“A Level Maths”
Statistics 1
Venn Diagrams
Venn Diagrams
Venn
diagrams
Venn
diagrams can be used to represent probabilities.
The outcomes that satisfy event A
(card is a club) can be represented
by a circle.
The outcomes that satisfy event B
(card is a Queen) can be represented
by another circle.
A
B
The circles can be overlapped to represent outcomes
that satisfy both events. (Card is Queen of Clubs)
3 of 32
© Boardworks Ltd 2005
Addition properties
Two events A and B are called mutually exclusive if they
cannot occur at the same time.
For example, if a card is picked at random from a
standard pack of 52 cards, the events “the card is a club”
and “the card is a diamond” are mutually exclusive.
If A and B are mutually exclusive,
then:
P(A B ) = P(A) + P(B )
A
B
In Venn diagrams
representing mutually
exclusive events, the circles
do not overlap.
This symbol means
‘union’ or ‘OR’
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© Boardworks Ltd 2005
The probability of A or B
𝑃 𝐴 ∪ 𝐵 not mutually exclusive.
𝑷 𝑪∪𝑸 =
𝟏𝟑 𝟒
𝟏
𝟏𝟔
+
−
=
𝟓𝟐 𝟓𝟐 𝟓𝟐 𝟓𝟐
P(C  Q) = P(C) + P(Q) – P(C  Q)
This symbol means “and”
Addition properties
Example: A card is picked at random from a pack of cards.
Find the probability that it is either a club or a queen or both.
Card is a club = event C
P(A B) = P(A) + P(B)
Card is a queen = event Q
1
P(C ) =
4
Note: A pack
has 52 cards.
P(Q ) =
4
1
=
52 13
1
P(C Q ) =
52
1
1
1
4
+
–
=
So, P(C Q ) =
4
13
52 13
6 of 32
© Boardworks Ltd 2005
This addition rule for finding P(A  B) is not true when A
and B are not mutually exclusive.
This represents the
other 3 queens.
Card is a club = event C
Card is a queen = event Q
This area
represents the
12 clubs that
are not queens.
This represents the
queen of clubs.
The more general rule for finding P(A  B) is:
P(A  B) = P(A) + P(B) – P(A  B)
This symbol means “and”
7 of 32
© Boardworks Ltd 2005
Addition properties
Example: A card is picked at random from a pack of cards.
Find the probability that it is either a club or a queen or both.
Note: A pack
has 52 cards.
Card is a club = event C
Card is a queen = event Q
1
P(C ) =
4
4
1
P(Q ) =
=
52 13
1
P(C Q ) =
52
This represents the
other 3 queens.
This area
represents the
12 clubs that
are not queens.
1
1
1
4
+
–
=
So, P(C Q ) =
4
13
52 13
8 of 32
This represents the
queen of clubs.
© Boardworks Ltd 2005
Venn Diagrams
Venn diagrams show the probabilities of more than one
event and can be used instead of tree diagrams. They
are quick and easy to use.
Venn Diagrams
Example 1: Among a group of 20 students, 7 are taking
Maths and of these 3 are also taking Biology. 5 are
taking neither. What is the probability that a student
chosen at random is taking Biology?
Solution: The diagram shows the 20 students.
20
The “eggs” show Maths
and Biology
3 do both
5 do neither
B
M
4
3
5
7 do Maths ( but we have 3 already )
Venn Diagrams
e.g.1 Amongst a group of 20 students, 7 are taking
Maths and of these 3 are also taking Biology. 5 are
taking neither. What is the probability that a student
chosen at random is taking Biology?
Solution: The diagram shows the 20 students.
20
The final number ( doing
Biology but not Maths )
is given by
20  4  3  5  8
B
M
4
3
3  8 11

