 100 --- 6 25

---Example

10

In Example 10, the probability of getting a red marble on the first draw is P(R)= =

The probability of getting a red marble on the second draw is P(R)= =

The probability of two red marbles can be worked out by multiplication: P(RR)= × =

Similarly, P(G)= =

The probability of getting green then red is P(GR)= × =

In Example 10, the second draw was not affected by the first, because the first marble was

replaced. The probabilities of green and red were not affected by the result of the first draw.

However, if we did not replace the first marble, the probabilities of green and red on the second draw would be affected by the result of the first draw, as was shown in Example 6.

4 10 --- 2 5 ---4 10 --- 2 5 ---2 5 --- 2 5 --- 4 25 ---6 10 --- 3 5 ---3 5 --- 2 5 --- 6 25

---Independent and dependent events

Two events are independent if the probability of one is not affected by the occurrence of the other.

Dependent events are not independent. The conditional probability of the second event is the probability of its occurrence, assuming the first event has occurred.

The combined probability of two events is the product of their probabilities. This is called the multiplication principle. In the case of dependent events, the probability of the first is multiplied by the probability of the second.

!

Three cards are dealt together from a well-shuffled standard pack of 52 cards. What is the probability that exactly 2 cards are aces?

Solution

We are only interested in drawing an ace, so we will regard drawing an ace as a success (S) and everything else as a failure (F). Construct a tree diagram as before with branches for an ace (S) and not an ace (F). Calculate the probability for each branch. Label each branch with its probability. The cards dealt are not independent.

‘Exactly 2 aces’ is SSF or SFS or FSS. Multiply probabilities on the branches of the tree diagram as required.

P(SSF) = × × ≈ 0.004 34

P(SFS) = × × ≈ 0.004 34

P(FSS) = × × ≈ 0.004 34

Add up the probabilities. P(exactly 2 aces) ≈ 0.013

S = SSS

F =SSF

S =SFS

F = SFF

S =FSS

F = FSF

S = FFS

F = FFF

1st card 2nd card 3rd card

Working with probability

Exercise 10.2

Finding probabilities

1 A bag contains three $1 coins and two $2 coins. A coin is randomly selected, its value is noted and it is then returned to the bag. This is done twice more. Assuming that the coins are equally likely to be selected, calculate the probability that:

a all 3 coins selected are $2 coins b at least two $2 coins are selected.

2 A jar contains 4 blue and 2 yellow balls. Another jar contains 6 blue and 3 yellow balls. A jar is randomly selected and a ball is chosen at random from that jar. Find the probability that it is yellow.

3 Two cards are drawn without replacement from a well-shuffled normal deck of 52 playing cards. Find the probability that the cards are a king and an ace.

4 A container holds 12 identical pieces of equipment of which 4 are defective. Three pieces of equipment are randomly selected from the container, one after the other and without replacement. Find the probability that:

a all 3 pieces of equipment are defective

b at least 1 of the pieces of equipment is defective.

5 There are 3 containers of light bulbs:

Container 1 has 10 light bulbs of which 4 are defective. Container 2 has 6 light bulbs of which 1 is defective. Container 3 has 8 light bulbs of which 3 are defective.

Use a tree diagram to find the probability that a light bulb selected at random is defective.

6 A counter is placed at zero on a number line. A die is rolled and the counter is moved according to the following rules:

■ 1 unit to the left for an odd number on the die ■ 1 unit to the right for an even number on the die.

The game is over after 5 rolls or when the counter reaches 3 or −2.

a Draw a tree diagram to describe all the possible paths that the counter may take. b Find the probability that the game concludes in under 5 rolls.

7 In the diagram below, the letters A, B, C, D, E and F represent benches in a park and the lines represent paths connecting them. A woman begins at A and walks from bench to bench. She stops to eat her lunch when she cannot continue to walk without going over the same path twice.

a Use a tree diagram to find the number of ways that she can walk before eating lunch.

b Find the probability that the woman eats her lunch on the bench at F.

