“Teach A Level Maths”
Statistics 1
Tree Diagrams
© Christine Crisp
Tree Diagrams
Statistics 1
AQA
EDEXCEL
MEI/OCR
OCR
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Tree Diagrams
This presentation reminds you about the tree
diagrams you used in GCSE.
We will use tree diagrams with some extra
notation in the next presentation to solve some
conditional probability problems.
Tree Diagrams
We use the word event in statistics to refer to a possible
result from a trial or experiment.
e.gs. A seed growing into a red flower.
A six showing when we throw a die.
A tree diagram shows the probabilities of 2 or more events.
e.g. There are 5 chocolates left in a box all looking the
same. 3 are raspberry creams (R), 2 are nougats (N)
(a) Draw a tree diagram to show the probabilities if
2 chocolates are taken at random.
(b) Find the probabilities of
(i) both being nougats
(ii) the 1st being a raspberry cream and the
2nd being a nougat
(iii) one of each type
Tree Diagrams
5 chocolates: 3Rs, 2Ns
Solution:
1st chocolate
2nd chocolate
R
3
5
2
5
R
Tip: To get to this
branch,
we’ve are
come up
3
of1st
the
5 chocolates
The
chocolate
can be R Rs
or
here . . .
N so we need 2 branches
N
2 of the 5 chocolates are Ns
Tree Diagrams
5 chocolates: 3Rs, 2Ns
Solution:
1st chocolate
2nd chocolate
R
3
5
2
5
R
N
Tip: To get to this
branch, we’ve come up
here . . .
So an R has gone.
Tree Diagrams
5 chocolates: 3Rs, 2Ns
Solution:
1st chocolate
2nd chocolate
2
4
3
5
R
R
N
2
5
N
Tree Diagrams
5 chocolates: 3Rs, 2Ns
Solution:
1st chocolate
3
5
2nd chocolate
2
4
R
2
4
N
R
Both the Ns are left
2
5
N
Tree Diagrams
5 chocolates: 3Rs, 2Ns
Solution:
1st chocolate
3
5
2
5
2nd chocolate
2
4
R
2
4
N
3
4
R
R
N
To reach here, an N
has gone so all 3 Rs are
left.
Tree Diagrams
5 chocolates: 3Rs, 2Ns
Solution:
1st chocolate
3
5
2
5
2nd chocolate
2
4
R
2
4
N
3
4
R
R
N
1
4
N
and only 1 N is left
Tree Diagrams
5 chocolates: 3Rs, 2Ns
Solution:
1st chocolate
2nd chocolate
2
4
3
5
R
2
4
2
5
N
3
4
1
4
R
We need to multiply to
find the probability of
reaching
the end of
N
each branch.
R
N
Tree Diagrams
5 chocolates: 3Rs, 2Ns
Solution:
1st chocolate
3
5
2
5
2nd chocolate
2
4
R
2
4
N
3
4
R
R
N
1
4
N
3 2
6
 
5 4 20
Tree Diagrams
5 chocolates: 3Rs, 2Ns
Solution:
1st chocolate
2nd chocolate
R
3 2
6
 
5 4 20
2
4
N
3 2
6
 
5 4 20
3
4
R
2
4
3
5
2
5
R
N
1
4
N
Tree Diagrams
5 chocolates: 3Rs, 2Ns
Solution:
1st chocolate
2nd chocolate
R
3 2
6
 
5 4 20
2
4
N
3 2
6
 
5 4 20
3
4
R
2 3
6
 
5 4 20
2
4
3
5
2
5
R
N
1
4
N
Tree Diagrams
5 chocolates: 3Rs, 2Ns
Solution:
1st chocolate
2nd chocolate
R
3 2
6
 
5 4 20
2
4
N
3 2
6
 
5 4 20
3
4
R
2 3
6
 
5 4 20
N
2
2 1
 
5 4 20
2
4
3
5
2
5
R
N
1
4
Tree Diagrams
5 chocolates: 3Rs, 2Ns
Solution:
1st chocolate
2nd chocolate
R
3 2
6
 