So, P( student takes Biology ) =
20
20
8
5
Venn Diagrams
e.g.2 In a class of 30 students, 3 out of the 16 girls
and 6 out of the 14 boys, are left-handed. Draw a
Venn diagram and find the probability that a student
chosen at random is a boy or left-handed.
Solution: The diagram needs to show the numbers for
Left-handedness and Boys.
30
B
L
There are 6 Lefthanded Boys . . .
so there are 8 boys
who are not.
There are 3 lefthanded girls . . .
and 13 who are right-handed.
3
6
8
13
Venn Diagrams
e.g.2 In a class of 30 students, 3 out of the 16 girls
and 6 out of the 14 boys, are left-handed. Draw a
Venn diagram and find the probability that a student
chosen at random is a boy or left-handed.
Solution: The diagram needs to show the numbers for
Left-handedness and Boys.
30
B
L
3
6
8
13
368
17

P( boy or a left-hander ) =
30
30
Venn Diagrams
Exercise
1. Customers at a restaurant are offered a choice of chips
or jacket potato to go with either lasagna or pizza.
Out of a group of 16, 11 have the lasagna. 7 choose
chips to accompany their meal. 5 of those who choose
chips have the lasagna. What is the probability that
one chosen at random has neither chips nor lasagna?
Solution:
L: lasagna
C: chips
L
C
Venn Diagrams
Exercise
1. Customers at a restaurant are offered a choice of chips
or jacket potato to go with either lasagna or pizza.
Out of a group of 16, 11 have the lasagna. 7 choose
chips to accompany their meal. 5 of those who choose
chips have the lasagna. What is the probability that
one chosen at random has neither chips nor lasagna?
Solution:
L: lasagna
C: chips
16
L
C
Venn Diagrams
Exercise
1. Customers at a restaurant are offered a choice of chips
or jacket potato to go with either lasagna or pizza.
Out of a group of 16, 11 have the lasagna. 7 choose
chips to accompany their meal. 5 of those who choose
chips have the lasagna. What is the probability that
one chosen at random has neither chips nor lasagna?
Solution:
L: lasagna
C: chips
16
C
L
5
Venn Diagrams
Exercise
1. Customers at a restaurant are offered a choice of chips
or jacket potato to go with either lasagna or pizza.
Out of a group of 16, 11 have the lasagna. 7 choose
chips to accompany their meal. 5 of those who choose
chips have the lasagna. What is the probability that
one chosen at random has neither chips nor lasagna?
Solution:
L: lasagna
C: chips
16
C
L
6
5
Venn Diagrams
Exercise
1. Customers at a restaurant are offered a choice of chips
or jacket potato to go with either lasagna or pizza.
Out of a group of 16, 11 have the lasagna. 7 choose
chips to accompany their meal. 5 of those who choose
chips have the lasagna. What is the probability that
one chosen at random has neither chips nor lasagna?
Solution:
L: lasagna
C: chips
16
C
L
6
5
2
Venn Diagrams
Exercise
1. Customers at a restaurant are offered a choice of chips
or jacket potato to go with either lasagna or pizza.
Out of a group of 16, 11 have the lasagna. 7 choose
chips to accompany their meal. 5 of those who choose
chips have the lasagna. What is the probability that
one chosen at random has neither chips nor lasagna?
Solution:
L: lasagna
C: chips
16
C
L
6
5
2
3
Venn Diagrams
Exercise
1. Customers at a restaurant are offered a choice of chips
or jacket potato to go with either lasagna or pizza.
Out of a group of 16, 11 have the lasagna. 7 choose
chips to accompany their meal. 5 of those who choose
chips have the lasagna. What is the probability that
one chosen at random has neither chips nor lasagna?
Solution:
L: lasagna
16
C: chips
P(no chips, no lasagna)
3

