probability ii practice questions 19

Destination Maths Chapter 19 Probability II

1

Probability II

Practice Questions 19.1 1. An ice cream shop has vanilla, chocolate or strawberry ice cream

and you can have either chocolate or caramel sauce on top.

(i) Represent this information on a tree diagram.

(ii) Use the tree diagram to answer the following questions:

What is the probability that a customer chosen at random will order

(a) a chocolate ice cream?

Total number of outcomes: 6

Outcomes of InterestProbabilityTotal Outcomes

=

P Chocolat 2e( )613

=

=

(b) an ice cream with caramel sauce?

P Caramel 3( )

612

=

=

(c) a strawberry ice cream with caramel sauce?

1

(Strawberry with Caramel)6

P =


2

2. You are buying a jumper from a shop. You have to choose a size, colour

and style.

You can choose between small, medium or large for size; blue, pink, red

or brown for colour; and round-neck or V-neck for style.

(i) Draw a tree diagram to represent the sample space.

(ii) Use the tree diagram to answer the following questions:

(a) How many different options do you have?

By counting from the tree diagram: 24 options in total

Using FPoC: Size × Colour × Style

3 × 4 × 2

= 24 options in total


3

(b) What is the probability that you pick a brown jumper?

Outcomes of InterestProbabilityTotal Outcomes

=

6P(Brown)2414

=

=

(c) What is the probability that you pick a jumper that has a V-neck?

12P(V neck)2412

=

=

(d) What is the probability that you pick a large, red, round-neck jumper?

1

P(Large and Red and Round)24

=

(e) What is the probability that the jumper you pick doesn’t have a round

neck?

P(no round) = 1 – P(Round)

1

1

– 12

2

=

=


4

3. A fair die is thrown and a fair coin is flipped at the same time.

(i) Draw a two-way table to represent this sample space.

Coin H T

Die

1 1H 1T

2 2H 2T

3 3H 3T

4 4H 4T

5 5H 5T

6 6H 6T

(ii) Use the two-way table to calculate the probability of getting:

Total number of outcomes = 12

Outcomes of Interest

Probability :Total Outcomes

(a) a head

6P(Head)

1212

=

=

(b) a 3

2P(3)

1216

=

=


5

(c) a head and a 3

There are two ways to calculate this probability:

From the outcomes in the tree diagram: 1P(Head and 3)12

=

By multiplying along the branches: P(H) and P(3)

P(H) × P(3)

1 12 6

112

×

=

(d) a tail and an even number


From the outcomes in the tree diagram:

3P(Tail and Even)1214

=

=

By multiplying along the branches:

P T P Even1 12

( ) (

14

)

2

×

= ×

=

(e) a tail or a prime number.

Prime numbers are 2, 3 and 5.

P(Tail or Prime)

Count the numbers of outcomes that contain a tail or a prime number

Outcomes that contain a tail or a prime number = 9

P(Tail or Prime)

191234

=

=


6

4. Suppose we flip a coin and spin a spinner with four coloured sections of equal size

at the same time.

(i) What is the sample space for the coin?

Coin = {head, tail}

(ii) What is the sample space for the spinner?

Spinner = {red, green, yellow and blue}

(iii) Draw a sample space diagram and use it to write out the sample space for

when we throw both the coin and spin the spinner.

Coin

H T

Spin

ner

Red RH RT

Green GH GT

Yellow YH YT

Blue BH BT

The sample is {RH, RT, GH, GT, YH, YT, BH, BT}

(iv) Using your sample space diagram, calculate the probability of getting:

(a) a tail on the coin

4( )P T il

1

a8

2

=

=


7

(b) red on the spinner

2(Red)P

814

=

=

(c) a tail on the coin and red on the spinner

1

( ) and P(RedP ail )8

T =

(d) a tail on the coin or red on the spinner

5

( ) or P(RedP ail )8

T =

* be careful not to the RT twice

(e) the colour on the spinner is not red

P(not Red) 1 P(Red)1

14

34

= −

= −

=

(f) green on the spinner

2P(Green)

814

=

=

(g) a head on the coin and either blue or yellow on the spinner.

P(head, blue) or P(head, yellow)

1 18 82814

= +

=

=


8

5. A spinner with four equal segments, numbered 1 to 4, is spun twice.

(i) Represent the sample space for this experiment using a

two-way table.

2nd Spin

1 2 3 4

1st S

pin

1 1, 1 1, 2 1, 3 1, 4

2 2, 1 2, 2 2, 3 2, 4

3 3, 1 3, 2 3, 3 3, 4

4 4, 1 4, 2 4, 3 4, 4

(ii) Use your two-way table to calculate the probability that:

(a) the same number comes up on both spins

Total Outcomes = 16

P(Same number) = {(1, 1), (2, 2), (3, 3), (4, 4)}

41614

=

=

(b) both numbers are odd

P(Both odd) = {(1, 1), (1, 3), (3, 1), (3, 3)}

41614

=

=

(c) the number on both the first and second spinner is even

P(Both even) = {(2, 2), (2, 4), (4, 4), (4, 2)}

41614

=

=


9

(d) the sum of the two numbers is 6

P(Sum is 6) = {(2, 4), (3, 3), (4, 2)}

3

16=

(e) 1 doesn’t come up on either spin.


P(no one) = {(2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4), (4, 2), (4, 3), (4, 4)}

916

=

Alternatively: P(no one) 1 P(one)

7116

916

= −

= −

=

6. Sean and Colin are playing a game involving throwing two fair dice one after the other.

Sean wins if both dice are odd and Colin wins if either die is a five.

(i) Draw a sample space diagram to represent this information.

Die 2

1 2 3 4 5 6

Die

1

1 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6

2 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6

3 3, 1 3, 2 3, 3 3, 4 3, 5 3, 6

4 4, 1 4, 2 4, 3 4, 4 4, 5 4, 6

5 5, 1 5, 2 5, 3 5, 4 5, 5 5, 6

6 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6


10

(ii) How many different outcomes are there?

6 × 6 = 36 possible outcomes

Use your sample space diagram to answer the following questions:

(iii) What is the probability that Sean wins?

P(Sean) = P (Both odd)

These outcomes are highlighted in green below:

Die 2

1 2 3 4 5 6

Die

1

1 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6

2 2, 1 2, 2 2, 3 2, 4 2,5 2, 6

3 3, 1 3, 2 3, 3 3, 4 3, 5 3, 6

4 4, 1 4, 2 4, 3 4, 4 4, 5 4, 6

5 5, 1 5, 2 5, 3 5, 4 5, 5 5, 6

6 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6

= 9

36

= 14


11

(iv) What is the probability that Colin wins?

P(Colin) = P(Either 5)

These outcomes are highlighted in blue below:

Die 2

1 2 3 4 5 6

Die

1

1 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6

2 2, 1 2, 2 2, 3 2, 4 2,5 2, 6

3 3, 1 3, 2 3, 3 3, 4 3, 5 3, 6

4 4, 1 4, 2 4, 3 4, 4 4, 5 4, 6

5 5, 1 5, 2 5, 3 5, 4 5, 5 5, 6

6 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6

P (Colin wins) = 1136

(v) Is the game fair? Explain your answer.

The game is not fair as the probability of a 5 occurring and Colin winning is higher

1136

than the probability of both dice being odd and Sean winning 936

.

(vi) Will any of the outcomes result in a tie?

The following outcomes: (3, 5), (5, 3), (1, 5), (5, 1), (5, 5) would all result in a tie.


12

7. A fair, six-sided die and four-sided die are thrown.

(i) Represent this information on a sample space diagram.

4 sided

1 2 3 4

6 si

ded

1 1, 1 1, 2 1, 3 1, 4

2 2, 1 2, 2 2, 3 2, 4

3 3, 1 3, 2 3, 3 3, 4

4 4, 1 4, 2 4, 3 4, 4

5 5, 1 5, 2 5, 3 5, 4

6 6, 1 6, 2 6, 3 6, 4

Total number of outcomes = 4 × 6 = 24

(ii) Use your sample space diagram to work out the probability that:

(a) the numbers on both dice are the same

These outcomes are highlighted in purple below:

4 sided

1 2 3 4

6 si

ded

1 1, 1 1, 2 1, 3 1, 4

2 2, 1 2, 2 2, 3 2, 4

3 3, 1 3, 2 3, 3 3, 4

4 4, 1 4, 2 4, 3 4, 4

5 5, 1 5, 2 5, 3 5, 4

6 6, 1 6, 2 6, 3 6, 4

P(both the same)

42416

=

=


13

(b) the numbers on both dice are even

These outcomes are highlighted in yellow below:

6P Both even

21

( )4

4

=

=

(c) the sum of the numbers on both dice is greater than six

These outcomes are highlighted in pink below:

4 sided

1 2 3 4

6 si

ded

1 1, 1 1, 2 1, 3 1, 4

2 2, 1 2, 2 2, 3 2, 4

3 3, 1 3, 2 3, 3 3, 4

4 4, 1 4, 2 4, 3 4, 4

5 5, 1 5, 2 5, 3 5, 4

6 6, 1 6, 2 6, 3 6, 4

10P Sum greater th( an 62

1

)4

52

=

=

4 sided 1 2 3 4

6 si

ded

1 1, 1 1, 2 1, 3 1, 4 2 2, 1 2, 2 2, 3 2, 4 3 3, 1 3, 2 3, 3 3, 4 4 4, 1 4, 2 4, 3 4, 4 5 5, 1 5, 2 5, 3 5, 4 6 6, 1 6, 2 6, 3 6, 4


14

(d) a four appears on at least one die


4 sided

1 2 3 4 6

side

d 1 1, 1 1, 2 1, 3 1, 4

2 2, 1 2, 2 2, 3 2, 4

3 3, 1 3, 2 3, 3 3, 4

4 4, 1 4, 2 4, 3 4, 4

5 5, 1 5, 2 5, 3 5, 4

6 6, 1 6, 2 6, 3 6, 4

at least on9

P( )4

3

e 24

8

=

=

(e) the sum of the two numbers is a multiple of 3. These outcomes are highlighted in blue below:

4 sided

1 2 3 4

6 si

ded

1 1, 1 1, 2 1, 3 1, 4

2 2, 1 2, 2 2, 3 2, 4

3 3, 1 3, 2 3, 3 3, 4

4 4, 1 4, 2 4, 3 4, 4

5 5, 1 5, 2 5, 3 5, 4

6 6, 1 6, 2 6, 3 6, 4

P(Sum is multiple of 3)

82413

=

=


15

8. A spinner with three equal sectors coloured red, green and blue is spun

three times.

(i) Represent the sample space on a tree diagram.