Additional Exercise

10.2

C

F

D

E B A

8 Three yachts, Alpha, Bravo and Charlie, often race, and it is known that their probabilities of winning are , and respectively. A contest involving 2 races is held. Find the probability that Alpha does not win either race.

9 Max is approaching his end-of-semester exam period and he knows from his many past performances that his chances of passing Maths, History and English are , and respectively. Max’s performance in any one subject has no impact on his performance in any other subject. Calculate the probability that he passes:

a only 1 subject

b at least Maths and English.

Modelling and problem solving

10 Monica and Ernesto, who are of roughly equal ability, play each other in a squash

tournament. The first person to win 2 games in a row or who wins a total of 3 games wins the tournament. What is the probability that Monica wins the tournament in 3 games or less?

11 An electric current may flow from A to B in the circuit shown. The current will flow from A to B only if the circuit is completed. That is, the switch at x, plus at least one of the switches at y and z, must be down.

The switches all operate independently and are just as likely to be up as down.

a Given any arrangement of switches at random, what is the probability that: i the current will flow from A to B? ii the current will not flow from A to B? b Draw another circuit from A to B so that the probability that the current will flow from one

side to the other is .

12 In the game of parcheesi, a player may move a token out of the starting base if, when a pair of dice is rolled, either at least one die shows a five or the sum of the uppermost faces is equal to 5. What is the probability that a player can move a token out of the starting base on the first roll of the dice?

13 A pair of fair 6-sided dice is rolled. Find the probability that the total of the 2 numbers on the uppermost faces of the dice is greater than 4.

14 Faith can’t remember Lee’s house number. She does remember that the houses in the street are numbered from 1 to 39. Calculate the probability that she guesses Lee’s house number by first guessing the correct tens digit followed by the correct units digit.

1 2 --- 1

3

--- 1

6

---3 4 --- 3

5

--- 1

2

---A B

y

z x

7 8

Working with probability

10.3

Using binomial probability

In many probability situations, there are only two possible outcomes, such as head/tail, boy/girl, etc. In other situations, we can classify the outcomes so that there are only two possibilities, such as drawing an ace or not drawing an ace from a pack of cards. A trial in which there are only two possible outcomes is known as a Bernoulli trial, after Jacob Bernoulli (1654–1705).

The results of successive Bernoulli trials may be found using tree diagrams to count the number of ways of obtaining a particular number of successes.

Extra Material

Venn diagrams and the addition rule

A Bernoulli trial has the following characteristics:

■ There are only two possible outcomes, usually called success and failure.

■ The probability of success (p) and the probability of failure (q) do not change.

■ p + q = 1

■ Successive trials are independent, so the result of one trial has no effect on other trials.

!

A card is drawn from a well-shuffled deck, checked to see whether it is a heart and then returned to the deck. What is the probability of obtaining 2 hearts from 3 such draws?

Solution

Draw a tree with the probabilities on it.

The draws are Bernoulli trials, with success = a heart.

Write the probabilities of success and failure. p = and q =

List all the ways of giving 2 successes. 2 successes= SSF, SFS, FSS

Calculate the probability of one way. P(SSF)= × ×

=

Write the number of ways. There are 3 ways to get 2 successes.

1 4

--- 3

4

---S = SSS

F =SSF

S =SFS

F = SFF

S =FSS

F = FSF

S = FFS

F = FFF

1st card 2nd card 3rd card

S F 1 4 3 4 1 4 3 4 1 4 3 4 S F 1 4 3 4 1 4 3 4 1 4 3 4 S F 1 4 3 4 1 4 --- 1 4 --- 3 4 ---3 64

---Example

12

It is so common to be concerned with a number of successes from Bernoulli trials that we give the probabilities a special name.

Work out the final probability. P(2 hearts)= 3 ×

=

Write the answer. The probability of obtaining 2 hearts

from 3 such draws is . 3 64

---9 64

---9 64

---Binomial probabilities arise from successive trials where:

■ there are a fixed number of Bernoulli trials

■ the probability of a particular number of successes is required

■ the order of the successes does not matter.

