Transcript discrete_maths_show_teachers

LEAVING CERTIFICATE.
ARRANGEMENTS,
SELECTIONS
AND
PROBABILITY
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This presentation will emphasise:
• The advantages of the counting method
• The importance of learning by doing
• Clear thinking over blind application of rules
• The advantages of an experimental/investigative
approaches
• The suitability of PowerPoint as a teaching tool
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Why Discrete Maths?
• Short syllabus
• Two Questions on Examination Paper
(Possible third)
• Least popular questions attempted by
students: Q.6 - 68%
Q.7- 57%
• Important at Third Level
Business - Science –Psychology – Medicine etc
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Higher Syllabus
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FUNDAMENTAL PRINCIPLE OF COUNTING
Fundamental Principle of Counting: If one task can be
accomplished in x different ways, and following this another task
can be accomplished in y different ways, then the first task
followed by the second can be accomplished in xy different ways.
Example: A car manufacturer makes 3 models of cars, a mini, a saloon
and an estate. These cars are all available in a choice of 5 colours; red,
green, blue, black and orange. How many different cars are available ?
3

5
= 15 different versions
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Example: A car manufacturer makes 3 models of cars, a mini, a saloon and an estate. These cars are all
available in a choice of 5 colours; red, green, blue, black and orange. How many different cars are
available ?
A tree diagram can be used to
illustrate this example
15 different versions
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Example. In a restaurant you can have 6 types of muffin, 8 varieties
of sandwich, and 5 drinks ( coffee, tea, coke, fanta, 7 up).
Lunch is either a muffin and a hot drink or a sandwich and
a cold drink. How many different choices of lunch are
possible?
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PERMUTATIONS (ARRANGEMENTS)
In how many ways can we arrange the letters a,b,c,d,e
taking the letters two at a time?
Method 1.
Write out all the arrangements
(a,b) (a,c) (a,d) (a,e) (b,c) (b,d) (b,e) (c,d) (c,e) (d,e)
Total = 20
(b,a) (c,a) (d,a) (e,a) (c,b) (d,b) (e,b) (d,c) (e,c) (e,d)
Method 2.
Box Method
Possibilities
5
4
Total = 5x4 = 20
Method 3.
Use the P notation
This is called FACTORIAL 5
5!
5! 5.4.3.2.1
5
P2 
 
 5.4  20
(5  2)! 3!
3.2.1
The most common method : Box Method. 8
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NOTES ON THE P NOTATION
Formally n Pr  (n  0)(n  1)(n  2).........[n  (r  1)]
n
6
Pr  n(n  1)(n  2).........(n  r  1)
n!
n
Pr 
(n  r )!
6!
P2 
 65
(6  2)!
8
On calculator
P4  8  7  6  5  1680
6
P
2
=
Answer is 30
SPECIAL ONES
5
P5  5  4  3  2 1  120 This is called FACTORIAL 5  5!
4
P4  4! 4  3  2 1  24
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n! n(n  1)(n  2)..........3  2 1
(n  3)! (n  3)(n  4)(n  5)...........3  2 1
Ex . 1
Simplify
3! (n  1)!
n!
3.2.1.(n  1)(n)(n  1)...............3  2 1
n(n  1)(n  2).............3  2 1
6( n  1 )
6n  6
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Note 0!  1 Why?
5 !  120
4 !  24
3!  6
2!  2
1!  1
0!  1
120  5  24
24  4  6
632
2  2 1
1 1  1
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Ex.1 Q.7(a) 2001 Paper 2
(i) How many different four letter arrangements
can be made from the letters of the word FRIDAY
if each letter is used no more than once in each
arrangement ?
(ii) How many of the above arrangements begin with the
letter D and also end with a vowel ?
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Ex.2 Q.6(a) 2000 Paper 2
(i) A Bank gives each of it's customers a four digit
pin number which is formed from the digits 0 to 9.
Examples are : 2475, 0865 and 3422.
(i) How many different pin numbers can the bank use?
(ii) If the bank decides not to use pin numbers that begin
with 0, how many different numbers can it then use?
The examples given are very important.
NOTE : 0865 means that you can start with 0.
NOTE : 3422 means that you can repeat the digits.
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Ex. 3 Q.6(a) 1998 Paper 2
(i) In how many ways can the letters of the word IRELAND
be arranged if each letter is used exactly once in each
arrangment?
(ii) In how many of these arrangments do the three vowels
come together?
(iii) In how many arrangments do the three vowels not
come together?
(i)
Possibilities
7
6
5
4
3
2
1
7  6  5  4  3  2 1  7!ways  5040
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( ii )
POSSIBILITIES
5
4
3
VOWELS
2
1
Treat the vowels as one unit
5 ! . 3!  120.6  720 ways
( iii ) Number of ways in which the vowels do not come together
 The total arrangemen ts - The number of ways they are together
5040  720  4320
This last part is a negative statement.
 Total - Positive statement
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ARRANGING THINGS IN A CIRCLE
RULE : The number of arrangment s of n objects in a circle
is (n - 1)!
A circular permutation is not considered to be distinct from another
unless one object is preceded or succeeded by a different object
in both permutations
Ex . In how many ways can we arrange six people
around a circular table?
Answer is (6 - 1)!  5! ways
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B
A
A
C
B
D
D
D
C
A
C
B
B
C
D
A
Are these different arrangements?
B
A
A
C
D
D
B
D
C
C
A
C
B
B
D
A
NO.
ALL FOUR ARE REPRESENTATIONS OF THE SAME ARRANGEMENT
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COMBINATIONS (SELECTIONS)
In how many ways can we select the letters a,b,c,d,e
two at a time.
Method 1.
Write out all the selections
(a,b) (a,c) (a,d) (a,e) (b,c) (b,d) (b,e) (c,d) (c,e) (d,e)
Total = 10
NOTE : (b,a) is the same selection as (a,b)
Method 2.
Use the C notation
Formal Definition
5
 