16

(ii) Use the tree diagram to work out the probability of getting:

(a) the same colour on all three spins

P(3 same colour):


RRR, GGG, BBB

P(3 same colour)

32719

=

=


17

(b) the same colour on two of the three spins

P(2 Same colour):

These outcomes are shown in orange below:

RRG, RRB, RGG, GRR, BRR, BGG, RBB, GGR, GGB, BBR, BBG, RGR,

RBR, GRG, GBG, BRB, BGB, GBB.

P(2 same colours)

182723

=

=


18

(c) a green and two reds in any order

P(GRR any order):

These outcomes are shown in green below:

GRR, RGR, RRG

3P(GRR in any order)

2719

=

=


19

(d) a different colour on all three spins

P(3 different colours):

These outcomes are highlighted in blue below:

RGB, RBG, GRB, GBR, BGR, BRG

6P(3 different colours)

2729

=

=


20

(e) not blue on any of the spins.

P(No blue):


RRR, RRG, RGR, RGG, GRR, GRG, GGR, GGG

8

P(no blue)27

=

9. When Joselyn comes home from school she has a snack, which is either a ham sandwich

or a chicken wrap. After that she either does her homework or studies. Then she has her

dinner. After dinner she either watches TV, plays on her Xbox or reads a book.

(i) To list all of Joselyn’s options, is it better to use a two-way table or a tree

diagram? Justify your answer.

It is better to use a tree diagram as there are more than two events.


21

(ii) Use either of the methods above to list all the possible ways Joselyn could spend

her evening.

List of Outcomes:

HHT, HHX, HHB,

HST, HSX, HSB,

CHT, CHX, CHB,

CST,CSX, CSB

(iii) With the help of your sample space diagram, determine the probability that

Joselyn:

(a) has a ham sandwich, does her homework and then reads a book


( )1

P HHB12

=

(b) watches TV

4P TV

1

3

( )2

1

=

=


22

(c) eats a chicken wrap and plays on her Xbox

2P

12(CX

6

)

1

=

=

(d) doesn’t study.

(No study

6P

12

2

)

1

=

=

10. Anita and Rebecca both think of a number between 1 and 10 inclusive.

Draw a two-way table to list all the possible outcomes and use it to determine the probability that:

Rebecca

1 2 3 4 5 6 7 8 9 10

Ani

ta

1 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6 1, 7 1, 8 1, 9 1, 10

2 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6 2, 7 2, 8 2, 9 2, 10

3 3, 1 3, 2 3, 3 3, 4 3, 5 3, 6 3, 7 3, 8 3, 9 3, 10

4 4, 1 4, 2 4, 3 4, 4 4, 5 4, 6 4, 7 4, 8 4, 9 4, 10

5 5, 1 5, 2 5, 3 5, 4 5, 5 5, 6 5, 7 5, 8 5, 9 5, 10

6 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6 6, 7 6, 8 6, 9 6, 10

7 7, 1 7, 2 7, 3 7, 4 7, 5 7, 6 7, 7 7, 8 7, 9 7, 10

8 8, 1 8, 2 8, 3 8, 4 8, 5 8, 6 8, 7 8, 8 8, 9 8, 10

9 9, 1 9, 2 9, 3 9, 4 9, 5 9, 6 9, 7 9, 8 9, 9 9, 10

10 10, 1 10, 2 10, 3 10, 4 10, 5 10, 6 10, 7 10, 8 10, 9 10, 10



23

(i) they both chose the same number These outcomes are highlighted in red below:

Rebecca

1 2 3 4 5 6 7 8 9 10

Ani

ta

1 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6 1, 7 1, 8 1, 9 1, 10

2 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6 2, 7 2, 8 2, 9 2, 10

3 3, 1 3, 2 3, 3 3, 4 3, 5 3, 6 3, 7 3, 8 3, 9 3, 10

4 4, 1 4, 2 4, 3 4, 4 4, 5 4, 6 4, 7 4, 8 4, 9 4, 10

5 5, 1 5, 2 5, 3 5, 4 5, 5 5, 6 5, 7 5, 8 5, 9 5, 10

6 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6 6, 7 6, 8 6, 9 6, 10

7 7, 1 7, 2 7, 3 7, 4 7, 5 7, 6 7, 7 7, 8 7, 9 7, 10

8 8, 1 8, 2 8, 3 8, 4 8, 5 8, 6 8, 7 8, 8 8, 9 8, 10

9 9, 1 9, 2 9, 3 9, 4 9, 5 9, 6 9, 7 9, 8 9, 9 9, 10

10 10, 1 10, 2 10, 3 10, 4 10, 5 10, 6 10, 7 10, 8 10, 9 10, 10

10P Same number

100( )

110

=

=


24

(ii) Anita chooses an even number and Rebecca chooses an odd number These outcomes are highlighted in blue below:

Rebecca

1 2 3 4 5 6 7 8 9 10

Ani

ta

1 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6 1, 7 1, 8 1, 9 1, 10

2 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6 2, 7 2, 8 2, 9 2, 10

3 3, 1 3, 2 3, 3 3, 4 3, 5 3, 6 3, 7 3, 8 3, 9 3, 10

4 4, 1 4, 2 4, 3 4, 4 4, 5 4, 6 4, 7 4, 8 4, 9 4, 10

5 5, 1 5, 2 5, 3 5, 4 5, 5 5, 6 5, 7 5, 8 5, 9 5, 10

6 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6 6, 7 6, 8 6, 9 6, 10

7 7, 1 7, 2 7, 3 7, 4 7, 5 7, 6 7, 7 7, 8 7, 9 7, 10

8 8, 1 8, 2 8, 3 8, 4 8, 5 8, 6 8, 7 8, 8 8, 9 8, 10

9 9, 1 9, 2 9, 3 9, 4 9, 5 9, 6 9, 7 9, 8 9, 9 9, 10

10 10, 1 10, 2 10, 3 10, 4 10, 5 10, 6 10, 7 10, 8 10, 9 10, 10

(Even, Odd)

25P

10014

=

=


25

(iii) at least one of them chooses a prime number Prime numbers are: 2, 3, 5 and 7 These outcomes are highlighted in green below:

Rebecca

1 2 3 4 5 6 7 8 9 10

Ani

ta

1 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6 1, 7 1, 8 1, 9 1, 10

2 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6 2, 7 2, 8 2, 9 2, 10

3 3, 1 3, 2 3, 3 3, 4 3, 5 3, 6 3, 7 3, 8 3, 9 3, 10

4 4, 1 4, 2 4, 3 4, 4 4, 5 4, 6 4, 7 4, 8 4, 9 4, 10

5 5, 1 5, 2 5, 3 5, 4 5, 5 5, 6 5, 7 5, 8 5, 9 5, 10

6 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6 6, 7 6, 8 6, 9 6, 10

7 7, 1 7, 2 7, 3 7, 4 7, 5 7, 6 7, 7 7, 8 7, 9 7, 10

8 8, 1 8, 2 8, 3 8, 4 8, 5 8, 6 8, 7 8, 8 8, 9 8, 10

9 9, 1 9, 2 9, 3 9, 4 9, 5 9, 6 9, 7 9, 8 9, 9 9, 10

10 10, 1 10, 2 10, 3 10, 4 10, 5 10, 6 10, 7 10, 8 10, 9 10, 10

64P One or more prime

10

25

)0

(

16

=

=


26

(iv) Anita chooses a multiple of two and Rebecca chooses a multiple of three Multiples of two: 2, 4, 6, 8, 10 Multiples of three: 3, 6, 9 These outcomes are shown below in yellow:

Rebecca

1 2 3 4 5 6 7 8 9 10

Ani

ta

1 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6 1, 7 1, 8 1, 9 1, 10

2 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6 2, 7 2, 8 2, 9 2, 10

3 3, 1 3, 2 3, 3 3, 4 3, 5 3, 6 3, 7 3, 8 3, 9 3, 10

4 4, 1 4, 2 4, 3 4, 4 4, 5 4, 6 4, 7 4, 8 4, 9 4, 10

5 5, 1 5, 2 5, 3 5, 4 5, 5 5, 6 5, 7 5, 8 5, 9 5, 10

6 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6 6, 7 6, 8 6, 9 6, 10

7 7, 1 7, 2 7, 3 7, 4 7, 5 7, 6 7, 7 7, 8 7, 9 7, 10

8 8, 1 8, 2 8, 3 8, 4 8, 5 8, 6 8, 7 8, 8 8, 9 8, 10

9 9, 1 9, 2 9, 3 9, 4 9, 5 9, 6 9, 7 9, 8 9, 9 9, 10

10 10, 1 10, 2 10, 3 10, 4 10, 5 10, 6 10, 7 10, 8 10, 9 10, 10

15P(Multiple of 2, Multiple of 3)

1003

20

=

=


27

(v) neither of them choose 5.

The outcomes where either girl chooses a 5 are highlighted in orange below: Rebecca

1 2 3 4 5 6 7 8 9 10

Ani

ta

1 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6 1, 7 1, 8 1, 9 1, 10

2 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6 2, 7 2, 8 2, 9 2, 10

3 3, 1 3, 2 3, 3 3, 4 3, 5 3, 6 3, 7 3, 8 3, 9 3, 10

4 4, 1 4, 2 4, 3 4, 4 4, 5 4, 6 4, 7 4, 8 4, 9 4, 10

5 5, 1 5, 2 5, 3 5, 4 5, 5 5, 6 5, 7 5, 8 5, 9 5, 10

6 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6 6, 7 6, 8 6, 9 6, 10

7 7, 1 7, 2 7, 3 7, 4 7, 5 7, 6 7, 7 7, 8 7, 9 7, 10

8 8, 1 8, 2 8, 3 8, 4 8, 5 8, 6 8, 7 8, 8 8, 9 8, 10

9 9, 1 9, 2 9, 3 9, 4 9, 5 9, 6 9, 7 9, 8 9, 9 9, 10

10 10, 1 10, 2 10, 3 10, 4 10, 5 10, 6 10, 7 10, 8 10, 9 10, 10

P(one of chooses a 5) 19100

=

P(no 5) 1 P(5)19

1100

81100

= −

= −

=


28

11. A lunch menu has ‘made to order’ sandwiches. You can pick one

type of bread, one meat and one other item.

There are two choices of bread: brown or white; three choices of

meat: ham, chicken or turkey; and four other items: coleslaw,

cheese, tomato or lettuce.

Draw a sample space diagram and use it to work out the probability that a customer

chosen at random purchased:



29

(i) a brown bread sandwich

These outcomes are highlighted in pink below:

12P(Brown)

2412

=

=


30

(ii) a sandwich with chicken or coleslaw

These outcomes are highlighted in brown below:

12P(Chicken or Coleslaw)

2412

=

=


31

(iii) a sandwich with cheese on it


6P(Cheese)

2414

=

=


32

(iv) a sandwich with turkey and lettuce


2P(Turkey, Lettuce)

241

12

=

=


33

(v) a white bread sandwich with ham and tomato. These outcomes are highlighted in red below:

1

P(White, Ham, Tomato)24

=

Practice Questions 19.2 1. Decide whether these events are independent or dependent.