We write P(success) = p and P(failure) = q The binomial probability of x successes from n trials is given by:

P(x successes) = (number of ways of obtaining x successes) × px× _qn − x

!

A multiple-choice test has 4 questions, each question having 5 alternative answers A, B, …, E. Only one answer is correct for any question. Find the probability of guessing exactly 2 correct answers on the test.

Solution

Draw a tree diagram to show the outcomes.

The guesses are Bernoulli trials, with P(success) = P(getting the question right) = p.

Write the probabilities of success and failure. p = 1 = 0.2 and q = = 0.8 5

--- 4

5

---S = SSSS F = SSSF S = SSFS

F =SSFF

S = SFSS

F =SFSF

S =SFFS

F = SFFF S = FSSS

F =FSSF

S =FSFS

F = FSFF

S =FFSS

F = FFSF S = FFFS F = FFFF

1st question 2nd question 4th question

S

F

S

F

S

F

S

F

S

F S

F S

F

Working with probability

The number of successes can be worked out using Pascal’s Triangle or combinations instead of drawing tree diagrams. However, binomial probabilities are so common that they may be looked up in tables or found using a graphics calculator.

A table of binomial probabilities is included as an appendix on pages 398–401.

When you are using the table for probabilities greater than 0.50, the table is read from the bottom and right side instead of the top and left.

Write the number of ways. There are 6 ways to get exactly 2 successes.

Write the formula. P(x successes) = (number of ways) × px× _qn − x

Work out the final probability using

n = 4 and x = 2.

P(2 successes) = 6 × 0.22× _0.84 − 2

= 6 × 0.22× _0.82

= 0.1536

Write the answer. The probability of guessing exactly 2 correct

answers on the test is 0.1536 or 15.36%.

A bag of 20 marbles contains 7 green ones. A marble is taken out at random and replaced. Use the binomial probability table to find the probability that 3 green marbles are obtained from 10 such trials.

Solution

Find the value from the table.

Find p. p = = 0.35

Write n and x. n = 10 and x = 3

Write the answer. The probability of 3 green marbles is about

0.2522 or 25.22%. 7

20

---p

x 0.01 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

n = 10 0 1 2 3 4 5 0.9044 0.0914 0.0042 0.0001 0.5987 0.3151 0.0746 0.0105 0.0010 0.0001 0.3487 0.3874 0.1937 0.0574 0.0112 0.0015 0.1969 0.3474 0.2759 0.1298 0.0401 0.0085 0.1074 0.2684 0.3020 0.2013 0.0881 0.0264 0.0563 0.1877 0.2816 0.2503 0.1460 0.0584 0.0282 0.1211 0.2335 0.2668 0.2001 0.1029 0.0135 0.0725 0.1757 0.2522 0.2377 0.1536 0.0060 0.0403 0.1209 0.2150 0.2508 0.2007 0.0025 0.0207 0.0763 0.1665 0.2384 0.2340 0.0010 0.0098 0.0439 0.1172 0.2051 0.2461 10 9 8 7 6 5

Example

14

Find the binomial probability of 8 successes from 14 trials, where the probability of success is 0.75.

Solution

Read the table from the bottom and right.

Write n, p and x. n = 14, p = 0.75 and x = 8

Write the answer. The probability of 8 successes is about

0.0734 or 7.34%.