16
C
L
6
5
2
3
Venn Diagrams
We’ll use the example about left-handed students to
illustrate a law of probability.
30
B
L
3
6
8
13
I’ll change this to A as we usually use A when we state
the law.
Venn Diagrams
We’ll use the example about left-handed students to
illustrate a law of probability.
30
B
A
3
6
8
13
Venn Diagrams
We’ll use the example about left-handed students to
illustrate a law of probability.
30
B
A
3
6
8
13
The number in A or B is anything shaded.
N.B. A or B in probability includes both.
Venn Diagrams
We’ll use the example about left-handed students to
illustrate a law of probability.
30
B
A
n( A or B ) = 17
3
6
8
13
We’ll use n for “ the number in . . .”, so
n( A or B ) = 3 + 6 + 8 = 17
Venn Diagrams
We’ll use the example about left-handed students to
illustrate a law of probability.
30
B
A
3
6
n( A or B ) = 17
n(A) = 9
8
13
n(A) = 9
n(B) =
Venn Diagrams
We’ll use the example about left-handed students to
illustrate a law of probability.
30
B
A
3
6
n( A or B ) = 17
n(A) = 9
8
n(B) = 14
13
n(A) = 9
n(B) = 14
Venn Diagrams
We’ll use the example about left-handed students to
illustrate a law of probability.
30
B
A
3
6
n( A or B ) = 17
n(A) = 9
8
n(B) = 14
13
n(A and B) = 6
The part with both types of shading gives the number
in A and B.
So, n( A and B ) = 6
Venn Diagrams
We’ll use the example about left-handed students to
illustrate a law of probability.
30
B
A
6
3
n( A or B ) = 17
n(A) = 9
8
n(B) = 14
13
We now get
n(A and B) = 6
n( A or B ) = n(A) + n(B) - n( A and B )
17
=
9
+ 14 -
6
N.B. The part with both types of shading is in A and in
B so it has been counted twice. We subtract one lot.
Venn Diagrams
We’ll use the example about left-handed students to
illustrate a law of probability.
30
B
A
3
6
n( A or B ) = 17
n(A) = 9
8
n(B) = 14
13
We now get
n(A and B) = 6
n( A or B ) = n(A) + n(B) - n( A and B )
Dividing by 30 gives probabilities, so
P( A or B ) = P(A) + P(B) – P(A and B)
Venn Diagrams
If A and B don’t overlap, P(A and B) = 0
A
B
( the intersection is empty )
So,
P( A or B ) = P(A) + P(B) – P(A and B)

P( A or B ) = P(A) + P(B)
A and B are said to be mutually exclusive events. ( If A
happens, B cannot or if B happens, A cannot. )
Venn Diagrams
Notation
We can write P(A or B) as P(A
B)
I remember the notation by thinking of this symbol as
a cup which can hold anything in A or B.
and
so,
P(A and B) as P(A
B)
P( A or B ) = P(A) + P(B) – P(A and B)
becomes
P( A
B ) = P(A) + P(B) – P(A
B)
Also, P( A/ )
means “ the probability that event A does not occur ”
Venn Diagrams
SUMMARY
If A and B are 2 events
P(A or B) can be written as P(A
B)
P(A and B) can be written as P(A
B)
P( A or B ) = P(A) + P(B) – P(A and B)
If P(A and B ) = 0, A and B are mutually exclusive
and then,
P( A or B ) = P(A) + P(B)
Venn Diagrams
17
2

,
e.g.1 Events A and B are such that P(A)
P(B)
25
5
4
and P(A or B)  . Find P(A and B).
5
Solution:
Þ
P(A or B) = P(A) + P(B) – P(A and B)
17
4
2


 P(A and B)
25
5
5
17 2 4
P(A and B) 
 