(i) You have a jar with 24 chocolates and 14 jellies. You take one sweet at random

from the jar, eat it, and then take a second sweet at random from the jar.

Dependent: The sweet is not replaced so there are fewer sweets to choose

from on the next pick.

(ii) Derek has a blue, a red, and a green tie. He also has a blue and a green shirt. He

chooses a random tie and shirt for work today.

Independent: The choice of tie does not affect the choice of shirts.


34

(iii) Amy plays card games. She picks a card at random. Then without putting the first

card back, she picks a second card at random.

Dependent: The card is not replaced so there are fewer cards to choose from

on the next pick.

(iv) Ian has 14 coins. He takes three of them at random, then he puts these back, and

then picks two more coins at random.

Independent: The coins are replaced so there are the same numbers of coins

to pick from each time.

(v) Jeff has three children. His first two children are boys. His last child is a girl.

Independent: Each child’s gender is not affected by the gender of the

others.

(vi) A plant has four red flowers and two blue flowers. Brenda picks one flower from

the plant. After some time, her sister picks a flower from the same plant.

Dependent: The flowers aren’t replaced, so there are fewer flowers to

choose from the next pick.

Questions 2 to 5 are based on independent events

2. You roll a fair six-sided die twice.

What is the probability the first roll shows a five and the second roll shows a six?

P(5 on 1st Roll and 6 on 2nd Roll)

P(5 and 6) ‘and’ means multiply

= P(5) × P(6)

1 16 6136

= ×

=


35

3. There are nine mugs in Tom’s cupboard, four blue and

five green.

He randomly selects one to use on Monday, washes it

and replaces it. He randomly selects another on

Tuesday.

What is the probability that:

(i) he uses a blue mug both days?

There are 4 blue mugs and 5 green mugs giving a total of 9 mugs.

P(Blue on Monday and Blue on Tuesday)

P(Blue and Blue)

= P(Blue) × P(Blue)

4 49 91681

= ×

=

(ii) he uses a blue mug on Monday and a green mug on Tuesday?

P(Blue on Monday and Green on Tuesday)

P(Blue and Green)

= P(Blue) × P(Green)

4 59 92081

= ×

=

4. Shannon spins a spinner numbered 1 to 7, and tosses a coin.

What is the probability she gets:

(i) an odd number on the spinner?

P(Odd)

47

=


36

(ii) an odd number on the spinner and a tail on the coin?

P(Odd and Tail)

= P(Odd) × P(Tail)

4 17 24

1427

= ×

=

=

5. You flip a coin and then roll a fair six-sided die.

(i) Draw a tree diagram to represent this situation.

Write the probabilities on each branch.

Use the tree diagram to calculate the probability that:

(ii) the coin lands heads-up and the die shows a three?

P(Head and 3)

= P(Head) × P(3)

1 12 61

12

= ×

=


37

(iii) the coin lands tails-up and the die shows a number greater than four?

P(Tail and 5, 6)

= P(Tail) × P(5 or 6)

1 22 62

1216

= ×

=

=

(iv) the coin lands heads-up and the die shows an even number?

P(Head and 2, 4, 6)

= P(Head) × P(2 or 4 or 6)

1 32 63

1214

= ×

=

=

(v) the coin lands heads-up and the die shows a prime number?

P(Head and 2, 3, 5)

= P(Head) × P(2 or 3 or 5)

1 32 63

1214

= ×

=

=


38

Questions 6 to 9 are based on dependent events

6. The names of 9 boys and 11 girls from your class are put into a hat. Two names are

selected without replacement.

(i) Draw a tree diagram to represent this situation.



(ii) the two names chosen will both be girls?

There are 9 boys and 11 girls so there are 20 students in total.

P(1st is a girl and 2nd is a girl)

= P(Girls) × P(Girls)

1 girl gone11 101 person gone20 19

1138

→= ×

→

=

(iii) the first name will be a boy and the second will be a girl?

P(1st is a boy and 2nd is a girl)

= P(Boys) × P(Girls)

9 11

1 person gone20 1999

380

= ×→

=


39

(iv) both names will be boys?

P(1st is a boy and 2nd is a boy)

= P(Boys) × P(Boys)

1 boy gone9 81 person gone20 19

1895

→= ×

→

=

7. A card is chosen at random from a standard deck of 52 playing cards. Without replacing

it, a second card is chosen.

What is the probability that the first card chosen is a queen and the second card chosen is

a jack?

P(1st is a Queen and 2nd is a Jack)

= P(Queen) × P(Jack)

4 4

1 card gone52 514

663

= ×→

=

8. A bowl of fruit is on the table. It contains four apples, six oranges and three bananas.

Alan and Kenneth come home from school and randomly grab one fruit each. What is the

probability that both grab apples?

There are 4 apples, 6 oranges and 3 bananas so there are 13 pieces of fruit in total.

P(Apple and Apple)

= P(Apple) × P(Apple)

1 apple gone4 31 piece of fruit less13 12

113

→= ×

→

=


40

9. Your drawer contains 11 red socks and six blue socks. It’s too dark to see which are

which, but you grab two anyway.

(i) Draw a tree diagram to represent this information. Write the probabilities on

each branch.


(ii) the first sock is red and the second is blue?

P(1st is Red and 2nd is Blue)

= P(Red) × P(Blue)

11 6

1 sock less17 1633

136

= ×→

=

(iii) the first sock is blue and the second is red?

P(1st is Blue and 2nd is Red)

= P(Blue) × P(Red)

6 1117 1633

136

= ×

=


41

(iv) both socks are blue?

P(Blue and Blue)

= P(Blue) × P(Blue)

6 517 1615

136

= ×

=

(v) both sock are red?

P(Red and Red)

= P(Red) × P(Red)

11 1017 1655

136

= ×

=

10. The game of backgammon uses two standard dice, each with the

numbers one through six.

You need to roll double twos to win the game.

What is the probability you will get that result on your next roll?

P(2 and 2)

= P(2) × P(2)

1 16 6136

= ×

=

11. I roll three dice.

(i) Are these events dependent or independent?

These events are independent because the result of rolling each die does not

influence the next die.


42

(ii) What is the probability that I get three ones?

P(1 and 1 and 1)

= P(1) × P(1) × P(1)

1 1 16 6 61

216

= × ×

=

12. I draw a card from a standard deck, replace it, and draw another.


These events are independent because the card is replaced so there are the same

number of cards to pick from on the second draw.

(ii) What is the probability I get two aces?

P(Ace and Ace)

= P(Ace) × P(Ace)

4 452 5216

27041

169

= ×

=

=

13. Four cards are chosen at random from a deck of 52 cards, one after the other, without

replacement.


These events are dependent because the cards are not replaced so there are fewer

cards to pick from each time.


43

(ii) What is the probability of choosing a ten, a nine, an eight and a seven in that

order?

P(10 and 9 and 8 and 7)

= P(10) × P(9) × P(8) × P(7)

4 4 4 452 51 50 49

32812175

× × ×=

=

14. Henry has three black shirts and seven blue shirts in his wardrobe. Two shirts are drawn

without replacement from the wardrobe. What is the probability that both of the shirts are

black?

There are 3 black shirts and 7 blue shirts so a total of 10 shirts.

P(Black and black)

= P(Black) × P(Black)

3 210 96

901

15

= ×

=

=

15. An archer always hits a circular target with each arrow shot.

He hits the bullseye two out of every five shots on average.

If he takes three shots at the target, calculate the probability that

he hits the bullseye:

(i) every time

Every time : P(Bullseye and Bullseye and Bullseye)

= P(Bullseye) × P(Bullseye) × P(Bullseye)

2 25 5

8125

25× ×

=

=


44

(ii) the first two times but not the third

P(Bullseye and Bullseye and no Bullseye)

= P(Bullseye) × P(Bullseye) × P(No Bullseye)

P(No Bullseye) = 1 – P(Bullseye)

215

35

= −

=

P(Bullseye and Bullseye and no Bullseye)

2 35 5

2512125

× ×

=

=

(iii) on no occasion.

P(No Bullseye and no Bullseye and no Bullseye)

= P(no Bullseye) × P(no Bullseye) × P(no Bullseye)

3 35 5

3527

125

× ×

=

=


45

16. A class has 14 boys and 16 girls. Three students are chosen at random to represent the

class at a school event.


(i) they are all girls?

There are 14 boys and 16 girls so 30 students in total.

P(all girls)

= P(Girl) × P(Girl) × P(Girl)

15 1429

12

630 84

29

× ×=

= (Dependent events)

(ii) they are all boys?

P(all boys)

= P(Boy) × P(Boy) × P(Boy)

13 1229

1430 2813

145

× ×

=

=

17. Emma says that it is very difficult to get five heads in a row when you flip a coin. Brian

disagrees and says he has done it loads of times.

(i) Which person do you agree with?

I agree with Emma as the probability is low of this happening. 132

Neither of them have a coin to try it so they decide to work the chances out

mathematically.


46

(ii) Show the calculations they would use to do this.

5 heads in a row : P(Head and Head and Head and Head and Head)

= P(Head) × P(Head) × P(Head) × P(Head) × P(Head)

1 1 1 1 12 2 2 2 2132

× × ×

=

×=

(iii) Do you want to change your answer to (i) above? Justify your answer using the

calculations in (ii) above.

The probability of this happening is only 132

which means once in 32 attempts.

Practice Questions 19.3 1. Decide whether these events are mutually exclusive or not. Explain.

(i) Roll a die; get an even number and a number less than 3.

Not mutually exclusive: 2 is an even number less than 3

Both events can occur at the same time. i.e. Roll a 2

(ii) Roll a die; get a prime number and an odd number.

Not mutually exclusive: 3 and 5 are prime and odd

Both events can occur at the same time. i.e. Roll a 3 or 5

(iii) Roll a die; get a number greater than 3 and a number less than 3.

Mutually exclusive: a number can’t be greater than 3 and less than 3 at the same

time

The events cannot occur at the same time.

(iv) Select a student in the classroom; student has blond hair and blue eyes.

Not mutually exclusive: A student can have blue eyes and blond hair.

Both events can occur at the same time.


47

(v) Select a student in your school; the student is a second year and does business.

Not mutually exclusive: A student can be in 2nd year and study Business.

Both events can occur at the same time.

(vi) Select a card from a standard deck; the card is red and a king.

Not mutually exclusive: A king can be red.

Both events can occur at the same time i.e. king of hearts.

2. A shop decides to pick a month for its annual sale.

Calculate the probability that:

(i) it is a month beginning with J

P(Month beginning with J) 312

= (January, June, July)

14

=

(ii) it is a month with 30 days

P(Month with 30 days) 412

= (Sept, Apr, June, Nov)

13

=

(iii) it is either September or March

P(September or March) Mutually exclusive

= P(September) + P(March)

1 112 122

1216

= +

=

=


48

(iv) it has less than five letters or has an ‘A’ in it.