5 4 20
2
4
N
3 2
6
 
5 4 20
3
4
R
2 3
6
 
5 4 20
2
4
3
5
2
5
R
N
1
4
N
Tip: The four probabilities added
together equal 1
2
2 1
 
5 4 20
20
20
1
Tree Diagrams
5 chocolates: 3Rs, 2Ns
Solution:
1st chocolate
2nd chocolate
R
3 2
6
 
5 4 20
2
4
N
3 2
6
 
5 4 20
3
4
R
2 3
6
 
5 4 20
N
2
2 1
 
5 4 20
2
4
3
5
2
5
R
N
1
4
Tree Diagrams
(b) Find the probabilities of (i) both being Ns (ii) the 1st
being R and the 2nd N:
1st chocolate
2nd chocolate
2
4
3
5
N
3
4
1
4
Solution:
N
6
3 2
 
20
5 4
6
2 3
 
20
5 4
R
2
4
2
5
R
3 2
6
 
20
5 4
(i) P ( N and N ) 
R
N
2
2 1
 
5 4 20
Tree Diagrams
(b) Find the probabilities of (i) both being Ns (ii) the 1st
being R and the 2nd N:
1st chocolate
2nd chocolate
2
4
3
5
N
3
4
1
4
Solution:
N
6
3 2
 
20
5 4
6
2 3
 
20
5 4
R
2
4
2
5
R
3 2
6
 
20
5 4
21 1
(i) P ( N and N ) 

2010 10
R
N
2
2 1
 
5 4 20
(ii) P ( R then N ) 
Tree Diagrams
(b) Find the probabilities of (i) both being Ns (ii) the 1st
being R and the 2nd N (iii) one of each type
1st chocolate
2nd chocolate
2
4
3
5
N
3
4
1
4
Solution:
N
6
3 2
 
20
5 4
6
2 3
 
20
5 4
R
2
4
2
5
R
3 2
6
 
20
5 4
21 1
(i) P ( N and N ) 

2010 10
R
N
2
2 1
 
5 4 20
6 3 3
(ii) P ( R then N ) 

20 10 10
Tree Diagrams
(b) (iii) probability of one of each type
Solution: There are 2 ways to get one of each type:
1st chocolate
2nd chocolate
2
4
3
5
N
3 2
 
5 4
2 3
 
5 4
R
2
4
2
5
R
3 2
6
 
20
5 4
N
3
4
1
4
R
N
6
20
6
20
2 1
2
 
5 4 20
Tree Diagrams
(b) (iii) probability of one of each type
Solution: There are 2 ways to get one of each type:
1st chocolate
2nd chocolate
2
4
3
5
N
3 2
 
5 4
2 3
 
5 4
R
2
4
2
5
R
3 2
6
 
20
5 4
N
3
4
1
4
R
N
6
20
6
20
2 1
2
 
5 4 20
Tree Diagrams
(b) (iii) probability of one of each type
Solution: There are 2 ways to get one of each type:
1st chocolate
2nd chocolate
2
4
3
5
N
3 2
 
5 4
2 3
 
5 4
R
2
4
2
5
R
3 2
6
 
20
5 4
N
3
4
R
6
20
6
20
1
4
2 1 the
2
We need to add
N to find
 
20 one
5 4
probability of moving
along
branch or another.
6 6 12 3 3
P ( R then N or N then R) 
 

20 20 20 5 5
+
Tree Diagrams
SUMMARY
•
A tree diagram shows the probabilities of more than
one event.
•
To find probabilities for a branch always assume the
event on the branch leading to it has occurred.
e.g.
5
9
•
A
4
8
B
Assumes A has occurred
To find the probability of 2 events, multiply the
individual probabilities.
4
e.g.
8
5
A
B
9
5 4 5
P( A then B)   
9 8 18
continued
Tree Diagrams
SUMMARY
•
To find the probability of 1 event or another, add
the individual probabilities.
2
e.g.
3 2
6
4


