Contributed by:

This ppt includes the following topics:-

Dependent

Independent Events

Characteristics of a tree diagram

Question and answers.

Dependent

Independent Events

Characteristics of a tree diagram

Question and answers.

1.
powerpointmaths.com 3:2

Quality resources for the mathematics classroom

Reduce your workload and cut down planning

Enjoy a new teaching experience

Watch your students interest and enjoyment grow

Key concepts focused on and driven home

Over 100 files available with many more to come

1000’s of slides with nice graphics and effects.

powerpointmaths.com Get ready to fly!

© Powerpointmaths.com All rights reserved.

Quality resources for the mathematics classroom

Reduce your workload and cut down planning

Enjoy a new teaching experience

Watch your students interest and enjoyment grow

Key concepts focused on and driven home

Over 100 files available with many more to come

1000’s of slides with nice graphics and effects.

powerpointmaths.com Get ready to fly!

© Powerpointmaths.com All rights reserved.

2.
Probability (Tree Diagrams)

Tree diagrams can be used to help solve problems involving both

dependent and independent events.

The following situation can be represented by a tree diagram.

Peter has ten coloured cubes in a bag. Three of the cubes are red and 7 are

blue. He removes a cube at random from the bag and notes the colour

before replacing it. He then chooses a second cube at random. Record the

information in a tree diagram.

First Choice Second Choice

3 3 3 9

10 red P(red and red) = x

10 10 100

3 red

10 7 3 7 21

10 blue P(red and blue) = x

10 10 100

3 7 3 21

7 10 red P(blue and red) = x

10 10 100

10 blue

Independent 7

10 blue P(blue and blue) =

7

x

7

10 10 100

49

Tree diagrams can be used to help solve problems involving both

dependent and independent events.

The following situation can be represented by a tree diagram.

Peter has ten coloured cubes in a bag. Three of the cubes are red and 7 are

blue. He removes a cube at random from the bag and notes the colour

before replacing it. He then chooses a second cube at random. Record the

information in a tree diagram.

First Choice Second Choice

3 3 3 9

10 red P(red and red) = x

10 10 100

3 red

10 7 3 7 21

10 blue P(red and blue) = x

10 10 100

3 7 3 21

7 10 red P(blue and red) = x

10 10 100

10 blue

Independent 7

10 blue P(blue and blue) =

7

x

7

10 10 100

49

3.
Probability (Tree Diagrams)

Characteristics of a tree diagram

First Choice Second Choice

3 3 3 9

10 red P(red and red) = x

10 10 100

3 red

10 7 3 7 21

10 blue P(red and blue) = x

10 10 100

3 7 3 21

7 10 red P(blue and red) = x

10 10 100

10 blue

7 7 7 49

The blue P(blue and blue) = x

10 10 10 100

probabilities

for each event

are shown along Ends of first and

the arm of Probabilities are

second level multiplied along

each branch branches show each arm.

and they sum

to 1.

Characteristics

the different

outcomes.

Characteristics of a tree diagram

First Choice Second Choice

3 3 3 9

10 red P(red and red) = x

10 10 100

3 red

10 7 3 7 21

10 blue P(red and blue) = x

10 10 100

3 7 3 21

7 10 red P(blue and red) = x

10 10 100

10 blue

7 7 7 49

The blue P(blue and blue) = x

10 10 10 100

probabilities

for each event

are shown along Ends of first and

the arm of Probabilities are

second level multiplied along

each branch branches show each arm.

and they sum

to 1.

Characteristics

the different

outcomes.

4.
Probability (Tree Diagrams)

Question 1 Rebecca has nine coloured beads in a bag. Four of the beads are

black and the rest are green. She removes a bead at random from the bag

and notes the colour before replacing it. She then chooses a second bead.

(a) Draw a tree diagram showing all possible outcomes. (b) Calculate the

probability that Rebecca chooses: (i) 2 green beads (ii) A black followed

by a green bead.

First Choice Second Choice

4 4 4 16

9 black P(black and black) = x

9 9 81

4 black

9 5 4 5 20

9 green P(black and green) = x

9 9 81

4

5 4 20

5 9 black P(green and black) = x

9 9 81

9 green

5 5 5 25

9 green P(green and green) = x

9 9 81

Q1 beads

Question 1 Rebecca has nine coloured beads in a bag. Four of the beads are

black and the rest are green. She removes a bead at random from the bag

and notes the colour before replacing it. She then chooses a second bead.

(a) Draw a tree diagram showing all possible outcomes. (b) Calculate the

probability that Rebecca chooses: (i) 2 green beads (ii) A black followed

by a green bead.

