# Probability: Tree Diagrams

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This ppt includes the following topics:-
Dependent
Independent Events
Characteristics of a tree diagram
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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
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.
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
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
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
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
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
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
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
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
12. Probability (Tree Diagrams) 3 Independent
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
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
15. Probability (Tree Diagrams) 3 Dependent
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
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
18. Probability (Tree Diagrams)
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
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
21. Probability (Tree Diagrams)
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
23. Worksheet 2
24. Worksheet 3
25. Worksheet 4