n = 14 0 0.8687 0.4877 0.2288 0.1028 0.0440 0.0178 0.0068 0.0024 0.0008 0.0002 0.0001 14 1 0.1229 0.3593 0.3559 0.2539 0.1539 0.0832 0.0407 0.0181 0.0073 0.0027 0.0009 13 2 0.0081 0.1229 0.2570 0.2912 0.2501 0.1802 0.1134 0.0634 0.0317 0.0141 0.0056 12 3 0.0003 0.0259 0.1142 0.2056 0.2501 0.2402 0.1943 0.1366 0.0845 0.0462 0.0222 11 4 0.0037 0.0349 0.0998 0.1720 0.2202 0.2290 0.2022 0.1549 0.1040 0.0611 10 5 0.0004 0.0078 0.0352 0.0860 0.1468 0.1963 0.2178 0.2066 0.1701 0.1222 9 6 0.0013 0.0093 0.0322 0.0734 0.1262 0.1759 0.2066 0.2088 0.1833 8 7 0.0002 0.0019 0.0092 0.0280 0.0618 0.1082 0.1574 0.1952 0.2095 7 8 0.0003 0.0020 0.0082 0.0232 0.0510 0.0918 0.1398 0.1833 6

9 0.0003 0.0018 0.0066 0.0183 0.0408 0.0762 0.1222 5

10 0.0003 0.0014 0.0049 0.0136 0.0312 0.0611 4

11 0.0002 0.0010 0.0033 0.0093 0.0222 3

12 0.0001 0.0005 0.0019 0.0056 2

13 0.0001 0.0002 0.0009 1

14 0.0001 0

0.99 0.95 0.90 0.85 0.80 0.75 0.70 0.65 0.60 0.55 0.50 x p

Example

15

Fifteen binomial trials are conducted with a probability of success of 0.63. Use a graphics calculator to find:

a the probability of 6 successes b the probability of at most 6 successes c the probability of at least 6 successes.

Solution

Graphics calculators have both probability density functions and cumulative density functions.

The binomial probability density function is abbreviated to Bpd, binompdf or pdfbin and gives the probability of x successes.

The binomial cumulative density function is abbreviated to Bcd, binomcdf or cdfbin and gives the sum of the probabilities from 0 to x successes:

(X = r) (i.e. at most x successes)

P

r=0 15

∑

Working with probability

Calculator Instructions

a We want the probability density function with n = 15, p = 0.63 and x = 6.

Casio fx-9860G AU

Use the STAT menu. Press to access DIST, the

distributions.

Press again to access BINM, the binomial

distribution.

Press to access Bpd, the density function.

Press to access Var, the variable data type.

Now enter the values and press to find the value.

Texas Instruments TI-84

Press VARS to use the DISTR menu.

Scroll down to A: binompdf and press .

Complete the function by entering 15, 0.63, 6)

and press .

b The method for all calculators is the same except that the cumulative density function is selected.

Casio fx-9860G AU Texas Instruments TI-84

c We want (X = r) for 6 or more successes. The cumulative density function will

give us the value of the complementary event (X = r).

Sharp EL-9900

See the instructions given on the CD-ROM.

Write the answer. The probability of 6 successes is about 0.0407 or 4.07%.

Write the answer. The probability of at most 6 successes is about 0.0597 or 5.97%.

Use your graphics calculator. (X = r)= 1 − (X = r) ≈ 1 − 0.0190 = 0.9810 F5

F5

F1

F2

EXE

2nd

ENTER

ENTER

P

r=6 15

∑

P

r=0 15

∑

P

r=6 15

∑

P

r=0 15

Exercise 10.3

Using binomial probability

Use tree diagrams for questions 1 to 5 and the binomial tables in the Appendix (pages 398–401) or a graphics calculator for the rest of the questions.

1 A certain plant variety produces 25% red flowers. One of these plants has 3 flowers. If the flower colours form a binomial distribution:

a what are the values of n, p and q?

b what is the probability that all flowers are red? c what is the probability that no flowers are red? d what is the probability that at least 2 flowers are red?

2 The Brisbane Bandits have a probability of of winning whenever they play. If the team plays 4 games, find the probability that they win:

a exactly 2 games b at least 1 game c more than half of the games.

3 A family has 6 children. Find the probability that there are:

a 3 boys and 3 girls b fewer boys than girls.

4 The probability of Jacqui hitting a target is .

a If she fires 7 times, what is the probability of her hitting the target at least twice? b How many times must she fire so that the probability of hitting the target at least once is

greater than ?