First Choice Second Choice

4 4 4 16

9 black P(black and black) = x

9 9 81

4 black

9 5 4 5 20

9 green P(black and green) = x

9 9 81

4

5 4 20

5 9 black P(green and black) = x

9 9 81

9 green

5 5 5 25

9 green P(green and green) = x

9 9 81

Q1 beads

5.
Q2 Coins

Probability (Tree Diagrams)

Question 2 Peter tosses two coins. (a) Draw a tree diagram to show all

possible outcomes. (b) Use your tree diagram to find the probability of

getting (i) 2 Heads (ii) A head or a tail in any order.

First Coin Second Coin

1

2 head P(head and head) = 1 x 1 1

2 2 4

1 head

2 1 1 1 1

2 tail P(head and tail) = x

2 2 4

1

1 1 1

1 2 head P(tail and head) = x

2 2 4

2 tail

1 1 1 1

2 tail P(tail and tail) = x

2 2 4

P(2 heads) = ¼ P(head and a tail or a tail and a head) = ½

Probability (Tree Diagrams)

Question 2 Peter tosses two coins. (a) Draw a tree diagram to show all

possible outcomes. (b) Use your tree diagram to find the probability of

getting (i) 2 Heads (ii) A head or a tail in any order.

First Coin Second Coin

1

2 head P(head and head) = 1 x 1 1

2 2 4

1 head

2 1 1 1 1

2 tail P(head and tail) = x

2 2 4

1

1 1 1

1 2 head P(tail and head) = x

2 2 4

2 tail

1 1 1 1

2 tail P(tail and tail) = x

2 2 4

P(2 heads) = ¼ P(head and a tail or a tail and a head) = ½

6.
Probability (Tree Diagrams)

Q3 Sports

Question 3 Peter and Becky run a race and play a tennis match. The

probability that Peter wins the race is 0.4. The probability that Becky wins

the tennis is 0.7. (a) Complete the tree diagram below. (b) Use your tree

diagram to calculate (i) the probability that Peter wins both events. (ii) The

probability that Becky loses the race but wins at tennis.

Race Tennis

0.3 Peter 0.4 x 0.3 = 0.12

Peter Win

0.4 Win Becky

0.7 0.4 x 0.7 = 0.28

Win

0.3 Peter 0.6 x 0.3 = 0.18

0.6 Win

Becky

Win 0.7 Becky 0.6 x 0.7 = 0.42

Win

P(Win and Win) for Peter = 0.12 P(Lose and Win) for Becky = 0.28

Q3 Sports

Question 3 Peter and Becky run a race and play a tennis match. The

probability that Peter wins the race is 0.4. The probability that Becky wins

the tennis is 0.7. (a) Complete the tree diagram below. (b) Use your tree

diagram to calculate (i) the probability that Peter wins both events. (ii) The

probability that Becky loses the race but wins at tennis.

Race Tennis

0.3 Peter 0.4 x 0.3 = 0.12

Peter Win

0.4 Win Becky

0.7 0.4 x 0.7 = 0.28

Win

0.3 Peter 0.6 x 0.3 = 0.18

0.6 Win

Becky

Win 0.7 Becky 0.6 x 0.7 = 0.42

Win

P(Win and Win) for Peter = 0.12 P(Lose and Win) for Becky = 0.28

7.
Probability (Tree Diagrams)

Dependent Events

The following situation can be represented by a tree diagram.

Peter has ten coloured cubes in a bag. Three of the cubes are red and seven

are blue. He removes a cube at random from the bag and notes the colour

but does not replace it. He then chooses a second cube at random. Record

the information in a tree diagram.

First Choice Second Choice

2 3 2 6

9 red P(red and red) = x

10 9 90

3 red

10 7 3 7 21

9 blue P(red and blue) = x

10 9 90

3 7 3 21

7 9 red P(blue and red) = x

10 9 90

10 blue

Dependent 6

9 blue P(blue and blue) =

7 6 42

x

10 9 90

Dependent Events

The following situation can be represented by a tree diagram.

Peter has ten coloured cubes in a bag. Three of the cubes are red and seven

are blue. He removes a cube at random from the bag and notes the colour

but does not replace it. He then chooses a second cube at random. Record

the information in a tree diagram.

First Choice Second Choice

2 3 2 6

9 red P(red and red) = x

10 9 90

3 red

10 7 3 7 21

9 blue P(red and blue) = x

10 9 90

3 7 3 21

7 9 red P(blue and red) = x

10 9 90

10 blue

Dependent 6

9 blue P(blue and blue) =

7 6 42

x

10 9 90

8.
Probability (Tree Diagrams)

Dependent Events

Question 4 Rebecca has nine coloured beads in a bag. Four of the beads are

black and the rest are green. She removes a bead at random from the bag

and does not replace it. She then chooses a second bead. (a) Draw a tree

diagram showing all possible outcome (b) Calculate the probability that

Rebecca chooses: (i) 2 green beads (ii) A black followed by a green bead.

