“a level maths” - meyers'...
TRANSCRIPT
© 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.
© Boardworks Ltd 2005 4 of 32
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)
© Boardworks Ltd 2005 6 of 32
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
© Boardworks Ltd 2005 7 of 32
P(A B) = P(A) + P(B) – P(A B)
The more general rule for finding P(A B) is:
This symbol means “and”
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.
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.
© Boardworks Ltd 2005 8 of 32
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,
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.
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 diagram shows the 20 students.
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
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:
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
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?
16 C L Solution:
L: lasagna
C: chips
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?
16 C L
5
Solution:
L: lasagna
C: chips
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?
16 C L
5 6
Solution:
L: lasagna
C: chips
Venn Diagrams
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?
16 C L
5 2 6
Solution:
L: lasagna
C: chips
Exercise
Venn Diagrams
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?
16 C L
5 2 6
3
Solution:
L: lasagna
C: chips
Exercise
Venn Diagrams
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?
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
We’ll use the example about left-handed students to illustrate a law of probability.
30 A B
6 8 3
13
Venn Diagrams
We’ll use the example about left-handed students to illustrate a law of probability.
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
We’ll use the example about left-handed students to illustrate a law of probability.
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
We’ll use the example about left-handed students to illustrate a law of probability.
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
We’ll use the example about left-handed students to illustrate a law of probability.
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
We’ll use the example about left-handed students to illustrate a law of probability.
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
We’ll use the example about left-handed students to illustrate a law of probability.
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
We’ll use the example about left-handed students to illustrate a law of probability.
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) – P(A and 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
(a) We want to find P(P or Y)
Solution: Let P be event ”Petunia” and Y be event “Yellow”
15
6
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?
Solution: Let P be event ”Petunia” and Y be event “Yellow”
Venn Diagrams
(a) We want to find P(P or Y)
Solution: Let P be event ”Petunia” and Y be event “Yellow”
15
6
15
8
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?
Solution: Let P be event ”Petunia” and Y be event “Yellow”
Venn Diagrams
(a) We want to find P(P or Y)
Solution: Let P be event ”Petunia” and Y be event “Yellow”
15
6
15
8 3
15
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?
Solution: Let P be event ”Petunia” and Y be event “Yellow”
Venn Diagrams
15
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”
15
6
15
8 3
15
11
P( P or Y ) = P(P) + P(Y) – P(P and Y)
(a) We want to find P(P or Y)
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
Solution: Let P be event ”Petunia” and Y be event “Yellow”
Y P
(a) We want to find P(P or 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?
Venn Diagrams
15
Solution: Let P be event ”Petunia” and Y be event “Yellow”
Y P
(a) We want to find P(P or 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?
Venn Diagrams
15
Solution: Let P be event ”Petunia” and Y be event “Yellow”
Y P
3
(a) We want to find P(P or 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?
Venn Diagrams
15
Solution: Let P be event ”Petunia” and Y be event “Yellow”
Y P
3 3
(a) We want to find P(P or 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?
Venn Diagrams
15
Solution: Let P be event ”Petunia” and Y be event “Yellow”
Y P
3 5 3
(a) We want to find P(P or 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?
Venn Diagrams
15
Solution: Let P be event ”Petunia” and Y be event “Yellow”
Y P
3 5 3
4
P(P or Y) =
(a) We want to find P(P or 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?
Venn Diagrams
15
Solution: Let P be event ”Petunia” and Y be event “Yellow”
Y P
3 5 3
4
P(P or Y) = 15
11
(a) We want to find P(P or 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?
(b)
Venn Diagrams
15
Solution: Let P be event ”Petunia” and Y be event “Yellow”
Y P
3 5 3
4
P(P or Y) = 15
11
(a) We want to find P(P or 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?
(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
e.g.3 A, B are 2 events such that
A B
p p 50
Solution:
Let pP )( BA
,50)( AP
and .70)( BP
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
Venn Diagrams
A B
p p 50 p 70
Solution:
Let pP )( BA
,50)( AP
and .70)( BP
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
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
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
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 or B) = P(A) + P(B) – P(A and B)
P(A) + 8
1
4
1
8
5
P(A) =
Solutions:
Or: A B
8
1
1. Events A and B are such that P(B) = , P(A or B) = 8
5
2
1
Either:
Venn Diagrams
2
1
1. Events A and B are such that P(B) = , P(A or B) = 8
5
P(A or B) = P(A) + P(B) – P(A and B)
P(A) + 8
1
4
1
8
5
P(A) =
Solutions:
Or: A B
8
1
8
3
2
1
8
1Find P(A). and P(A and B) = .
Either:
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:
Or: A B
8
1
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) = .
8
1
8
3
Either:
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:
Or: A B
P(A) = 4
1
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) = .
8
3
8
1
8
1
Either:
Venn Diagrams
Solution:
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?
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:
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?
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
The diagram shows the 20 students.
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
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:
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
We’ll use the example about left-handed students to illustrate a law of probability.
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
P( A or B ) = P(A) + P(B) – P(A and B)
8 6