P(<5 letters or has on ‘A’) Not mutually exclusive

= P(<5 letters) + P(with ‘A’) – P(both)

3 1

P 5 letters1

(4

)2

< = = (May, June, July)

6 1

P with ‘A’( )12 2

= = (Jan, Feb, Mar, Apr, May, Aug)

1

P(both)12

= (May)

P(<5 letters) + P(with ‘A’) – P(both)

1 1 14 2 1223

= + −

=

3. A card is picked at random from a standard deck.

What is the probability that it is:

(i) a red card?

2P

6(R

51

ed)2

2

=

=

(ii) a picture card?

12( )P Pi

5ctur

1

e2

33

=

=


49

(iii) a red card or a picture card?

P(Red or Picture) Not mutually exclusive

= P(Red) + P(Picture) – P(Both)

P(Both) = 652

(J♥, Q♥, K♥, J◊, Q◊, K◊)

P(Red or Picture)

= P(Red) + P(Picture) – P(Both)

1 3 62 13 52

= + −

8

13=

(iv) a king or a black card?

P(King or Black) Not mutually exclusive

= P(King) + P(Black) – P(Both)

P(King) 452

= P(Black) 12

= P(Both) 252

= (K♠, K♣)

P(King or Black)

= P(King) + P(Black) – P(Both)

4 1 2

52 2 52= + −

7

13=


50

(v) a red queen or a black king?

P(Red Queen or Black King) Mutually exclusive: Can’t be both

= P(Red Queen) + P(Black King)

2 252 524

521

13

= +

=

=

4. A bag contains 20 chocolates; 5 are milk chocolate, 6 are dark chocolate, and 9 are white

chocolate.

If a chocolate is selected at random, what is the probability that the chocolate

chosen is:

(i) milk chocolate?

5P(Milk Chocolate)

2014

=

=

(ii) dark chocolate?

6P(Dark Chocolate)

203

10

=

=

(iii) either milk chocolate or white chocolate?

P(Milk or Dark) Mutually exclusive: Can’t be both

= P(Milk) + P(Dark)

1 34 101120

= +

=


51

5. The probability of a student owning a mobile phone is 0·83. The probability of a student

owning an iPad is 0·58. The probability of a student owning both devices is 0·46. What is

the probability that a student chosen at random will own either a phone or an iPad?

P(Mobile) = 0·83

P(iPad) = 0·58

P(Mobile and iPad) = 0·46

P(Mobile or iPad)

These events are not mutually exclusive because can have both a mobile and an iPad.

P(Mobile) + P(iPad) – P(Both)

= 0·83 + 0·58 – 0·46

= 0·95

6. A number between 1 and 10 is chosen at random.

What is the probability that it is:

(i) an even number?

5P(even)

1012

=

= (2, 4, 6, 8, 10)

(ii) a number greater than 3?

( )7

P 310

> = (4, 5, 6, 7, 8, 9, 10)


52

(iii) an even number or a number greater than 3?

P(even or >3) Not mutually exclusive

= P(even) + P(>3) – P(both)

P(both) = 410

(4, 6, 8, 10)

P(even or >3)

= P(even) + P(>3) – P(both)

1 7 42 10 10

= + −

45

=

(iv) a prime number or an odd number?

P(Prime or odd) Not mutually exclusive

4( )P p

1rim

2

e0

5

=

= (2, 3, 5, 7)

5( )dd

1

P o10

2

=

= (1, 3, 5, 7, 9)

P both3

( )10

= (3, 5, 7)

P(Prime or odd)

= P(Prime) + P(odd) – P(both)

2 1 35 2 1035

= + −

=


53

7. At a particular school with 400 students, 116 play football, 80 play basketball and 16 play

both.

What is the probability that a randomly selected student:

(i) plays football?

116P football =( )

40029

100=

(ii) plays basketball?

80P basketball =

40( )

015

=

(iii) plays basketball or football?

P(basketball or football) Not mutually exclusive

P(Basketball) + P(Football) – P(Both)

29 1 16100 5 40049 16

100 400920

= + −

= −

=


54

8. The probability that a man will be alive in 15 years is 45

and the probability that his wife

will also be alive is 34

.

What is the probability that in 15 years: (i) they will both be alive? P(Both) = P(man and woman alive) “and” means mutiply = P(man) × P(woman)

4 35 435

= ×

=

(ii) at least one of them will be alive? 1 1

Both die 5 4120

= ×

=

P(At least one alive) = 1 – P(Both die)

1120

1920

= −

=

(iii) neither of them will be alive? P(Neither) = [1 – P(Man alive)] × [1 – P(Woman alive)]

1 15 4120

= ×

=

(iv) only the wife will be alive? P(Wife alive and man not alive) = P(Wife) × [1 – P(Man alive)]

3 14 5320

= ×

=


55

9. A shop carried out a survey and found that 40% of the customers were male. They also

found that 75% of the males spent over €100 on an average visit to the shop and 55% of

the females spent over €100.

What is the probability that a customer chosen at random is:

(i) a female or a customer who spends less than €100 on average?

75% of 40% = 30% of total are male spending > €100

so 10% of total are male spending < €100

55% of 60% = 33% of total are female spending > €100

so 27% of total are female spending < €100

Male Female

< €100 10% 27%

> €100 30% 33%

P(female or < €100) = P(female) + P(< €100) – P(female and < €100)

= 60% + 37% – 27%

= 70%

(ii) a customer that spends more than €100?

P(Customer that spends > €100)

P(Male > €100 or Female > €100)

= (75% of 40%) + (55% of 60%)

= 30% + 33%

= 63%


56

10. A box contains nine green marbles, eight blue marbles and eleven yellow marbles. Dave

picks two marbles without looking.

(i) Draw a tree diagram to represent this information. Write the probabilities on each

branch.

9 Green, 8 Blue, 11 Yellow Total number of marbles: 28

Assume no replacement.

Use your tree diagram to work out the probability that:

(ii) the 1st will be blue and the 2nd will be yellow?

P(Blue) and P(Yellow)8 1128 2722

189

= ×

=

“and” means multiply


57

(iii) both will be green?

P(Green) and P(Green)

9 828 27221

= ×

=

(iv) the first will be green and the second will be blue?

P(Green) and P(Blue)

9 828 27221

= ×

=

(v) both will be yellow?

P(Yellow) and P(Yellow)

11 1028 2755

378

= ×

=

(vi) one will be green and the other will be yellow?

P(1st Green) and P(2nd Yellow) or P(1st Yellow) and P(2nd Green)

9 11 11 928 27 28 27

11 1184 841142

= × + ×

= +

=

“and” means multiply; “or” means add


58

(vii) at least one will be yellow?

P(At least one Yellow)

P(Green and Yellow) or P(Blue and Yellow) or P(Yellow and Green) or P(Yellow

and Blue) or P(Yellow and Yellow)

9 11 8 11 11 9 11 8 11 1028 27 28 27 28 27 28 27 28 27

11 22 11 22 5584 189 84 189 378121189

= × + × + × + × + ×

= + + + +

=

11. Fourteen coloured discs are placed in a bag. Six are red, five are blue and three are

yellow. Two discs are picked from the bag without replacement.

14 discs in total: 6 red, 5 blue, 3 yellow

Tree diagram:


59


(i) both discs are red?

P(Red) and P(Red)6 5

14 131591

= ×

=

(ii) one disc is blue and the other is yellow?

P(1st Blue) and P(2nd Yellow) or P(1st Yellow) and P(2nd Blue)

5 3 3 514 13 14 13

15 15182 18230

1821591

= × + ×

= +

=

=

(iii) both discs are blue?

P(Blue) and P(Blue)5 4

14 1320

1821091

= ×

=

=

(iv) the first is blue and the second is not blue?

st ndP(1 Blue) and P(2 not Blue)5 41

14 135 9

14 1345

182

= × −

= ×

=


60

(v) one is red and the other is blue?

st ndP(1 Red) and P(2 Blue)6 5

14 131591

= ×

=

st ndP(1 Blue) and P(2 Red)5 6

14 131591

= ×

=

P(RB) or P(BR)15 1591 913091

= +

=

(vi) at least one is yellow?

P(At least one is Yellow)

P(RY) or P(BY) or P(YY) or P(YR) or P(YB)

5 3 3 514 13 14 13

6 3 3 2 3 614 13 14 13 14 139 15 3 9 1591 182 91 91 1823691

× × = × + + × + × +

= + + + +

=

12. A jar contains coloured stones consisting of four pink stones, nine orange stones and five green stones.

Ryan picks one stone, records its colour and puts it back in the jar. Then he draws another stone.

4 Pink, 9 Orange, 5 Green Total number of stones: 18


61


These events are independent because the stone is replaced each time.

What is the probability of taking out:

(ii) an orange stone followed by the green stone?

P(1st Orange and 2nd Green)

9 518 185

36

= ×

=

(iii) two pink stones?

P(Pink and Pink)

4 418 18481

= ×

=

(iv) at least one green stone?

P(At least one green)

= [P(Green and Green)] or [P(Green and Pink)] or [P(Green and

Orange)] or [P(Pink and Green)] or [P(Orange and Green)]

5 5 5 4 5 9 4 5 9 518 18 18 18 18 18 18 18 18 18155324

× + × + × + × + ×

=

Practice Questions 19.4

1. On separate Venn diagrams of events A and B, shade the regions that represent the

following:

(i) P(A)


62

P(A)

(ii) P(B)

P(B)

(iii) P(A or B)

P(A or B)

(iv) P(A and B)

P(A and B)

(v) P(not A)

P(not A)

(vi) P(B only).


63

P(B only)

2. In a class of 30 students, 25 passed their Christmas Maths exam, 24 passed their Irish

exam and 23 passed both exams.

Draw a Venn diagram to represent this information and use it to calculate the probability

that a student chosen at random:

30 students 25 passed Maths, 24 passed Irish 23 passed both.