5 What proportion of families with exactly 4 children should be expected to have at least 2 girls?

6 Metal welds that are done with robotic devices have a 15% chance of containing a flaw. A pipeline contains 7 welds. Find the probability that: a all the welds have flaws

b at least 1 weld has a flaw.

7 For a binomial experiment, find all the values of

P(X = x) for:

a n = 6 and p = 0.05 b n = 8 and p = 0.10 c n = 7 and p = 0.50

8 It is a bank’s experience that 5% of the cheques received in an automatic teller will ‘bounce’. What is the probability that, for 8 cheques deposited:

a none will bounce? b exactly 2 will bounce? c at least 1 will bounce?

9 If, over a given period in Brisbane, rain falls at random on 4 days of every 10, find the probability that:

a the first 2 days of a given week will be wet and the remainder of the week fine b rain will fall on the first 2 days of the week

c at least 2 days in the week will be wet.

10 A multiple-choice test has 7 questions, each with 4 alternatives, only 1 of which is correct. What is the probability of guessing correct answers to:

a all of the 7 questions? b none of the 7 questions?

c 3 of the 7 questions? d at least 3 of the 7 questions?

Additional Exercise

10.3

2 3

---1 4

---Working with probability

11 A gardener plants 8 seeds. The probability that a seed will germinate is 0.85. What is the probability that at least 6 of the seeds will germinate?

12 A particular family has 3 children. It is known that the genetic makeup of the parents is such that there is a 20% chance that a child born will have curly hair.

a What are the values of n, p and q for this binomial situation? b What is the probability that all of the children have curly hair? c What is the probability that none of the children have curly hair? d What is the probability that at least 1 of the children has curly hair?

Modelling and problem solving

13 Hospital records confirm that, of patients suffering from a particular complaint, 80% recover during their hospital stay. What is the probability that 6 randomly selected patients suffering from the complaint will all recover?

14 A judge is scheduled to hear 10 appeals for traffic violations. Each appeal has a probability of 0.4 of being approved, independently of other appeals. Find the probability that less than half of the appeals are granted.

15 Boom Boom is the school’s best tennis player. From experience it is known that the

probability that Boom Boom serves an ace is 0.15. What is the probability that Boom Boom wins a game with just 4 serves?

16 After a certain drug in tablet form is stored above 30°C it is found that an average of one-fifth of the tablets become ineffective. A person who has been on a holiday in the tropics in summer kept 8 of these tablets in a suitcase where the temperature was consistently over 30°C. What is the probability that:

a exactly 3 of the tablets are now ineffective? b at least 1 of the tablets is now ineffective?

17 Find the probability that, of the first 6 people met in the street on a given day, at least 4 will have their birthday on a Sunday this year. Give your answer as a fraction.

18 A manufacturer finds that, in the long run, 15% of the manufactured articles are defective. If a sample of 10 articles is randomly selected, find the probability that:

a the sample contains 2 defective articles b the first 3 articles selected are defective

c the sample contains at least 3 defective articles.

19 Wendy owns 10 holiday units in a remote location and does not want to risk leaving VCRs permanently in the units, so she hires them out to occupants as they require them. She finds that about 80% of people who occupy the units want to hire VCRs, so she buys 7. Find the probability that, at a time when all the units are occupied, the demand for VCRs will exceed supply.

■ Experimental or empirical probability uses data to work out chances (probabilities). Each item of data, or every attempt at an experiment, is called a trial, with the result of each trial being called an outcome. The outcome we are checking is called the favourable outcome. The number of times a particular outcome occurs is called its frequency.

■ Empirical probabilities are calculated using the relative frequency, which is calculated by

Relative frequency =

■ The expected frequency of an outcome is calculated by:

Expected frequency = relative frequency × number of trials

■ Theoretical probability is calculated using a list of possible outcomes called the sample space. Each possible outcome is called an event point, a sample point or element. The number of elements in a sample space is written n(sample space).