First Choice Second Choice

3

8 black P(black and black) = 4 x 3 12

9 8 72

4 black

5

green P(black and green) = 4 x 5 20

9

8

9 8 72

4

5 4 20

5 8 black P(green and black) = x

9 8 72

9 green

4 5 4 20

Q4 beads 8 green P(green and green) = x

9

8 72

Dependent Events

Question 4 Rebecca has nine coloured beads in a bag. Four of the beads are

black and the rest are green. She removes a bead at random from the bag

and does not replace it. She then chooses a second bead. (a) Draw a tree

diagram showing all possible outcome (b) Calculate the probability that

Rebecca chooses: (i) 2 green beads (ii) A black followed by a green bead.

First Choice Second Choice

3

8 black P(black and black) = 4 x 3 12

9 8 72

4 black

5

green P(black and green) = 4 x 5 20

9

8

9 8 72

4

5 4 20

5 8 black P(green and black) = x

9 8 72

9 green

4 5 4 20

Q4 beads 8 green P(green and green) = x

9

8 72

9.
Probability (Tree Diagrams)

Dependent Events

Question 5 Lucy has a box of 30 chocolates. 18 are milk chocolate and the

rest are dark chocolate. She takes a chocolate at random from the box and

eats it. She then chooses a second. (a) Draw a tree diagram to show all the

possible outcomes. (b) Calculate the probability that Lucy chooses:

(i) 2 milk chocolates. (ii) A dark chocolate followed by a milk chocolate.

First Pick Second Pick

17 18 17 306

29 Milk P(milk and milk) = x

30 29 870

18 Milk

30 12 18 12 216

29 Dark P(milk and dark) = x

30 29 870

18

12 18 216

12 29 Milk P(dark and milk) = x

30 29 870

30 Dark

11 12 11 132

Q5 Chocolates 29 Dark P(dark and dark) = x

30 29 870

Dependent Events

Question 5 Lucy has a box of 30 chocolates. 18 are milk chocolate and the

rest are dark chocolate. She takes a chocolate at random from the box and

eats it. She then chooses a second. (a) Draw a tree diagram to show all the

possible outcomes. (b) Calculate the probability that Lucy chooses:

(i) 2 milk chocolates. (ii) A dark chocolate followed by a milk chocolate.

