“a level maths” - meyers'...

Venn Diagrams

“A Level Maths” Statistics 1

Venn Diagrams

© Boardworks Ltd 2005 3 of 32

Venn diagrams

A B

The circles can be overlapped to represent outcomes

that satisfy both events. (Card is Queen of Clubs)

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.


In Venn diagrams

representing mutually

exclusive events, the circles

do not overlap.

Two events A and B are called mutually exclusive if they

cannot occur at the same time.

A B

If A and B are mutually exclusive,

then:

Addition properties

This symbol means

‘union’ or ‘OR’

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.

P( ) = P( ) + P( )A B A B

The probability of A or B 𝑃 𝐴 ∪ 𝐵 not mutually exclusive.

P(C Q) =

𝑷 𝑪 ∪ 𝑸 = 𝟏𝟑

𝟓𝟐+𝟒

𝟓𝟐−𝟏

𝟓𝟐=𝟏𝟔

𝟓𝟐

This symbol means “and”

P(C) + P(Q) – P(C Q)


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.

Addition properties

1P( ) =

4C

4 1P( ) = =

52 13Q

1P( ) =

52C Q

Card is a club = event C

Card is a queen = event Q

1 1 1 4P( ) = + – =

4 13 52 13C QSo,

Note: A pack

has 52 cards. P( ) = P( ) + P( )A B A B


P(A B) = P(A) + P(B) – P(A B)

The more general rule for finding P(A B) is:

This symbol means “and”



This area

represents the

12 clubs that

are not queens.

This represents the

queen of clubs.

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.


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.

Addition properties

1P( ) =

4C

4 1P( ) = =

52 13Q

1P( ) =

52C Q



1 1 1 4P( ) = + – =

4 13 52 13C QSo,

This area

represents the

12 clubs that

are not queens.

This represents the

queen of clubs.

This represents the

other 3 queens.

Note: A pack

has 52 cards.

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.

0 Contents.ppt#4. Slide 4

Venn Diagrams

20

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 “eggs” show Maths M

B

5

4 3

The diagram shows the 20 students.

and Biology

3 do both

5 do neither

7 do Maths ( but we have 3 already )


Venn Diagrams

20

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:

M B

5

4 3 8


The final number ( doing Biology but not Maths ) is given by

20 4 3 5 8

So, P( student takes Biology ) = 20

11

20

83


Venn Diagrams

30

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.

L B

6 8 3

13

There are 6 Left-handed Boys . . .

so there are 8 boys who are not.

There are 3 left-handed girls . . . and 13 who are right-handed.


Venn Diagrams

30


Solution:

L B

6 8 3

13

P( boy or a left-hander ) = 30

863

30

17

The diagram needs to show the numbers for Left-handedness and Boys.


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?

C L Solution:

L: lasagna

C: chips


Venn Diagrams

Exercise


16 C L Solution:

L: lasagna

C: chips


Venn Diagrams

Exercise


16 C L

5

Solution:

L: lasagna

C: chips


Venn Diagrams

Exercise


16 C L

5 6

Solution:

L: lasagna

C: chips


Venn Diagrams


16 C L

5 2 6

Solution:

L: lasagna

C: chips

Exercise


Venn Diagrams


16 C L

5 2 6

3

Solution:

L: lasagna

C: chips

Exercise


Venn Diagrams


16 C L

5 2 6

3

Solution:

L: lasagna

C: chips

P(no chips, no lasagna)

16

3

Exercise


Venn Diagrams

We’ll use the example about left-handed students to illustrate a law of probability.

30 L B

6 8 3

13

I’ll change this to A as we usually use A when we state the law.


Venn Diagrams


30 A B

6 8 3

13


Venn Diagrams


30 A B

6 8 3

13

The number in A or B is anything shaded.

N.B. A or B in probability includes both.


Venn Diagrams


30 A B

6 8 3

13

We’ll use n for “ the number in . . .”, so

n( A or B ) = 3 + 6 + 8 = 17

n( A or B ) = 17


Venn Diagrams


30 A B

6 8 3

13

n(A) = 9

n(B) =

n( A or B ) = 17

n(A) = 9


Venn Diagrams

n( A or B ) = 17


30 A B

6 8 3

13

n(B) = 14

n(A) = 9

n(B) = 14

n(A) = 9


Venn Diagrams

30 A B

6 8 3

13

n( A or B ) = 17


The part with both types of shading gives the number in A and B.

n(B) = 14

n(A) = 9

So, n( A and B ) = 6

n(A and B) = 6


Venn Diagrams


30 A B

6 8 3

13

n( A or B ) = 17

n(B) = 14

n(A) = 9

n(A) +

n(A and B) = 6

We now get n( A or B ) =

17 =

n( A and B ) n(B) -

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


30 A B

6 8 3

13

n( A or B ) = 17

n(B) = 14

n(A) = 9

n(A) +

n(A and B) = 6

We now get n( A or B ) = n( A and B ) n(B) -

Dividing by 30 gives probabilities, so

P( A or B ) = P(A) + P(B) – P(A and B)


Venn Diagrams

A B

If A and B don’t overlap, P(A and B) = 0

( the intersection is empty )

P( A or B ) = P(A) + P(B) – P(A and B) So,

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)