■ An event is a subset of the sample space. The number of elements in an event is n(event).

■ The theoretical probability of an event is written as P(event) or P(E) and calculated using the formula

P(event) =

■ Probabilities range from 0 (impossible) to 1 (certain). More likely events have a higher probability.

■ A tree diagram shows successive stages of a compound sample space as branches. At each stage, multiple forks are shown to indicate the possible outcomes. The final ‘twigs’ show the final outcomes of the sample space.

■ A table shows the sample space as a two-dimensional grid. The first row headings show one set of outcomes, the first column shows the other, and the cells show the elements of the space. ■ An event E, and its complement E′, make up the sample space: P(E) + P(E′) = 1

■ Two events are independent if the probability of one is not affected by the occurrence of the other.

■ Dependent events are not independent. The conditional probability of the second event is the probability of its occurrence, assuming the first event has occurred.

■ According to the multiplication principle the combined probability of two events is the product of their probabilities. In the case of dependent events, the probability of the first is multiplied by the probability of the second.

■ Bernoulli trials have the characteristics that:

– there are only two possible outcomes, usually called success and failure

– the probability of success, p, and the probability of failure, q, do not change p + q = 1 – successive trials are independent, so the result of one trial has no effect on other trials. ■ Binomial probabilities arise from successive trials where:

– there are a fixed number of Bernoulli trials

– the probability of a particular number of successes is required – the order of the successes does not matter.

■ The binomial probability of x successes from n trials is given by

P(x successes) = (number of ways of obtaining x successes) × px× _qn − x

■ Binomial probabilities can be calculated using tree diagrams, tables or graphics calculators. frequency of favourable outcome

number of trials

---n(event) n(sample space)

---A = AA

B = AB

A = BA

B = BB

‘Twig’ A

Chapter review

Knowledge and procedures

Questions 1 to 3 use the following information:

When people in a shopping mall were asked their reason for being there, 20 said to buy food, 15 said to buy clothes, 25 said they were ‘just looking’ and 10 said were passing through.

1 The number of trials was:

A 20 B 15 C 60 D 70 E 50

2 The relative frequency of ‘just looking’ was about:

A 0.357 B 2.8 C 0.286 D 3.5 E 0.214

3 If another 30 people were asked, how many of these would you expect to be buying food?

A 6 B 7 C 8 D 9 E 11

4 List the sample space for selecting a marble from a bag containing 3 pink, 5 green and 2 orange marbles.

5 A survey of voters about their voting intentions at the next election gave these results.

a What was the number of trials?

b If another 500 people were surveyed, how many of these would you expect to prefer the Liberals?

The ‘two-party preferred’ vote is calculated by reallocating the numbers for minor parties. In this case, 70% of ‘Others’ is assigned to the Liberals and 60% of ‘Democrats’ is assigned to the ALP.

c What is the ‘two-party preferred’ result for the ALP from this survey? d What is the probability of a Liberal win on the ‘two-party preferred’ basis?

6 A bag contains 4 red and 6 yellow marbles. The probability of selecting a yellow marble at random is:

A 4 B 6 C 10 D 0.4 E 0.6

Questions 7 to 9 refer to a bag containing 3 red, 5 green and 4 yellow marbles. A single marble is selected at random from the bag.

7 The probability that it is green is about:

A 5 B 0.25 C 0.42 D 0.33 E 0.71

8 The probability that it is yellow is about:

A 0.5 B 0.25 C 0.42 D 0.33 E 0.4

9 The probability that it is black is:

A 1 B 0 C 0.083 D 0.10 E 0.11

10 A coin is chosen at random from a purse containing 4 twenty-cent coins, 6 $1 coins and 5 fifty-cent coins. What is the probability that a fifty-cent coin is selected?

11 The ‘float’ at a market stall has 5 × $2 coins, 10 × $1 coins, 10 × 50c coins, 20 × 20c coins, 10 × 10c coins and 10 × 5c coins. A coin is selected at random. What is the probability that: a a 20c coin is selected? b the coin is of value greater than 20c? c the coin is of value less than 20c? d the coin is of value greater than 2c?