First Pick Second Pick

17 18 17 306

29 Milk P(milk and milk) = x

30 29 870

18 Milk

30 12 18 12 216

29 Dark P(milk and dark) = x

30 29 870

18

12 18 216

12 29 Milk P(dark and milk) = x

30 29 870

30 Dark

11 12 11 132

Q5 Chocolates 29 Dark P(dark and dark) = x

30 29 870

10.
Probability (Tree Diagrams) 3 Independent

Events

First Choice Second Choice

4 red

20

11

20 blue

red

4 5

20 20 yellow

4 red

11 20 11

20 20 blue

blue

5

20

yellow

5 4 red

20 20 11

yellow 20

blue

5

20 yellow 3 Ind

Events

First Choice Second Choice

4 red

20

11

20 blue

red

4 5

20 20 yellow

4 red

11 20 11

20 20 blue

blue

5

20

yellow

5 4 red

20 20 11

yellow 20

blue

5

20 yellow 3 Ind

11.
Probability (Tree Diagrams) 3 Independent

Events

First Choice Second Choice

red

red blue

4

20 yellow

red

11

20 blue

blue

yellow

5 red

20

yellow blue

3 Ind/Blank

yellow

Events

First Choice Second Choice

red

red blue

4

20 yellow

red

11

20 blue

blue

yellow

5 red

20

yellow blue

3 Ind/Blank

yellow

12.
Probability (Tree Diagrams) 3 Independent

Events

First Choice Second Choice

3 Ind/Blank/2

Events

First Choice Second Choice

3 Ind/Blank/2

13.
Probability (Tree Diagrams) 3 Dependent

Events

First Choice Second Choice

3 red

19

11

19 blue

red

4 5

20 19 yellow

4 red

11 19 10

20 19 blue

blue

5

19 yellow

5 4 red

20 19 11

yellow 19

blue

4

19 yellow 3 Dep

Events

First Choice Second Choice

3 red

19

11

19 blue

red

4 5

20 19 yellow

4 red

11 19 10

20 19 blue

blue

5

19 yellow

5 4 red

20 19 11

yellow 19

blue

4

19 yellow 3 Dep

14.
Probability (Tree Diagrams) 3 Dependent

Events

First Choice Second Choice

red

red blue

4

20 yellow

red

11

20 blue

blue

yellow

5 red

20

yellow blue

3 Dep/Blank

yellow

Events

First Choice Second Choice

red

red blue

4

20 yellow

red

11

20 blue

blue

yellow

5 red

20

yellow blue

3 Dep/Blank

yellow

15.
Probability (Tree Diagrams) 3 Dependent

Events

First Choice Second Choice

3 Dep/Blank/2

Dep/Blank

Events

First Choice Second Choice

3 Dep/Blank/2

Dep/Blank

16.
Probability (Tree Diagrams)

First Choice Second Choice Third Choice

3 red

10

2 Independent Events.

3 Selections 3 blue

red 7

10

10

3

3 red

red 10

10

7

blue blue

10 7

3

3 10

7 red 10

10 red

10 blue

7

10 3 blue

7

10

10 blue red

3 Ind/3

7

Select 10

blue

First Choice Second Choice Third Choice

3 red

10

2 Independent Events.

3 Selections 3 blue

red 7

10

10

3

3 red

red 10

10

7

blue blue

10 7

3

3 10

7 red 10

10 red

10 blue

7

10 3 blue

7

10

10 blue red

3 Ind/3

7

Select 10

blue

17.
Probability (Tree Diagrams)

First Choice Second Choice Third Choice

red

2 Independent Events.

3 Selections blue

red

3 red

10 red

blue blue

7 red

red

10 blue

blue

blue red

3 Ind/3 Select/Blank blue

First Choice Second Choice Third Choice

red

2 Independent Events.

3 Selections blue

red

3 red

10 red

blue blue

7 red

red

10 blue

blue

blue red

3 Ind/3 Select/Blank blue

18.
Probability (Tree Diagrams)

First Choice Second Choice Third Choice

2 Independent Events.

3 Selections

3 Ind/3 Select/Blank2

First Choice Second Choice Third Choice

2 Independent Events.

3 Selections

3 Ind/3 Select/Blank2

19.
Probability (Tree Diagrams)

First Choice Second Choice Third Choice

1 red

8

2 Dependent Events.

3 Selections 2 blue

red 7

9

8

2

3 red

red 8

10

7

blue blue

9 6

2

3 8

7 red 8

red

9

10 blue

6

8 3 blue

6 8

9 blue red

3 Dep/3 Select 5

8

blue

First Choice Second Choice Third Choice

1 red

8

2 Dependent Events.

3 Selections 2 blue

red 7

9

8

2

3 red

red 8

10

7

blue blue

9 6

2

3 8

7 red 8

red

9

10 blue

6

8 3 blue

6 8

9 blue red

3 Dep/3 Select 5

8

blue

20.
Probability (Tree Diagrams)

First Choice Second Choice Third Choice

red

2 Dependent Events.

3 Selections blue

red

3 red

10 red

blue blue

7 red red

10 blue

blue

blue red

3 Dep/3

3 Dep/3 Select/Blank

Select blue

First Choice Second Choice Third Choice

red

2 Dependent Events.

3 Selections blue

red

3 red

10 red

blue blue

7 red red

10 blue

blue

blue red

3 Dep/3

3 Dep/3 Select/Blank

Select blue

21.
Probability (Tree Diagrams)

First Choice Second Choice Third Choice

2 Dependent Events.

3 Selections

3 Dep/3 Select/Blank2

3 Dep/3 Select

First Choice Second Choice Third Choice

2 Dependent Events.

3 Selections

3 Dep/3 Select/Blank2

3 Dep/3 Select

22.
Probability (Tree Diagrams)

Tree diagrams can be used to help solve problems involving both

dependent and independent events.

The following situation can be represented by a tree diagram.

Peter has ten coloured cubes in a bag. Three of the cubes are red and 7 are

blue. He removes a cube at random from the bag and notes the colour

before replacing it. He then chooses a second cube at random. Record the

information in a tree diagram.

Worksheet

1

Tree diagrams can be used to help solve problems involving both

dependent and independent events.

The following situation can be represented by a tree diagram.

Peter has ten coloured cubes in a bag. Three of the cubes are red and 7 are

blue. He removes a cube at random from the bag and notes the colour

before replacing it. He then chooses a second cube at random. Record the

information in a tree diagram.

Worksheet

1

23.
Worksheet 2

24.
Worksheet 3

25.
Worksheet 4