Only Maths = 25 – both

= 25 – 23

= 2

Only Irish = 24 – both

= 24 – 23

= 1

Neither

Total – (Maths + Irish + both)

= 30 – (2 + 1 + 23)

= 30 – 26

= 4

(i) didn’t pass maths


64

1 4P(not Maths) =305

3016

+

=

=

(ii) passed maths or Irish but not both

P(Maths only) or P(Irish only)2 1=

30 303

301

10

+

=

=

(iii) passed both exams

P(both)

23=30

(iv) passed maths only P(Maths only)

2=

301

15=

(v) didn’t pass either exam. P(Neither)

4=302

15=

3. Twenty-six people were surveyed about their choice of mobile phones. The survey finds

that 14 people have Apple iPhones, 10 have Samsungs and five have Nokias. Four have


65

Apple iPhones and Samsungs, three have Apple iPhones and Nokias and one has a Samsung and a Nokia. No one has all three types of phone. Represent this information on a Venn diagram and use it to calculate the probability that a person chosen at random has:

26 people 14 iPhones, 10 Samsungs, 5 Nokias 4 iPhones + Samsungs, 3 iPhones + Nokia; 1 Samsung + Nokia 0 All 3.

iPhone only

14 – (4 + 3) = 14 – 7 = 7

Samsung only

10 – (4 + 1)

= 10 – 5

= 5

Nokia only

5 – (1 + 3)

= 5 – 4

= 1

None

26 – (7 + 4 + 3 + 1 + 5 + 1)

= 26 – 21

= 5

(i) an iPhone


66

14P(iPhone)

267

13

=

=

(ii) a Samsung or a Nokia

5 1 1 3 4P(Samsung or Nokia)26

14267

13

+ + + +=

=

=

(iii) two phones

P iPhone and Samsung or P iPhone and Nokia or P Samsung and Nok

P(Two phones)( ) ( )( )

3 4 126

8264

a

13

i

+ +=

=

=

=

(iv) no phone

5P(no phone)26

=

(v) no iPhone.

5 1 1 5P(no iPhone)26

12266

13

+ + +=

=

=


67

4. A group of 20 people are waiting at a bus stop one morning. Nine of them

have an umbrella, six have a raincoat and three have both an umbrella and

a raincoat.

Represent this information on a Venn diagram and use it to calculate the

probability that a person chosen at random has:

20 people: 9 Umbrella, 6 raincoat, 3 both.

Umbrella only

= 9 – 3

= 6

Raincoat only

= 6 – 3

= 3

None

20 – (6 + 3 + 3)

= 20 – 12

= 8

(i) an umbrella

9P(Umbrella)20

=

(ii) an umbrella and a raincoat

( )P Umbrella and Raincoat 320

=


68

(iii) an umbrella or a raincoat P(Umbrella or Raincoat) = P(Umbrella) + P(Raincoat) − P(Umbrella) and Raincoat)

9 6 320 20 20122035

= + −

=

=

(iv) neither.

8P(neither)2025

=

=

5. A group of students were asked if they preferred soccer, rugby or Gaelic football. The results are shown on the Venn diagram.

Calculate the probability that a randomly chosen students likes: (i) soccer or rugby

Total : 7 14 12 9 15 6 20 19 102

P(Soccer) or P(Rugby) = Rugby + Soccer

7 14 12 9 15 6= 102

631022134

+ + + + + + + =

+ + + + +

=

=


69

(ii) rugby or Gaelic football

P(Rugby) or P(Gaelic)Rugby Gaelic20 9 15 6 14 12

10276

1023851

= ++ + + + +

=

=

=

(iv) soccer and Gaelic football but not rugby

P(Soccer and Gaelic but not Rugby)

7 9 20102

361026

17

+ +=

=

=

(iv) only Gaelic football

20P(Only Gaelic)

1021051

=

=

(v) none of these sports.

19

P(None)102

=

6. Two events A and B have the following probabilities: P(A) = 0·4 P(B) = 0·6 P(A ∩ B) = 0·2 Draw a Venn diagram to represent these probabilities and use it to calculate:

P(A\B)= 0·4 – 0·2 P(B\A)= 0·6 – 0·2

= 0·2 = 0·4


70

(i) P(A ∪ B)

P(A ∪ B)

= P(A) + P(B) − P(A ∩ B)

= 0·6 + 0·4 − 0·2

= 0·8

(ii) P(A ∪ B)′

P(A ∪ B)′ = 0·2

7. Let A and B be independent events, where P(A) = 0·4 and P(B) = 0·7.

(i) Find P(A ∩ B)

P(A ∩ B) = P(A and B)

= P(A) × P(B)

= 0·4 × 0·7

= 0·28

(ii) Find P(A ∪ B)

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

= 0·4 + 0·7 – 0·28

= 0·82

(iii) Draw a Venn diagram and shade the region that represents A ∩ B′

A ∩ B′ = A\B


71

(iv) Find P(A ∩ B′).

P(A ∩ B′) = P(A/B)

= P(A only)

= 0·4 – 0·28

= 0·12

8. A driving test consists of a practical test and theory test. One day everyone who took the

test passed at least one section. 64% passed the practical section and 78% passed the

theory section.

64% Practical (Pr)

78% Theory (T)

64 + 78 = 142%

Pr ∩ T = 142% − 100%

= 42%

(i) Represent this information on a Venn diagram showing the probabilities of

candidates in each section of the diagram.

Pr′ = 64 – 42

= 22%

T′ = 78 – 42

= 36%

One person is chosen at random from all the people who took the test that day. What is

the probability that this person:

(ii) passed the practical section and the theory section?

P(Theory and Practical) = 42% (both = Pr ∩ T)


72

(iii) passed the theory section only?

P(Theory only) = 36%


1. Jamie is playing a game where he must toss a coin until he gets a head. What is the

probability that he gets a head for the first time on:

(i) the first throw?

Success = Flipping a head

1 1 1P(Success) P(Failure) 1

2 2 21

P(Head on first flip)2

= = −

=

(ii) the second throw?

1 1P(FS)

2 214

= ×

=

(iii) the third throw?

1 1 1P(FFS)

2 2 218

= × ×

=

(iv) his fifth throw?

44 1 1P(F) P(S)

2 2132

× = ×

=


73

2. Ciaran is playing a game using a fair die. He must throw the die until he gets a 6.

What is the probability that he throws a 6 in:

Success = throwing a 6

1P(S) P(F) 1 P(S)6

1 16

5 6

= = −

−

(i) one throw?

1

P(S)6

=

(ii) two throws?

5 1P(FS)

6 65

36

= ×

=

(iii) three throws?

5 5 1P(FFS)

6 6 625

216

= × ×

=

(iv) his tenth throw?

P(Success on 10th throw)

9

9

P(F) P(S)

5 16 6

0 0323

= ×

= ×

= ⋅


74

3. A fair die is thrown three times.

What is the probability that a 2 or a 3 is thrown:

2 1Success 2 or 3 P(S)6 3

P(F) 1 P(S)1 = 13

2 3

= = =

= −

−

=

(i) for the first time on the third throw?

P(FFS)2 2 13 3 34

27

= × ×

=

(ii) once in three throws?

P(FFS) or P(SFF) or P(FSF)4 1 2 2 2 1 2

27 3 3 3 3 3 349

= + × × + × ×

=

4. A spinner has five equal sections.

Three of them are green and two of them are yellow. Chloe is playing a

game where she has to spin the spinner until it lands on yellow.

What is the probability that she lands on yellow for the first time on:

3 2 2P(Green) P(Yellow) P(S) P(Yellow)5 5 5

3P(F) P(Green)5

= = = =

= =


75

(i) the first spin?

P(Yellow) = P(S) = 25

(ii) the third spin?

P(GGY)=P(FFS)

3 3 25 5 518

125

= × ×

=

(iii) the sixth spin?

5

5

P(F) P(S)

3 25 5486

15625

×

= ×

=

(iv) the ninth spin?

Give your answer in scientific notation.

P(F)8 × P(S)

8

3

3 25 5

6 718464 10−

= ×

= ⋅ ×

5. Zoe is the best free-throw taker on her basketball team. Her average probability

of scoring from a free shot is 0·78.

If she takes three free throws in a game, what is the probability that she will:

P(S) = 0·78 P(F) = 1 – P(S)

= 1 – 0·78

= 0·22


76

(i) score all three?

P(SSS) = 0·78 × 0·78 × 0·78

= 0·474552

(ii) score two out of three?

Scores 2 out of 3.

P(SSF) or P(SFS) or P(FSS)

= (0·78 × 0·78 × 0·22) + (0·78 × 0·22 × 0·78) + (0·22 × 0·78 × 0·78)

= 0·401544

(iii) score only on the final throw?

Score on final throw

P(FFS)

= (0·22 × 0·22 × 0·78)

= 0·037752

(iv) not score at all?

No score

P(FFF)

= 0·22 × 0·22 × 0·22

= 0·010648

(v) score at least once?

Scores at least once

= Scores once or twice or three times

= P(SFF) or P(FSF) or P(FFS) or P(SSF) or P(SFS) or P(FSS) or P(SSS)

= (0·22 × 0·22 × 0·78) + (0·78 × 0·22 × 0·22) + (0·22 × 0·78 × 0·22) + 0·401544

+ 0·474552

= 0·037752 + 0·037752 + 0·037752 + 0·401544 + 0·474552

= 0·989352


77

6. At the end of Maths class, a teacher gives her students a three-question multiple-choice

test to check if they have understood the lesson. Each question has three options. A

student guesses on all questions.

What is the probability that he gets:

Success = Guessing the correct answer out of the 3 options

1 2

P(S) (3 choices) P(F)3 3

= =

(i) all questions correct?

1 1 1P(SSS)3 3 3127

= × ×

=

(ii) two questions correct?

P(2Q correct) P(SSF) or P(SFS) or P(FSS)1 1 2 1 2 1 2 1 13 3 3 3 3 3 3 3 32 2 2

27 27 2762729

=

= × × + × × + × ×

= + +

=

=

(iii) no questions correct?

2 2 2P(FFF)3 3 3827

= × ×

=

(iv) the first two correct and the final question incorrect?

1 1 2P(SSF)3 3 32

27

= × ×

=


78

7. In a survey of her class, Olivia found that 35% of the students walk to school. Three

students are selected at random from the class. Giving your answers as a decimal, what is

the probability that:

35% walk Success = A student walks

P(S) = 0·35 P(F) = 1 – 0·35

= 0·65

(i) all three students walk to school?

P(SSS) = 0·35 × 0·35 × 0·35

= 0·042875

(ii) none of the three students walk to school?

P(FFF) = 0·65 × 0·65 × 0·65

= 0·274625

(iii) two out of the three students walk to school?

P(SSF) or P(SFS) or P(FSS)

= (0·35 × 0·35 × 0·65) + (0·35 × 0·65 × 0·35) + (0·65 × 0·35 × 0·35)

= 0·079625 + 0·079625 + 0·079625

= 0·238875

8. Charlie is the goalkeeper on the school soccer team.

His average success rate for saving goals is 5 .8

If there

are three scoring opportunities in a match, what is the probability that:

5 5P(S) P(F) 18 8

3 8

= = −

=


79

(i) he saves all three?

5 5 5P(SSS)8 8 8125512

= × ×

=

(ii) he doesn’t save any of them?

3 3 3P(FFF)8 8 827

512

= × ×

=

(iii) he saves the first two but not the third?

5 5 3P(SSF)8 8 875

512

= × ×

=

(iv) he saves one?

P(one success) P(SFF) or P(FSF) or P(FFS)5 3 3 3 5 3 3 3 58 8 8 8 8 8 8 8 845 45 45

512 512 512135512

=

= × × + × × + × ×

= + +

=

9. Stephen comes to school by car. On his way to school he passes three

sets of traffic lights. The probability that the lights will be red is 47

.