P( A or B ) = P(A) + P(B) – P(A and B) so,

I remember the notation by thinking of this symbol as a cup which can hold anything in A or B.

and P(A and B) as P(A B)

becomes

P( A B ) = P(A) + P(B) – P(A B)

means “ the probability that event A does not occur ”

Also, P( A/ )


Venn Diagrams

If A and B are 2 events

SUMMARY

If P(A and B ) = 0, A and B are mutually exclusive and then,

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)


Venn Diagrams

Solution: P(A or B) = P(A) + P(B) – P(A and B)

25

17

5

2

5

4

e.g.1 Events A and B are such that P(A) P(B) ,25

17

5

2

and .5

4 Find P(A and B).

25

7

P(A or B)

P(A and B)

25

201017

P(A and B) 5

4

5

2

25

17Þ


Venn Diagrams

(a) We want to find P(P or Y)

Solution:

P( P or Y ) = P(P) + P(Y) – P(P and Y)

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?

Let P be event ”Petunia” and Y be event “Yellow”


Venn Diagrams


Solution: Let P be event ”Petunia” and Y be event “Yellow”

15

6





Venn Diagrams



15

6

15

8





Venn Diagrams



15

6

15

8 3

15





Venn Diagrams

15



15

6

15

8 3

15

11



A Venn diagram can also be used to answer the question.

(b) We must subtract another P(P and Y). ANS: 15

8


Venn Diagrams


Y P




Venn Diagrams

15


Y P




Venn Diagrams

15


Y P

3




Venn Diagrams

15


Y P

3 3




Venn Diagrams

15


Y P

3 5 3




Venn Diagrams

15


Y P

3 5 3

4

P(P or Y) =




Venn Diagrams

15


Y P

3 5 3

4

P(P or Y) = 15

11



(b)


Venn Diagrams

15


Y P

3 5 3

4

P(P or Y) = 15

11



(b) 15

8


Venn Diagrams

A B

p

e.g.3 A, B are 2 events such that

Draw a Venn diagram and use it to help you to find )( BAP

,50)( AP 70)( BP and 20)( /B/AP

Solution:

Let pP )( BA

,50)( AP


Venn Diagrams


A B

p p 50

Solution:

Let pP )( BA

,50)( AP

and .70)( BP



,50)( AP 70)( BP and 20)( /B/AP


Venn Diagrams

A B

p p 50 p 70

Solution:

Let pP )( BA

,50)( AP

and .70)( BP



,50)( AP 70)( BP and 20)( /B/AP


Venn Diagrams

A B

p p 50 p 70

20

Solution:

Let pP )( BA

,50)( AP

and .70)( BP

Finally,

20)( /B/AP

( probability of not in A and not in B )

The total probability is 1, so

141 p

1207050 ppp

40 p



,50)( AP 70)( BP and 20)( /B/AP


Venn Diagrams

,2

11. Events A and B are such that P(B) P(A or B)

and P(A and B)

8

5

.8

1 Find P(A).

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?

Exercise


Venn Diagrams

2

1

P(A or B) = P(A) + P(B) – P(A and B)

P(A) + 8

1

4

1

8

5

P(A) =

Solutions:

Either:

Or: A B

1. Events A and B are such that P(B) = , P(A or B) = 8

5

2

1

8

1Find P(A). and P(A and B) = .


Venn Diagrams

2

1

and P(A and B) = . 8

1Find P(A).


P(A) + 8

1

4

1

8

5

P(A) =

Solutions:

Or: A B

8

1


5

2

1

Either:


Venn Diagrams

2

1


5


P(A) + 8

1

4

1

8

5

P(A) =

Solutions:

Or: A B

8

1

8

3

2

1

8


Either:


Venn Diagrams

2

1


P(A) + 8

1

4

1

8

5

P(A) =

Solutions:

Or: A B

8

1


5

2

1

8


8

1

8

3

Either:


Venn Diagrams

2

1


P(A) + 8

1

4

1

8

5

P(A) =

Solutions:

Or: A B

P(A) = 4

1


5

2

1

8


8

3

8

1

8

1

Either:


Venn Diagrams

Solution:


20

8

20

11

20

6

20

13

Let M be event “plays music” and S “is in sports team”

P(M or S) = P(M) + P(S) – P(M and S) Either:

Exercise


Venn Diagrams

20 S M

Solution:

6 5 2

7

Let M be event “plays music” and S “is in sports team”

20

13P(M or S)

Exercise

Or:



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

20

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 “eggs” show Maths M

B

5

4 3


and Biology

So, P( student takes Biology ) = 20

11

The number doing Biology but not Maths is given by

20 4 3 5 8

8

Venn Diagrams

30


Solution:

L B

6 8 3

13

P( boy or a left-hander ) = 30

863

30

17

The diagram needs to show the numbers for Left-handedness and Boys.

There are 6 left-handed boys so 8 boys are not.

There are 3 left-handed girls and 13 right-handed

Venn Diagrams


30 A B

3

13

n( A or B ) = 17

n(B) = 14

n(A) = 9

n(A) +

n(A and B) = 6

We now get n( A or B ) = n( A and B ) n(B) -

Dividing by 30 gives probabilities, so


8 6

Venn Diagrams

If A and B are 2 events

SUMMARY

If P(A and B ) = 0, A and B are mutually exclusive and then,

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)

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