Independent

red

blue

First Choice Second Choice3 3

P(red and red) =10 10

9100

x

710

red

blue

red

blue

310 3 7

P(red and blue) =10 10

21100

x

7 3P(blue and red) =

10 1021

100x

7 7P(blue and blue) =

10 1049100

x

310

710

310

710

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.

Probability (Tree Diagrams)

Characteristics

Probability (Tree Diagrams)

The probabilities

for each event are shown

along the arm of each

branch and they sum to 1.

red

blue

First Choice Second Choice3 3

P(red and red) =10 10

9100

x

710

red

blue

red

blue

310 3 7

P(red and blue) =10 10

21100

x

7 3P(blue and red) =

10 1021

100x

7 7P(blue and blue) =

10 1049100

x

310

710

310

710

Ends of first and second level

branches show the different outcomes.

Probabilities are multiplied along

each arm.

Characteristics of a tree diagram

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.

Probability (Tree Diagrams)

black

green

First Choice

59

49

black

green

First Choice Second Choice

59

black

green

black

green

49

49

59

49

59

black

green

First Choice Second Choice

59

black

green

black

green

49 4 5

P(black and green) =9 9

2081

x

5 5P(green and green) =

9 92581

x

49

59

49

59

4 4P(black and black) =

9 91681

x

5 4P(green and black) =

9 92081

x

Q2 CoinsQuestion 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.

Probability (Tree Diagrams)

head

tail

First Coin

12

12

Second Coin

head

tail

head

tail

12

12

12

12

1 1P(head and tail)

21

=2 4

x

1 1P(tail and tail)

21

=2 4

x

1 1P(head and head)

21

=2 4

x

1 1P(tail and head)

21

=2 4

x

P(2 heads) = ¼

P(head and a tail or a tail and a head) = ½

Q3 SportsQuestion 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.

Probability (Tree Diagrams)

Race Tennis

0.4 x 0.3 = 0.12

0.4 x 0.7 = 0.28

0.6 x 0.3 = 0.18

0.6 x 0.7 = 0.42

P(Win and Win) for Peter = 0.12

P(Lose and Win) for Becky = 0.28

Peter Win0.4

Becky Win

0.7

Becky Win

0.6

Becky Win

Peter Win

Peter Win

0.3

0.3

0.7

Dependent

red

blue

First Choice Second Choice3 2

P(red and red) =10 9

690

x

710

red

blue

red

blue

310 3 7

P(red and blue) =10 9

2190

x

7 3P(blue and red) =

10 92190

x

7 6P(blue and blue) =

10 94290

x

29

79

39

69

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.

Probability (Tree Diagrams)

Dependent Events

Q4 beads

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.

Probability (Tree Diagrams)

Dependent Events

black

green

First Choice Second Choice

59

black

green

black

green

49 4 5

P(black and green) =9 8

2072

x

5 4P(green and green) =

9 82072

x

38

58

48

48

4 3P(black and black) =

9 81272

x

5 4P(green and black) =

9 82072

x

Q5 Chocolates

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.

Probability (Tree Diagrams)

Dependent Events

Milk

Dark

First Pick Second Pick

1230

Milk

Dark

Milk

Dark

1830

18 17P(milk and milk) =

30687030 29

x 1729

1229

1829

1129

18 12P(milk and dark) =

21687030 29

x

12 18P(dark and milk) =

21687030 29

x

12 11P(dark and dark) =

13287030 29

x

3 Ind/Blank

Probability (Tree Diagrams)

red

yellow

First Choice Second Choicered

blue

blue

yellowred

blue

yellowred

blue

yellow

3 Independent Events

520

420

1120

3 Ind

Probability (Tree Diagrams)

red

yellow

First Choice Second Choice

520

red

420

420

520

blue

1120

blue

yellowred

blue

yellowred

blue

yellow

1120

420

520

1120

420

520

1120

3 Independent Events

3 Ind/Blank/2

Probability (Tree Diagrams)

First Choice Second Choice

3 Independent Events

3 Dep/Blank/23 Dep/Blank

Probability (Tree Diagrams)

First Choice Second Choice

3 Dependent Events

3 Dep/Blank

Probability (Tree Diagrams)

red

yellow

First Choice Second Choicered

blue

blue

yellowred

blue

yellowred

blue

yellow

3 Dependent Events

520

420

1120

3 Dep

Probability (Tree Diagrams)

red

yellow

First Choice Second Choice

520

red

420

319

519

blue

1120

blue

yellowred

blue

yellowred

blue

yellow

1119

419

519

1019

419

419

1119

3 Dependent Events

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.

Probability (Tree Diagrams)

Worksheet 1

3 Ind/3 Select/Blank2

Probability (Tree Diagrams)

First Choice Second Choice

2 Independent Events. 3 Selections

Third Choice

3 Ind/3 Select/Blank

Probability (Tree Diagrams)

red

First Choice Second Choice

710

red310

blue

2 Independent Events. 3 Selections

red

red

red

red

blue

blue

blue

blue

red

blue

blue

Third Choice

3 Ind/3 Select

Probability (Tree Diagrams)

red

First Choice Second Choice

710

red310

blue

2 Independent Events. 3 Selections

red

red

red

red

blue

blue

blue

blue

310

310

710310

710

710

310

710

310

710 3

10

710

red

blue

blue

Third Choice

3 Ind/3 Select/Blank2

Probability (Tree Diagrams)

First Choice Second Choice Third Choice

2 Dependent Events. 3 Selections

3 Dep/3 Select/Blank3 Dep/3 Select

Probability (Tree Diagrams)

red

First Choice Second Choice

710

red310

blue

2 Dependent Events. 3 Selections

red

red

red

red

blue

blue

blue

blue

red

blue

blue

Third Choice

3 Dep/3 Select

Probability (Tree Diagrams)

red

First Choice Second Choice

710

red310

blue

2 Dependent Events. 3 Selections

red

red

red

red

blue

blue

blue

blue

29

18

79

39

69

78

28

68

28

68 3

8

58

red

blue

blue

Third Choice