4 4P(Red) P(S) P(F) 17 7

3 7

= = = −

=


80

(i) the first set of lights will be green?

3

P(F)7

=

(ii) the first time the lights are red is at the last set of lights?

3 3 4P(FFS)7 7 736343

= × ×

=

(iii) all three sets of lights are red?

4 4 4P(SSS)7 7 764343

= × ×

=

(iv) at least one set of lights are red? P(At least one S) P(FFS) or P(FSF) or P(SFF) or

P(FSS) or P(SFS) or P(SSF) or P(SSS)3 3 4 3 4 3 4 3 37 7 7 7 7 7 7 7 73 4 4 4 3 4 4 4 3 4 4 47 7 7 7 7 7 7 7 7 7 7 7

36 36343

=

= × × + × × + × × + × × + × × + × × + × ×

= +36 48 48 48 64

343 343 343 343 343 343316343

+ + + + +

=

10. Suppose a student takes a multiple choice test.

The test has 10 questions, each of which has four possible answers (only one correct).

(i) If the student guesses the answer to each question, do the questions form a

sequence of Bernoulli trials? Explain your answer.

Yes. Each question is independent and has two options; success or failure.

Therefore, the test forms a series of Bernoulli trials.


81

(ii) List the possible outcomes.

Possible outcomes:

1. Success – the guess is correct

2. Failure – the guess is incorrect

(iii) Write the probability associated with each outcome.

1 1P(Success) P(Failure) 14 4

3 4

= = −

=


1. A goalkeeper has a probability of 35

of saving a penalty. How many penalties would you

expect him to save out of 50 penalties taken?

3P(Save) 50 Penalties5

Expected3 50530 Expect to save

n

p n

= =

= ×

= ×

=

2. A batch of 1,600 items is examined. The probability that an item from this batch is

defective is 0·04. How many items from this batch are defective?

1600 items = n P(Defective) = 0·04

Expected = p × n

= 0·04 × 1600

= 64 Expect defective


82

3. In an experiment, a standard six-sided die was rolled 72 times. The results are shown in

the table.

Which number on the die was obtained the expected number of times?

Number on the die 1 2 3 4 5 6

Frequency 11 8 12 15 16 10

1P(One number on a die) 72 rolls6

Expected1 72612

n

p n

= =

= ×

= ×

=

Therefore, 3 was rolled the expected number of times

4. In a random survey of the voting intentions of a local electorate, the

following results were obtained:

Politician A Politician B Politician C

165 87 48

Total votes = 165 + 87 + 48

= 300

(i) Calculate the probability that a randomly selected voter will vote for:

(a) Politician A

165P(A)3001120

=

=


83

(b) Politician B

87P(B)30029

100

=

=

(c) Politician C

48P(C)300425

=

=

(ii) If there are 7,500 people in the electorate, how many votes would you expect to be

for:

7,500 people = n

Expected = p × n

(a) Politician A?

11Expected 7,500204125

= ×

=

(b) Politician B?

29Expected 7,5001002175

= ×

=

(c) Politician C?

4Expected 7,500251200

= ×

=


84

5. In a raffle, 250 tickets are sold at €1 each for three prizes of €100, €50 and €10.

You buy one ticket.

250 tickets sold at €1 prizes: €100, €50, €10

(i) What is the expected value of this raffle.

Expected value: Σx.p(x)

Outcome (x) p(x) x.p(x)

1st €100 1250

1 2

100250 5

× =

2nd €50 1250

1 1

50250 5

× =

3rd €10 1250

1 1

10250 25

× =

Lose 0 247250

247

0 0250

× =

2 1 1( )5 5

.25

16250 64

x p x = + +

=

= ⋅

∑

(ii) Does it represent a good investment of €1?

Explain your answer.

It does not represent a good investment of €1 as the expected value is 0·64

which means if a large number of people bought tickets the average win is

below the pay in amount. You would expect to lose 0·36.

Net “winnings” = 0·64 – 1

= – 0·36


85

6. Jennifer is playing a game at an amusement park.

There is a 0·1 probability that she will score 10 points, a 0·2 probability that she

will score 20 points, and a 0·7 probability that she will score 30 points.

How many points can Jennifer expect to receive by playing the game?

Expected value = Σx.p(x)

x p(x) x.p(x)

10 0·1 10 × 0·1 = 1

20 0·2 20 × 0·2 = 4

30 0·7 30 × 0·7 = 21

Σx.p(x) = 1 + 4 + 21

= 26 points

7. You take out a fire insurance policy on your home.

The annual premium is €300.

In case of fire, the insurance company will pay you €200,000.

The probability of a house fire in your area is 0·0002.

X p(x) x.p(x)

Fire 200,000 0·0002 200,000 × 0·0002 = 40

No Fire 0 [1 – 0·002] 0·9998 0 × 0·9998 = 0

(i) What is the expected value?

Expected value = Σx·p(x)

= 40 + 0

= 40


86

(ii) What is the insurance company’s expected value?

The expected value is the same for the insurance company and customer.

However, the perspective is changed. The customer is down €260 since they paid

€300 for the policy, the company is up €260.

(iii) Suppose the insurance company sells 100,000 of these policies. What can the

company expect to earn?

Profit on 1 policy: = €260

100,000 policies : 100,000 × €260 = €26,000,000

8. A hundred tickets are sold for a movie at the cost of €10 each. Some tickets have cash

prizes as a part of a promotional campaign: one prize of €50, three prizes of €25 and five

prizes of €20.

What is the expected value if you buy one ticket?

Expected value = Σx.p(x)

x p(x) x.p(x)

€50 1100

1 150

100 2× =

€25 3100

3 325

100 4× =

€20 5100

520 1

100× =

0 91100

910 0

100× =

1 3. ( ) 1 02 4

9 4

€2 ·25 Expected value.

Σ = + + +

=

=

x p x

You pay €10 so “nett winnings” = 2·25 – 10 = − €7∙75

9. You pay €10 to play the following game of chance.


87

There is a bag containing 12 balls, five are red, three are green and the rest are yellow.

You are to draw one ball from the bag.

You will win €14 if you draw a red ball and you will win €12 if you draw a green ball.

How much do you expect to win or lose if you play this game 100 times?

Outcome x Pay Out p(x) x.p(x)

Red €14 512

5 3514

12 6× =

Green €12 312

312 3

12× =

Yellow 0 412

40 0

12× =

35. ( ) 36

536

8 83

Σ = +

=

= ⋅

x p x

Pay €10 – Expected value €8·83 so “nett winnings” = 8 ⋅83 – 10 6

– 7=

7Loss6

100 Games 1 167 10

€11

0350

36 67 nett loss

= −

= −

⋅

⋅ ×

=

= −

= −

10. A sports club decides to hold a field day to raise money for local

charities. One of the games involves spinning the wheel shown on

the right.


88

You win the amount the pointer lands on. It costs €5 to play the game.

(i) What is the expected value of the game? Expected value: Σx.p(x)

x Outcome x p(x) x.p(x) Green €3 1

4

34

Yellow €7 18

78

Blue €10 18

10 58 4=

White €5 14

54

Pink €2 14

2 14 2=

3 7 5 5 1( ).4 8 4 4 2378

4 625

Σ = + + + +

=

= ⋅

p x x

€4·63 is the expected value of the game.

(ii) How could they adjust the wheel to ensure they make more money for the charities? In order to make more money for the charities they could lower the amount that is won in each/some segment. or

Otherwise you could enlarge the €2 segment and make the €5 segment smaller

11. A child asks his parents for some money. The parents make the following offers.

Father’s offer: The child flips a coin. If the coin lands heads-up, the father will give the

child €20. If the coin lands tails-up, the father will give the child nothing.


89

Mother’s offer: The child rolls a 6-sided die. The mother will give the child €3 for each

dot on the up side of the die.

(i) Which offer has the greater expected value?

Expected Value : Σx.p(x)

Father’s offer

x p(x) x.p(x)

Heads €20 12

20 102=

Tails 0 12

0

) 1. €( 0Σ =x p x


90

Mother’s offer

x p(x) x.p(x)

1 = €3 16

1 3 136 6 2× = =

2 = €6 16

1 66 16 6× = =

3 = €9 16

1 9 396 6 2× = =

4 = €12 16

1 1212 26 6× = =

5 = €15 16

1 15 5156 6 2× = =

6 = €18 16

1 1818 36 6× = =

1 3 5. ( ) 1 2 32 2 2212

01

1 05

€0

5

x p xΣ = + + + + +

=

=⋅

⋅=

€10·50 > €10

Therefore, the mother’s offer is greater.

(ii) Which offer would you choose if you were the child? Justify your answer.

I would choose the mother’s offer as over many times the expected outcome

would be higher.


91

12. (i) What is the expected value for each of the following investment packages for a €1,000 investment?

Speculative investment Conservative investment

• Complete loss: 40% chance • Complete loss: 1% chance

• No gain or loss: 15% chance • No gain or loss: 35% chance

• 100% gain: 15% chance • 10% gain: 59% chance

• 400% gain: 15% chance • 20% gain: 5% chance

• 900% gain: 15% chance

Expected Value : Σx.p(x) €1,000 investment

Speculative:

X p(x) x.p(x)

–1,000 440%

10= –400

0 15% 0

1,000 15% –700

3,000 15% 450

8,000 15% 1,200

€

. ( ) 400 0 150 450 1,2001,400

Expected payout 1,400 on a €1,000 investmen =

Expe

t

€2,400cted value =

x p xΣ = − + + + +=


92

Conservative:

X p(x) x.p(x)

–1,000 1 1%

100= –10

0 35% 0

100 59% 59

200 5% 10

Σx.p(x) = –10 + 0 + 59 + 10

= 59

Expected payout = €59 on a €1,000 investment

Expected value = €1,059

(ii) Which would you choose? Why?

There are two possible answers here:

I would choose:

The speculative investment because the expected value is higher so it should earn

more.

Or

The conservative because the risk of losing everything is lower.


93

13. A biased die is used in a fairground game. The probabilities of getting the six different

numbers on the die are shown in the table below.

Number 1 2 3 4 5 6

Probability 0·12 0·26 0·1 0·23 0·18 0·11

(i) Find the expected value of the random variable x, where x is the number thrown.

Expected Value : Σx.p(x)

x p(x) x.p(x)

1 0·12 0·12

2 0·26 0·52

3 0·1 0·3

4 0·23 0·92

5 0·18 0·9

6 0·11 0·66

Σx.p(x) = 0·12 + 0·52 + 0·3 + 0·92 + 0·9 + 0·66

= 3·42

Expected value = €3·42

(ii) It costs €3·50 to play the game. The player rolls a die once and wins the number of

euro shown on the die.