25 5 5
17  10  20

25
7

25
Venn Diagrams
e.g.2 In a group of 15 mixed plants, 6 are petunias, 8 are
yellow and 3 are both. What is the probability if a plant is
picked at random that it is (a) either a petunia or yellow
(b) either a petunia or yellow but not both?
Solution:
Let P be event ”Petunia” and Y be event “Yellow”
(a) We want to find P(P or Y)
P( P or Y ) = P(P) + P(Y) – P(P and Y)
Venn Diagrams
e.g.2 In a group of 15 mixed plants, 6 are petunias, 8 are
yellow and 3 are both. What is the probability if a plant is
picked at random that it is (a) either a petunia or yellow
(b) either a petunia or yellow but not both?
Solution:
Let P be event ”Petunia” and Y be event “Yellow”
(a) We want to find P(P or Y)
P( P or Y ) = P(P) + P(Y) – P(P and Y)
 6 
15
Venn Diagrams
e.g.2 In a group of 15 mixed plants, 6 are petunias, 8 are
yellow and 3 are both. What is the probability if a plant is
picked at random that it is (a) either a petunia or yellow
(b) either a petunia or yellow but not both?
Solution:
Let P be event ”Petunia” and Y be event “Yellow”
(a) We want to find P(P or Y)
P( P or Y ) = P(P) + P(Y) – P(P and Y)
 6  8 
15 15
Venn Diagrams
e.g.2 In a group of 15 mixed plants, 6 are petunias, 8 are
yellow and 3 are both. What is the probability if a plant is
picked at random that it is (a) either a petunia or yellow
(b) either a petunia or yellow but not both?
Solution:
Let P be event ”Petunia” and Y be event “Yellow”
(a) We want to find P(P or Y)
P( P or Y ) = P(P) + P(Y) – P(P and Y)
 6  8  3
15 15
15
Venn Diagrams
e.g.2 In a group of 15 mixed plants, 6 are petunias, 8 are
yellow and 3 are both. What is the probability if a plant is
picked at random that it is (a) either a petunia or yellow
(b) either a petunia or yellow but not both?
Solution:
Let P be event ”Petunia” and Y be event “Yellow”
(a) We want to find P(P or Y)
P( P or Y ) = P(P) + P(Y) – P(P and Y)
 6  8  3
15 15
15
11