C
3
A
5
5
2
5
B
1
4
P(A and C or B and D) 
D
4
20
2 1
2
 
5 4 20
6
2
2


20 20 5
Exercise
Tree Diagrams
1. There are 4 red and 6 blue biros in a drawer. Two
are taken out at random. Draw a tree diagram
showing the probabilities of the different events if
(a) The 1st is replaced before the 2nd is taken, and
(b) The 1st is not replaced before the 2nd is taken.
In each case find the probability of one red and one
blue.
2. Components for a machine are supplied by 3
factories A, B and C. 45% come from A, 25% from
B and the rest from C. Data shows that 4% of
those from A, 5% of those from B and 2% of those
from C are faulty.
Draw a tree diagram showing all the probabilities
and use it to find the probability that a randomly
selected component is faulty.
Tree Diagrams
1. There are 4 red and 6 blue pens in a drawer. Two
are taken out at random. Draw a tree diagram
showing the probabilities of the different events if
(a) The 1st is replaced before the 2nd is taken.
1st
4
10
6
10
2nd
4
10
R
B
6
4 10
10
6
10
R
4
4
16


10 10 100
B
4
6
24


10 10 100
R
6
4
24


10 10 100
B
6
6
36


10 10 100
P ( one red and one blue ) 
24
24
48 12 12

P (R, B )  P (B, R) 


100 100 10025 25
Tree Diagrams
1. There are 4 red and 6 blue pens in a drawer. Two
are taken out at random. Draw a tree diagram
showing the probabilities of the different events if
(b) The 1st is not replaced before the 2nd is taken.
2nd
1st
4
10
6
10
3
9
R
B
4
9
5
9
6
9
R
4
3 12
 
10 9 90
B
4 6 24
 
10 9 90
R
6 4 24
 
10 9 90
B
6 5 30
 
10 9 90
P ( one red and one blue ) 
24 24 488 8
P (R, B )  P (B, R) 



90 90 9015 15
Tree Diagrams
2. Components for a machine are supplied by 3
factories A, B and C. 45% come from A, 25% from
B and the rest from C. Data shows that 4% of
those from A, 5% of those from B and 2% of those
from C are faulty.
0  45
0  25
0 3
A
B
C
0  04
F
0  45  0  04  0  018
0  96
0  05
N
0  45  0  96  0  432
F
0  25  0  05  0  0125
0  95
N
0  25  0  95  0  2375
0  02
F
0  3  0  02  0  006
0  98
N
0  3  0  98  0  294
Tree Diagrams
A
0  45
0  25
0 3
B
C
0  04
F
0  45  0  04  0  018
0  96
0  05
N
0  45  0  96  0  432
F
0  25  0  05  0  0125
0  95
N
0  25  0  95  0  2375
0  02
F
0  3  0  02  0  006
0  98
N
0  3  0  98  0  294
P (F)  0  018  0  0125  0  006
 0  0365
The following slides contains the summary,
shown without colour, so that they can be
printed and photocopied.
Tree Diagrams
SUMMARY
•
A tree diagram shows the probabilities of more than
one event.
•
To find probabilities for a branch always assume the
event on the branch leading to it has occurred.
e.g.
5
9
•
A
4
8
B
Assumes A has occurred
To find the probability of 2 events, multiply the
individual probabilities.
4
e.g.
8
5
A
B
9
5 4 5
P( A then B)   
9 8 18
continued
Tree Diagrams
SUMMARY
•
To find the probability of 1 event or another, add
the individual probabilities.
2
e.g.
3 2
6
4


C
3
A
5
5
2
5
B
1
4
P( A and C or B and D) 
D
4
20
2 1
2
 
5 4 20
6
2 2


20 20 5