By doing the calculations required, complete the following sentence:

‘If you play the game many times with a fair die, you will win an average of

______ per game, but if you play with the biased die you will lose an average

of ______ per game.’


94

Playing with a fair die

x p(x) x.p(x)

1 16

1 116 6

× =

2 16

1 226 6

× =

3 16

1 336 6

× =

4 16

1 446 6

× =

5 16

1 556 6

× =

6 16

1 666 6

× =

1 2 3 4 5 6. ( )6 6 6 6 6 6216

3 5

Σ = + + + + +

=

= ⋅

x p x

Expected value = €3·50

Fair die nett winnings

= expected value – cost

= 3·50 – 3·50

= 0

Baised die nett winnings

Expected value – cost

= 3·42 – 3·50

= –0·08

If you play the game many times with a fair die, you will win an average of €0·00 per game, but

if you play with the biased die you will lose an average of €0·08 per game.


95

Revision and Exam Style Questions – Section A

1. The probability that Ben’s soccer team wins this weekend is 35

. The probability that his

rugby league team wins this weekend is 23

.

P(Soccer wins) = 35

P(Rugby wins) = 23

(i) What is the probability that neither team wins this weekend?

P(Neither wins) = P(Soccer loses and Rugby loses)

= P(Soccer loses) × P(Rugby loses)

= 2 15 3×

= 2

15

(ii) What is the probability that only one of the teams wins?

P(One team wins)

= P(Soccer wins and Rugby loses) or P(Soccer loses and Rugby wins)

= (P(Soccer wins) × P(Rugby loses)) + (P(Soccer loses) × P(Rugby wins))

= 3 1 2 25 3 5 3

× + ×

= 3 4

15 15+

= 7

15


96

2. A packet of lollipops contains eight red lollipops and 14 blue

lollipops. Two lollipops are selected at random without

replacement.

8 Red 14 Blue Total : 22

(i) Draw a tree diagram to show the possible outcomes.


Tree diagram: No replacement

(ii) What is the probability that the two lollipops are of different colours?

P(Two different) = P(1st Red and 2nd Blue) or P(1st Blue and 2nd Red)

= (P(Red) × P(Blue)) + (P(Blue) × P(Red))

= 8 14 14 822 21 22 21

× + ×

= 8 833 33

+

= 1633


97

(iii) What is the probability that the two lollipops are the same colour?

P(Two same) = P(Red and Red) or P(Blue and Blue)

= (P(Red) × P(Red)) or (P(Blue) × P(Blue))

= 8 7 14 1322 21 22 21

× + ×

= 4 1333 33

+

= 1733

3. The probability of winning a game is 58

. If the game is played three times, what is the

probability of winning:

P(winning) = 58

P(losing) = 5 318 8

− =

(i) three consecutive games?

P(WWW) = 5 5 58 8 8× × (‘and rule’)

= 125512

(ii) at least two games?

P(At least 2) = P(2 or 3 wins)

P(WWW) or P(WLW) or P(LWW) or P(WWL)125 5 3 5 3 5 5 5 5 3= 512 8 8 8 8 8 8 8 8 8125 75 75 75512 512 512 512350512175256

=

+ × × + × × + × ×

= + + +

=

=


98

4. A bag contains five yellow marbles and seven green marbles. Two marbles are selected at

random without replacement.

5 yellow 7 green Total: 12 No replacement

(i) What is the probability that at least one of the marbles selected is yellow?

P(At least 1 yellow)

= P(YY) or P(YG) or P(GY)

5 4 5 7 7 512 11 12 11 12 11 = × + × + ×

5 35 3533 132 132

= + +

1522

=

Alternatively

P(At least 1 yellow) = 1 – P(both green)

7 6112 117122

1522

= − ×

= −

=

(ii) If a marble is selected at random and replaced 72 times, how many times would

you expect to get a yellow marble?

572 times

5P(Y)1

2

2

721

30

= ×

=

=


99

5. Two boxes each contain green balls and yellow balls.

Box A contains four green and three yellow balls.

Box B contains four green and five yellow balls.

Chris randomly chooses one ball from each box.

Box A: 4 green 3 yellow

Total : 7

Box B : 4 green 5 yellow

Total : 9

(i) What is the probability that both balls are green?

4 4P(GG)7 91663

= ×

= (‘and’ rule)

(ii) What is the probability that at least one of the balls is yellow?

P(At least 1 yellow) = P(YY) or P(YG) or P(GY)

3 5 3 4 4 57 9 7 9 7 9

4 5 2021 21 634763

= × + × + ×

= + +

=

Alternatively

P(At least 1 yellow) = 1 – P(both green)

4 417 9

16163

4763

= − ×

= −

=


100

(iii) What is the probability that both balls are the same colour?

P(YY) or P(GG) = 3 5 4 47 9 7 9

× + ×

15 1663 633163

= +

=

6. Jake has four blue t-shirts and three white t-shirts.

On each of the three days, Friday, Saturday and Sunday, he selects one

t-shirt at random to wear. Jake wears each t-shirt that he selects only once.

4 Blue 3 White Total: 7 3 Days

(i) What is the probability that Jake wears a blue t-shirt on Friday?

P(B) = 47

(ii) What is the probability that Jake wears a t-shirt of the same colour on all three

days?

Wears once = No replacement.

P(Same colour) = P(BBB) or P(WWW)

= [P(B) × P(B) × P(B)] + [P(W) × P(W) × P(W)]

4 3 2 3 2 17 6 5 7 6 54 1

35 3553517

= × × + × ×

= +

=

=


101

(iii) What is the probability that Jake does not wear a t-shirt of the same colour on

consecutive days?

P(No same colour on consecutive days)

= P(BWB) or P(WBW)

= [P(B) × P(W) × P(B)] + [P(W) × P(B) × P(W)]

4 3 3 3 4 27 6 5 7 6 5

6 435 35103527

= × × + × ×

= +

=

=

7. There are 12 chocolates in a box. Four of the chocolates have caramel centres, four have

orange centres and four have strawberry centres. Abdul randomly selects two chocolates

and eats them.

12 chocolates

Total

4 caramel 4 Orange 4 strawberry

Eats chocolate = No replacement

(i) What is the probability that the two chocolates have orange centres?

P(OO) = 4 3

12 11×

= 111


102

(ii) What is the probability that the two chocolates have the same centre?

P(2 same) = P(OO) or P(CC) or P(SS)

4 3 4 3 4 312 11 12 11 12 111 1 1

11 11 113

11

= × + × × ×

= + +

=

(iii) What is the probability that the two chocolates have different centres?

P(2 different centres)

= P(OC) or P(OS) or P(CO) or P(CS) or P(SO) or P(SC)

4 4 4 4 4 4 4 4 4 4 4 412 11 12 11 12 11 12 11 12 11 12 11 = × + × + × + × + × + ×

4 6338

11

= ×

=

Alternatively

P(2 different centres) = 1 – P(same centres)

3111

811

= −

=


103

8. Sarah often oversleeps and arrives late for school. She has estimated the probability that

she will oversleep on Monday as 25% and on Tuesday as 35%.

The tree diagram shows these probabilities.

(i) Copy the tree diagram and fill in the probabilities that Sarah does not oversleep.

Monday

1 – 0.25 = 0·75 1 – 0·35 = 0·65

(ii) Calculate the probability that Sarah oversleeps on both Mondays and Tuesdays.

P(Oversleeps both) = P(Oversleeps) × P(Oversleeps)

= 0·25 × 0·35

= 0·0875


104

(iii) Calculate the probability that Sarah oversleeps on exactly one of the two days.

P(Oversleeps on one day) = P(OD) or P(DO)

= (0·25 × 0·65) + (0·75 × 0·35)

= 0·1625 + 0·2625

= 0·425

9. A card is taken from a standard deck. This card is replaced and a second card is drawn.

What is the probability that the cards drawn are:

52 cards Replacement

(i) a seven and a queen?

P(7 and Queen) = 4

52 4

52× (‘and’ rule)

= 1

169

(ii) two red tens?

P(Red 10 and Red 10) = 2 2

52 52×

= 1

676

(iii) a spade or an ace?

P(A spade or on Ace) Not mutually exclusive

= P(spade) + P(Ace) – P(Ace + spade)

13 4 152 52 52

= + −

17 152 5216524

13

= −

=

=


105

(iv) not picture cards?

P(Not picture) 40 4052 52

= ×

100169

=

10. If you draw three cards from a deck without replacement, what is the probability that:

No replacement 52 cards

(i) all three cards are red?

P(RRR) = 26 25 2452 51 50

× ×

2

17=

(ii) you don’t draw any spades?

P(No spades) = 13 spades ⇒ 52 – 13 = 39 not spades

39 38 3752 51 50703

1,700

= × ×

=

(iii) you draw a club, a heart and a diamond (in that order)?

P(♣♥♦) 13 13 1352 51 50

= × ×

169

10,200=


106

(iv) you draw a club, a heart and a diamond in any order?

(CHD, CDH, DCH, DHC, HCD, HDC)

13 13 13 652 51 50

169 6102001691700

× × ×

= ×

=

11. A survey of 154 people was carried out.

Participants were asked whether they liked tea or coffee. Some of the results are

recorded in the Venn diagram below.

(i) Of those surveyed, 21 people did not like either tea or coffee. Complete the

diagram.

154 people 21 liked neither

154 – (52 + 36 + 21) = coffee only

= 45


107

(ii) A person is chosen at random from those surveyed. What is the probability that the

person liked both coffee and tea?

P(Both Tea and coffee) = 52

154

= 2677

(iii) What percentage of the people surveyed liked one beverage only? Give your answer correct to one decimal place.

P(One beverage only) = P(Tea or Coffee)

= P(Tea only) + P(Coffee only)

36 45154 15481

154

= +

=

0 5259 10052 5952 6% (to one decimal place)

= ⋅ ×= ⋅= ⋅

Revision and Exam Style Questions – Section B

More challenging problems

1. Jane plays a game which involves two coins being tossed. The amounts to be won for the

different possible outcomes are shown below:

Win €6 for two heads

Win €1 for one head and one tail

Win €2 for two tails

It costs €4 to play one game.

Expected value : Σx.p(x)


108

Outcome Pay Out x p(x) x.p(x)

HH €6 1 1 12 2 4× =

1 6 364 4 2

× = =

HT €1 1 1 12 2 4× =

1 114 4

× =

TH €1 1 1 12 2 4× =

1 114 4

× =

TT €2 1 1 12 2 4× =

1 2 124 4 2

× = =

3 1 1 1. ( )2 4 4

€2·

25

502

Σ = + + +

=

=

x p x

(i) Will Jane expect a gain or a loss, and how much will it be? Justify your answer

with suitable calculations.

Costs €4 to play

€4 – €2·50 = €1·50 loss Jane will expect a loss of €1·50. (ii) Is this a fair game? Justify your answer.