15
(b) We must subtract another P(P and Y). ANS:
8
15
A Venn diagram can also be used to answer the question.
Venn Diagrams
e.g.2 In a group of 15 mixed plants, 6 are petunias, 8 are
yellow and 3 are both. What is the probability if a plant is
picked at random that it is (a) either a petunia or yellow
(b) either a petunia or yellow but not both?
Solution:
Let P be event ”Petunia” and Y be event “Yellow”
(a) We want to find P(P or Y)
P
Y
Venn Diagrams
e.g.2 In a group of 15 mixed plants, 6 are petunias, 8 are
yellow and 3 are both. What is the probability if a plant is
picked at random that it is (a) either a petunia or yellow
(b) either a petunia or yellow but not both?
Solution:
Let P be event ”Petunia” and Y be event “Yellow”
(a) We want to find P(P or Y)
15
P
Y
Venn Diagrams
e.g.2 In a group of 15 mixed plants, 6 are petunias, 8 are
yellow and 3 are both. What is the probability if a plant is
picked at random that it is (a) either a petunia or yellow
(b) either a petunia or yellow but not both?
Solution:
Let P be event ”Petunia” and Y be event “Yellow”
(a) We want to find P(P or Y)
15
Y
P
3
Venn Diagrams
e.g.2 In a group of 15 mixed plants, 6 are petunias, 8 are
yellow and 3 are both. What is the probability if a plant is
picked at random that it is (a) either a petunia or yellow
(b) either a petunia or yellow but not both?
Solution:
Let P be event ”Petunia” and Y be event “Yellow”
(a) We want to find P(P or Y)
15
Y
P
3
3
Venn Diagrams
e.g.2 In a group of 15 mixed plants, 6 are petunias, 8 are
yellow and 3 are both. What is the probability if a plant is
picked at random that it is (a) either a petunia or yellow
(b) either a petunia or yellow but not both?
Solution:
Let P be event ”Petunia” and Y be event “Yellow”
(a) We want to find P(P or Y)
15
Y
P
3
3
5
Venn Diagrams
e.g.2 In a group of 15 mixed plants, 6 are petunias, 8 are
yellow and 3 are both. What is the probability if a plant is
picked at random that it is (a) either a petunia or yellow
(b) either a petunia or yellow but not both?
Solution:
Let P be event ”Petunia” and Y be event “Yellow”
(a) We want to find P(P or Y)
15
Y
P
3
3
P(P or Y) =
5
4
Venn Diagrams
e.g.2 In a group of 15 mixed plants, 6 are petunias, 8 are
yellow and 3 are both. What is the probability if a plant is
picked at random that it is (a) either a petunia or yellow
(b) either a petunia or yellow but not both?
Solution:
Let P be event ”Petunia” and Y be event “Yellow”
(a) We want to find P(P or Y)
15
Y
P
3
3
11
P(P or Y) =
15
5
4
(b)
Venn Diagrams
e.g.2 In a group of 15 mixed plants, 6 are petunias, 8 are
yellow and 3 are both. What is the probability if a plant is
picked at random that it is (a) either a petunia or yellow
(b) either a petunia or yellow but not both?
Solution:
Let P be event ”Petunia” and Y be event “Yellow”
(a) We want to find P(P or Y)
15
Y
P
3
3
5
4
11
P(P or Y) =
15
(b) 8
15
Venn Diagrams
e.g.3 A, B are 2 events such that
P (A)  0  5 , P (B)  0  7 and P (A /  B / )  0  2
Draw a Venn diagram and use it to help you to find P (A  B)
Solution:
Let P (A  B)  p
P (A)  0  5 ,
B
A
p
Venn Diagrams
e.g.3 A, B are 2 events such that
P (A)  0  5 , P (B)  0  7 and P (A /  B / )  0  2
Draw a Venn diagram and use it to help you to find P (A  B)
Solution:
Let P (A  B)  p
P (A)  0  5 ,
and P (B)  0  7.
B
A
0 5  p
p
Venn Diagrams
e.g.3 A, B are 2 events such that
P (A)  0  5 , P (B)  0  7 and P (A /  B / )  0  2
Draw a Venn diagram and use it to help you to find P (A  B)
Solution:
Let P (A  B)  p
P (A)  0  5 ,
and P (B)  0  7.
B
A
0 5  p
p
0 7  p
Venn Diagrams
e.g.3 A, B are 2 events such that
P (A)  0  5 , P (B)  0  7 and P (A /  B / )  0  2
Draw a Venn diagram and use it to help you to find P (A  B)
Solution:
Let P (A  B)  p
P (A)  0  5 ,
and P (B)  0  7.
B
A
0 5  p
p
Finally,
P (A /  B / )  0  2
0 7  p
0 2
( probability of not in A and not in B )
The total probability is 1, so
0 5  p  p  0 7  p  0 2 1
1 4  p  1
 p  0 4
Venn Diagrams
Exercise
1
5

,

1. Events A and B are such that P(B) 2 P(A or B)
8
1
and P(A and B)  . Find P(A).
8
2. In a group of 20 students, 8 play music, 11 belong to
a sports team and 6 do both. What is the
probability that a student picked at random from the
group plays music or belongs to a sports team?
Venn Diagrams
Solutions:
5
1. Events A and B are such that P(B) = 1 , P(A or B) =
8
2
1
and P(A and B) =
. Find P(A).
8
Either:
P(A or B) = P(A) + P(B) – P(A and B)


Or:
1
1
5

 P(A) +
8
2
8
1
P(A) =
4
A
B
Venn Diagrams
Solutions:
5
1. Events A and B are such that P(B) = 1 , P(A or B) =
8
2
1
and P(A and B) =
. Find P(A).
8
Either:
P(A or B) = P(A) + P(B) – P(A and B)