Party ALP Liberals Democrats Others

Number of voters 421 439 126 74

Ex10.1

Ex10.1

Ex10.1

Ex10.1

Ex10.1

Ex10.1

Ex10.1

Ex10.1

Ex10.1

Ex10.1

Ex10.1

12 Three normal coins are tossed together. The probability of obtaining exactly 2 heads is:

A B C D E

13 Two 8-sided dice numbered 1–8 are thrown. What is the probability that the total shown on the dice is more than 5?

14 A box contains 3 blue, 1 black and 2 red biros. Two biros are taken at random. a Draw a tree diagram of the event space.

b What is n(event space)? c Find P(2 blue biros).

d Find P(2 different-coloured biros).

15 a Draw a tree diagram to show the results of tossing a normal coin 3 times. b Calculate P(3 tails).

c What is P(exactly 2 heads)? d What is P(at least 2 heads)?

16 A euchre pack of cards consists of the ace, 7–10, jack, queen and king of the 4 suits, together with a joker.

a What is the probability of being dealt 2 aces in the first 2 cards?

b What is the probability of being dealt 2 jacks of the same colour and the joker in the first 3 cards?

17 A carton of eggs contains three 55 g eggs, five 59 g eggs and two 52 g eggs. The probability of choosing two 55 g eggs at random is:

A 0.09 B 0.067 C 0.05 D 0.1 E 0.033

18 A coin is weighted so that the probability of a head is 0.7. What is the probability of getting at least 2 heads from 3 tosses of the coin?

19 A fair coin is tossed 12 times. What is the probability of obtaining 4 heads?

20 Find the following binomial probabilities using tables or a graphics calculator. a P(x = 5) when n = 15 and p = 0.3

b P(x = 7) when n = 19 and p = 0.45 c P(x 6) when n = 18 and p = 0.75 d P(x 4) when n = 12 and p = 0.6

Modelling and problem solving

21 A student factoring x2− _x − _{12 is not sure how to obtain the middle term, but knows that}

the binomial expression contains factors of −12. The student guesses that the factorisation is (x − 2)(x + 6).

a Is this student correct?

b How many different guesses could the student make? c What is the probability of making the correct guess?

The student then guesses the factorisation of 4x2+ _17x − _{15 by choosing factors of the first}

and last terms and ignoring the middle term.

d What is the probability of guessing the correct factorisation for 4x2+ _17x − _15?

e What is the probability of guessing the correct factorisation for x2+ _4x + _3?

Ex 10.2

3 10

--- 3

8

--- 5

8

--- 7

8

--- 2

3

---Ex 10.2

Ex 10.2

Ex 10.2

Ex 10.2

Ex 10.2

Ex 10.3

Ex 10.3

Ex 10.3

Chapter review

22 What is the probability of being dealt 4 fours from a well-shuffled 52-card standard pack of cards?

23 A target shooter knows from experience that she hits a target 4 times out of each 5 shots. If she is allowed 10 shots in a competition, what is the probability that she gets:

a 10 hits? b 9 hits?

c at least 7 hits?

24 The probability that children will be absent from school in a flu epidemic is 15%. What is the probability that from a class of 20 students:

a 7 will be away? b at least 7 will be away?

c at least 5 will be away?

25 A new vaccine is claimed to be 85% effective in immunising young children against a particular childhood disease. In a sample of 15 children who are vaccinated, what is the probability that:

a all become immune to the disease? b at least 2 have no immunity to the disease? c fewer than 4 have no immunity?

26 Five per cent of thermostats produced by a particular process are defective. The thermostats are packed 15 to a box for shipping purposes. When a box of thermostats is received by a spare parts dealer, what is the probability that:

a all are operative? b 2 are defective?

c at least 2 are defective? d no more than 2 are defective?

Ex10.2

Ex10.3

Ex10.3

Ex10.3

Ex10.3