It is not a fair game as the player will expect a loss. It is unfair for the player.


109

2. Conor says that the probability of getting two tails in three throws is higher than the

probability of getting three heads in a row.

(i) Do you agree with Conor? Justify your answer by showing the calculations for

each situation.

P(Two Tails in three throws) = P(TTH) or P(THT) or P(HTT)

1 1 1 1 1 1 1 1 12 2 2 2 2 2 2 2 2

1 1 18 8 838

= × × + × × + × ×

= + +

=

P(HHH) = 1 1 12 2 2× ×

= 18

3 18 8>

Conor is right and it is more likely to get 2 Tails in 3 throws than 3 Heads.

Georgia says that if you get three heads in a row you are more likely to get a tail on the

next throw because a tail is due.

(ii) Do you agree with Georgia? Explain your answer.

No, I don’t agree with Georgia.

Flipping a coin is an independent event. The previous throw does not effect the

next. The probability of getting a Tail will always be 1 .2


110

3. The table shows the probability of choosing each of the different coloured sweets from

packets containing red, green, yellow, brown and blue sweets.

Colour of sweet Red Green Yellow Brown Blue

Probability 0·3 0·15 0·26 0·18

(i) What is the probability of choosing a yellow sweet?

Probabilities always add up to 1.

1 – (0·3 + 0·15 + 0·26 + 0·18)

= 1 – 0·89

= 0·11 = P(yellow)

(ii) What is the probability of not choosing a red sweet?

P(Not red) = 1 – P(red)

= 1 – 0·3

= 0·7

(iii) If three sweets are chosen at the same time, what is the probability of choosing a

green, a blue and a brown sweet?

You may have solved this question by multiplying the three probabilities, however

this is incorrect.

Once one sweet is removed, the total number of sweets remaining changes and so

the probability of selecting a certain colour also changes.

These events are dependent.

Therefore, without knowing the number of each coloured sweet present at the

start, we cannot calculate the probabilities after each sweet is selected.


111

4. Weather records for a town suggest the following:

If it rains (R) on a given day, the probability it will rain the next day is 16

.

If it is sunny (S) on a given day, the probability of the next day being sunny is12

.

In a particular week Tuesday is sunny. The tree diagram shows the possible outcomes for

the next three days.

(i) Show that the probability of Thursday being sunny is 23

.

P(Thursday sunny) = P(SS) or P(RS)

1 1 1 52 2 2 61 54 12

23

= × + × = +

=


112

(ii) What is the probability of rain on both Thursday and Friday?

P(RR) = cannot ignore Wednesday

∴ P(RRR) or P(SRR)

1 1 1 1 1 12 6 6 2 2 6

1 172 241

18

= × × + × ×

= +

=

(iii) What is the probability of at least one of Thursday and Friday being sunny?

P(At least one of Thurs/Fri Sunny)

= P(RRS) or P(RSR) or P(RSS) or P(SSS) or P(SRS) or P(SSR)

1 1 5 1 5 1 1 5 12 6 6 2 6 2 2 6 21 1 1 1 1 5 1 1 1 2 2 2 2 2 6 2 2 25 5 5 1 5 172 24 24 8 24 81718

= × × + × × + × × + × × + × × + × ×

= + + + + +

=


113

5. Judy is playing a game in which she rolls a fair dice three times and tries to get ‘6’ as many times as she can.

(i) Using the options ‘6’ and ‘not a 6’, draw a tree diagram to represent this game. Fill

in the probabilities on the branches of the tree diagram.

Tree Diagram:

(ii) Calculate the probability that Judy gets a ‘6’ on all three tries.

P(666) = 1 1 16 6 6

× ×

=1

216

(iii) Calculate the probability that Judy gets a ‘6’ on at least two of her three tries.

P(At least 2 6s) = P(6, 6, Not 6) or P(6, Not 6, 6) or P(666) or P(Not 6, 6, 6)

1 1 5 1 5 1 1 5 1 16 6 6 6 6 6 216 6 6 65 5 1 5

216 216 216 216162162

27

= × × + × × + + × ×

= + + +

=

=


114

6. A bag contains 10 counters with the letters of the word STATISTICS written on them. A counter is chosen at random and not replaced before choosing another one.

10 counters STATISTICS No replacement

(i) Draw a tree diagram to represent this situation. Tree Diagram S, T, C = consonants = 7 letters

A, I = Vowels = 3 letters

(ii) What is the probability of getting two consonants?

P(2 consonants) = P(SS) or P(ST) or P(SC) or P(TS) or P(TT) or P(TC) or

P(CC) or P(CT) or P(CS)

3 2 3 3 3 1 3 3 3 210 9 10 9 10 9 10 9 10 9

3 1 1 1 3 1 3 3 010 9 10 10 9 10 9 90

6 9 3 9 6 3 390 90 90 90 90 90 9042907

15

= × + × + × + × + × + × + × + × + × +

= + + + + + +

=

=

or

P(consonant, consonant) = 7 6

10 9×

42907

15

=

=


115

(iii) What is the probability of getting a vowel at least once?

P(1 vowel at least) = P(vowel, vowel) or P(vowel, consonant) or

P(consonant, vowel)

3 2 3 7 7 310 9 10 9 10 96 21 21

90 90 9048908

15

= × + × + ×

= + +

=

=

(iv) What is the probability of getting exactly one vowel?

P(1 vowel) = P(vowel, consonant) or P(consonant, vowel)

21 2190 9042907

15

= +

=

=

(v) What is the probability of not getting exactly one vowel?

P(Not 1 vowel) = 1 – P(1 vowel)

7115

815

= −

=


116

7. In a certain game, you pick a card from a standard deck of 52 cards. If the card is a heart,

you win.

If the card is not a heart, you replace the card to the deck, reshuffle, and draw again.

What is the probability that you will pick the first heart on the third draw?

P(success) = P(Heart)

= 14

P(Failure) = 1 – P(Success)

P(not heart) = 114

−

= 34

P(not heart and not heart and heart) = P(FFS)

P(FFS) = 3 3 14 4 4× ×

964

=

8. A biased die is used in a game. The probabilities for each outcome

are given below.

Outcome 1 2 3 4 5 6

Probability 0·22 0·12 0·25 0·14 0·18 0·09

(i) The die is thrown once. What is the probability of getting:

(a) an even number?

P(Even) = P(2) or P(4) or P(6)

= 0·12 + 0·14 + 0·09

= 0·35


117

(b) a prime number or an odd number?

P(Prime or Odd)

= P(prime) + P(odd) – P(both) [Not Mutually exclusive]

= P(2, 3, 5) + P(1, 3, 5) – P(3, 5)

= (0·12 + 0·25 + 0·18) + (0·22 + 0·25 + 0·18) – (0·25 + 0·18)

= 0·55 + 0·65 – 0·43

= 0·77

(c) a 4 or less?

P(A 4 or less) = P(1, 2, 3, or 4)

= 0·22 + 0·12 + 0·25 + 0·14

= 0·73

(d) a multiple of 3?

P(A multiple of 3) = P(3 or 6)

= 0·25 + 0·09

= 0·34

(b) (ii) Calculate the expected value of the biased die above.

Expected Value: Σx.p(x)

Outcome x p(x) x.p(x)

1 0·22 0·22

2 0·12 0·24

3 0·25 0·75

4 0·14 0·56

5 0·18 0·9

6 0·09 0·54

Σx.p(x) = 0·22 + 0·24 + 0·75 + 0·56 + 0·9 + 0·54

= 3·21


118

This die was used in a game where the player rolls the die once and wins back the

number of euro shown on the die. The game costs €3 to play.

(iii) By doing the necessary calculations, show the difference between using the biased die and a fair die to play the game.

Cost €3 Expected win €3·21

3·21 – 3 = 21 c

Fair die:

Outcome p(x) x.p(x)

1 16

16

2 16

2 16 3=

3 16

3 16 2=

4 16

4 26 3=

5 16

56

6 16

6 16=

1 1 1 2 5. 16 3 2 3 67

(

5

)

23

Σ = + + + + +

=

= ⋅

x p x

Cost €3 Expected win €3·50 €3·50 – €3 = 50 c Therefore, in a larger number of games the average win in higher with a fair die. 50 c > 21 c


119

9. While shuffling a deck of cards, two cards are accidently dropped. What is the probability that the cards are:

No replacement (i) a king and a queen? P(K and Q) = P(KQ) or P(QK)

4 4 4 452 51 52 514 4

663 6638

663

= × + ×

= +

=

(ii) both red fours?

2 1P Red 4s52 51

( ) = ×

22652

11326

=

=

(iii) both of the same suit? P(Some Suit) = P(♦♦) or P(♥♥) or P(♣♣) or P(♠♠)

13 12 452 51

156 426524

17

= × ×

= ×

=


120

(iv) both black picture cards?

6 5P Black Picture Cards52 5

)1

( = ×

3026525

442

=

=

(v) one is a spade and the other is not a spade?

P(Spade and not spade) = P(♠?) or P(?♠)

13 39 39 1352 51 52 51

13 1368 6826681334

= × + ×

= +

=

=

(vi) at least one is a diamond?

P(At least 1 ♦) = P(♦♦) or P(♦?) or P(?♦)

13 12 13 39 252 51 52 51

156 132652 341534

= × + × ×

= +

=


121

10. A team from school A and a team from school B are in the finals of a Gaelic football competition.

In this competition the final round consists of a best-of-

three game series. Each game continues until one team

has a winning score. The team that wins two games wins the competition. Both

teams are equally likely to win a game.

Team A: Team B

1SucceP ss)2

( = 1SucceP ss)2

( =

1FailuP re)2

( = 1FailuP re)2

( =

(i) Find the probability that team A will win the competition at the end of the first

two games.

1 1P SS ( )2 2

= ×

14

=

(ii) Find the probability that no winner is decided until after three games have been played.

No winner until 3 games

Team A: P(SF) or P(FS)

1 1 1 12 2 2 2

1 14 412

= × + ×

= +

=


122

From an analysis of previous games, it is found that the probability of winning at home

for both teams is58

The first game is played at school A, the second at school B and the third at

school A.

(iii) Draw a tree diagram to represent all possible outcomes for the three matches.

Include the probabilities on each branch.


123

(iv) Use your tree diagram to investigate a claim that team A is more likely to win

the competition. Clearly justify your answer with mathematical calculations.

Team A win: P(AA) or P(ABA) or P(BAA)

5 3 5 5 5 3 3 5Team A win 8 8 8 8 8 8 8 815 125 4564 512 512

290512145256

= × + × × + × × = + +

=

=

Team B win: P(BB) or P(BAB) or P(ABB)

3 5 3 3 3 5 5 3Team B win8 8 8 8 8 8 8 815 27 7564 512 512

222512111256

= × + × × + × × = + +

=

=

145 111256 256

>

Therefore, Team A is more likely to win the competition

probability ii practice questions 19

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