Transcript Probability (Tree Diagrams)

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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
3
10
7
10
Second Choice
red
blue
3
3
9
x

10 10 100
3
10
red P(red and red) =
7
10
blue P(red and blue) =
3
10
P(blue and red) =
Independent
7
10
red
3
7
21
x

10 10 100
7
3
21
x

10 10 100
7
7
49
P(blue
and
blue)
=
x

blue
10 10 100
Probability (Tree Diagrams)
Characteristics of a tree diagram
First Choice
3
10
7
10
The
probabilities
for each event
are shown along
the arm of
each branch
and they sum
to 1.
Second Choice
red
blue
3
3
9
x

10 10 100
3
10
red P(red and red) =
7
10
blue P(red and blue) =
3
10
red
P(blue and red) =
7
10
7
7
49
P(blue
and
blue)
=
x

blue
10 10 100
Ends of first and
second level
branches show
the different
outcomes.
3
7
21
x

10 10 100
7
3
21
x

10 10 100
Probabilities are
multiplied along
each arm.
Characteristics
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
4
9
5
9
Second Choice
black
green
4 4 16
P(black and black) = x 
9 9 81
4
9
black
5
9
green P(black and green) = x
4
9
black P(green and black) = x
5
9
green P(green and green) = x
Q1 beads
4
9
5 20

9 81
5
9
4 20

9 81
5
9
5 25

9 81
Probability (Tree Diagrams)
Q2 Coins
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.
Second Coin
First Coin
1
2
1
2
head
tail
P(2 heads) = ¼
1
2
head P(head and head) = 1 x 1  1
1
2
tail
1 1 1
P(head and tail) = x 
2 2 4
1
2
head
1 1 1
P(tail and head) = x 
2 2 4
1
2
tail
2
2
1 1 1
P(tail and tail) = x 
2 2 4
P(head and a tail or a tail and a head) = ½
4
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.
Tennis
0.3
Race
0.4
Peter
Win
0.7
0.3
0.6
Becky
Win
0.7
P(Win and Win) for Peter = 0.12
Peter
Win
0.4 x 0.3 = 0.12
Becky
Win
0.4 x 0.7 = 0.28
Peter
Win
0.6 x 0.3 = 0.18
Becky
Win
0.6 x 0.7 = 0.42
P(Lose and Win) for Becky = 0.28
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
3
10
7
10
Second Choice
red
blue
Dependent
3 2
6
x 
10 9 90
2
9
red
7
9
blue P(red and blue) =
3
9
red
P(blue and red) =
6
9
7 6 42
P(blue
and
blue)
=
x 
blue
10 9 90
P(red and red) =
3 7 21
x 
10 9 90
7 3 21
x 
10 9 90
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
4
9
5
9
Second Choice
black
green
Q4 beads
3
8
black P(black and black) = 4 x 3  12
5
8
green P(black and green) = 4 x 5  20
9
9
8
8
72
72
4
8
black P(green and black) = 5 x 4  20
4
8
green P(green and green) = x
9
8
72
5
9
4 20

8 72
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.
Second Pick
First Pick
18
30
12
30
Milk
Dark
18 17 306
x

30 29 870
17
29
Milk P(milk and milk) =
12
29
Dark P(milk and dark) = 18 x 12  216
30
18
29
Milk P(dark and milk) =
11
29
Dark P(dark and dark) =
Q5 Chocolates
29
870
12 18 216
x

30 29 870
12 11 132
x

30 29 870
Probability (Tree Diagrams)
First Choice
Second Choice
4
20
red
4
20
11
20
5
20
4
20
blue
11
20
4
20
yellow
5
20
red
blue
yellow
red
11
20
5
20
5
20
3 Independent
Events
blue
yellow
red
11
20
blue
yellow
3 Ind
Probability (Tree Diagrams)
First Choice
Second Choice
red
4
20
11
20
3 Independent
Events
red
blue
yellow
red
blue
blue
yellow
red
5
20
yellow
blue
3 Ind/Blank
yellow
Probability (Tree Diagrams)
First Choice
Second Choice
3 Independent
Events
3 Ind/Blank/2
Probability (Tree Diagrams)
First Choice
Second Choice
3
19
red
4
20
11
20
5
19
4
19
blue
red
11
19
4
19
yellow
4
19
blue
yellow
red
10
19
5
19
5
20
3 Dependent
Events
blue
yellow
red
11
19
blue
yellow
3 Dep
Probability (Tree Diagrams)
First Choice
Second Choice
red
4
20
11
20
3 Dependent
Events
red
blue
yellow
red
blue
blue
yellow
red
5
20
yellow
blue
3 Dep/Blank
yellow
Probability (Tree Diagrams)
First Choice
Second Choice
3 Dependent
Events
3 Dep/Blank/2
Dep/Blank
Probability (Tree Diagrams)
First Choice
2 Independent Events.
3 Selections
3
10
7
10
Third Choice
Second Choice
3
10
3
10
red
7
10
red
blue
7
10
3
10
7
10
3 Ind/3
Select
blue
red
blue
3
10
7
10
3
10
7
10
blue
red
3
10
7
10
red
blue
red
blue
red
blue
Probability (Tree Diagrams)
First Choice
2 Independent Events.
3 Selections
3
10
Third Choice
Second Choice
red
red
red
red
blue
7
10
blue
blue
red
blue
red
blue
blue
3 Ind/3 Select/Blank
red
blue
Probability (Tree Diagrams)
First Choice
Second Choice
2 Independent Events.
3 Selections
3 Ind/3 Select/Blank2
Third Choice
Probability (Tree Diagrams)
First Choice
2 Dependent Events.
3 Selections
3
10
7
10
Third Choice
Second Choice
2
9
1
8
red
red
blue
7
8
2
8
red
blue
7
9
blue
3
9
red
6
9
6
8
2
8
6
8
3
8
blue
3 Dep/3 Select
5
8
red
blue
red
blue
red
blue
Probability (Tree Diagrams)
First Choice
2 Dependent Events.
3 Selections
3
10
7
10
Third Choice
Second Choice
red
red
red
red
blue
blue
blue
blue
red
red
blue
blue
red
3 Dep/3
Select/Blank
3 Dep/3
Select
blue
Probability (Tree Diagrams)
First Choice
Second Choice
Third Choice
2 Dependent Events.
3 Selections
3 Dep/3
Select/Blank2
3 Dep/3
Select
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
Worksheet 2
Worksheet 3
Worksheet 4