Or:
1
1
5

 P(A) +
8
2
8
1
P(A) =
4
B
A
1
8
Venn Diagrams
Solutions:
5
1. Events A and B are such that P(B) = 1 , P(A or B) =
8
2
1
and P(A and B) =
. Find P(A).
8
Either:
P(A or B) = P(A) + P(B) – P(A and B)


Or:
1
1
5

 P(A) +
8
2
8
1
P(A) =
4
B
A
1
8
3
8
Venn Diagrams
Solutions:
1. Events A and B are such that P(B) = 1 , P(A or B) = 5
8
2
1
and P(A and B) =
. Find P(A).
8
Either:
P(A or B) = P(A) + P(B) – P(A and B)


Or:
1
1
5

 P(A) +
8
2
8
1
P(A) =
4
B
A
1
8
1
8
3
8
Venn Diagrams
Solutions:
5
1. Events A and B are such that P(B) = 1 , P(A or B) =
8
2
1
and P(A and B) =
. Find P(A).
8
Either:
P(A or B) = P(A) + P(B) – P(A and B)


Or:
1
1
5

 P(A) +
8
2
8
1
P(A) =
4
1
P(A) =
4
B
A
1
8
1
8
3
8
Venn Diagrams
Exercise
2. In a group of 20 students, 8 play music, 11 belong to
a sports team and 6 do both. What is the
probability that a student picked at random from the
group plays music or belongs to a sports team?
Solution:
Let M be event “plays music” and S “is in sports team”
Either:
P(M or S) = P(M) + P(S) – P(M and S)
8
6
11



20
20
20
13

20
Venn Diagrams
Exercise
2. In a group of 20 students, 8 play music, 11 belong to
a sports team and 6 do both. What is the
probability that a student picked at random from the
group plays music or belongs to a sports team?
Solution:
Let M be event “plays music” and S “is in sports team”
Or:
20
S
M
2
6
13
P(M or S) 
20
5
7
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
Venn Diagrams
Venn Diagrams
Venn diagrams show the probabilities of more than one
event and can be used instead of tree diagrams. They
are quick and easy to use.
Those of you taking the Edexcel or MEI specifications
need to understand these diagrams. For the rest of
you their use is optional but I am going to use them to
illustrate some important laws of probability.
Venn Diagrams
e.g.1 Amongst a group of 20 students, 7 are taking
Maths and of these 3 are also taking Biology. 5 are
taking neither. What is the probability that a student
chosen at random is taking Biology?
Solution: The diagram shows the 20 students.
20
The “eggs” show Maths
and Biology
B
M
4
3
8
The number doing Biology
but not Maths is given by
20  4  3  5  8
11
So, P( student takes Biology ) =
20
5
Venn Diagrams
e.g.2 In a class of 30 students, 3 out of the 16 girls
and 6 out of the 14 boys, are left-handed. Draw a
Venn diagram and find the probability that a student
chosen at random is a boy or left-handed.
Solution: The diagram needs to show the numbers for
Left-handedness and Boys.
There are 6 lefthanded boys so 8
boys are not.
There are 3 lefthanded girls and 13
right-handed
30
B
L
3
6
368
17

P( boy or a left-hander ) =
30
30
8
13
Venn Diagrams
We’ll use the example about left-handed students to
illustrate a law of probability.
30
B
A
3
6
n( A or B ) = 17
n(A) = 9
8
n(B) = 14
13
We now get
n(A and B) = 6
n( A or B ) = n(A) + n(B) - n( A and B )
Dividing by 30 gives probabilities, so
P( A or B ) = P(A) + P(B) – P(A and B)
Venn Diagrams
SUMMARY
If A and B are 2 events
P(A or B) can be written as P(A
B)
P(A and B) can be written as P(A
B)
P( A or B ) = P(A) + P(B) – P(A and B)
If P(A and B ) = 0, A and B are mutually exclusive
and then,
P( A or B ) = P(A) + P(B)