 
2
 

n
n!
 
   r !(n  r )!
r
 
5!
2!(5-2)!
5.4

1.2
5.4
 10

2!
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Origin of Formal Definition

From n objects
select r objects

  Arrange the chosen r objects
From n objects
select r objects

  r!
From n objects
select r objects

  r!
From n objects
select r objects
We use this
symbol


n
r

 From n objects
arrange r objects
 pr
n
n!

 nr !
n!

r ! nr !
n!

r ! nr !
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
NOTES ON THE C NOTATION
6
 
6.5.4.3.2.1
6.5


 15
 
 4  (4.3.2.1)( 2.1) 2.1
 
8
   8.7.6.5  8.7.6.5  70
  1.2.3.4
4!
 4
 
n
   n(n1)(n2)(n3)...3.2.1  n(n1)(n2)
 
3.2.1.(n3)...3.2.1
3!
 3
 
n
   n(n1)(n2)..........(nr1)
 
r!
r
 
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TWO IMPORTANT RULES
RULE 1
 n
 
 1
0
 
RULE 2
 n  n 
  

 

 r  n r
  

Examples
5
 
 1
0
 
5 5
Examples     
   
 3  2
   
 20 
 
 1
0
 
 20   20 
   
  
 18   2 
   
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Ex . 1 (i) In how many ways can a committee of four be
choosen from nine people.
(ii) If a certain person is always to be included, in how
many ways can the committee be chosen.
(i ) From 9 we are choosing 4.
(ii )
9
  9.8.7.6
 126
 
 4  1.2.3.4
 
From 8 we are choosing 3, as one person is always included.
8
  8.7.6
 56
 
 3  1.2.3
 
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Ex .2 Q.7(a) 2000 Paper 2
The points a, b, c, d, e, and f lie on a cricle
(i) If these points are used as as vertices, how many
different quadrilaterals can be formed?
(ii) How many of these quadrilaterals will have
ab as one side?
(i ) Each 4-sided figure requires 4 points
 From 6 we are choosing 4.
(ii) If ab is one of the sides we now
have 4 points to choose 2.
a
f
b
e
c
d
6  6 
    6.5
 15
  
 4   2  1.2
   
4
  4.3
6
 
 2  1.2
 
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Ex .3 Q.6(a) 1999 Paper 2
(i) In how many ways can a group of five people
be selected from four women and four men?
(ii) In how many of these groups are there exactly three wom en?
( i ) We have a total of 8 people
from whom we want to select 5.
( ii ) In order to have exactly 3 women, we can have the following.
3 women
and
2 men
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Ex .4 Q.6(a) 1996 Paper 2
(i) In how many ways can a group of five people
be selected from ten people?
(ii) How many groups can be selected if two particular
people from the ten cannot be in the same group?
( i ) We have a total of 10 people
from whom we want to select 5.
 10 
  10.9.8.7.6
 252
 
 5  1.2.3.4.5
 
( ii ) This is a negative statment " cannot"
Cannot be included  Total - When they are Included
8
  8.7.6
252 
 56
 
 3  1.2.3
 
252 - 56  196
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Ex .5 Q.6(a) 2001 Paper 2
(i) How many different sets of three books or four books
can be selected from six different books?
(ii) How many of the above sets contain one particular book?
( i ) 3 BOOKS
or 4 BOOKS
6 
6 
 
 
    
 3
4
 
 
20  15  35 ways
( ii ) One book is always to be included
 we are choosing 2 books from 5 OR 3 books from 5.
5
5
 
 
      10  10  20 ways
 2
 3
 
 
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HEADS
OR
TAILS
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Socks Problem
Four distinguishable pairs of socks are in a
drawer. If two socks are selected at random
find the probability that a pair is chosen.
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Fundamental Basis of
Probability
The number of things we want (favourable outcomes)
P
The total number of things available (total outcomes)
Number of outcomes of interest
P
Number of possible outcomes
Number of possible outcomes is also called the Sample Space
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A bag contains five blue and six red discs. A disc is drawn at random from
the bag. Find the probability that the disc is red.
Favourable Outcomes :
Total Outcomes :
From 6 red choose 1
From 11 discs choose 1
6
 
Favourable 1  6
Probability =


Total
11 11
 
1
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A bag contains five blue and six red discs. Two discs are drawn at random
from the bag. Find the probability that both discs are red.
Favourable Outcomes :
Total Outcomes :
From 6 red choose 2
From 11 discs choose 2
6
 
Favourable  2  3
Probability =


Total
11 11
 
2
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A bag contains five blue and six red discs. Two discs are drawn
at random from the bag.
Find the probability that exactly one red disc is drawn.
Find the probability that one red and one blue disc are drawn
Favourable Outcomes :
}
Note: Both are
asking the same
question
From 6 red choose 1 and from 5 blue choose 1
 6  5
  
Favourable 1  1  6
Probability =


Total
11
11
 
2
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A bag contains five blue and six red discs. Three discs are drawn at random
from the bag. Find the probability that all three discs are red.
Favourable Outcomes :
Total Outcomes :
From 6 red choose 3
From 11 discs choose 3
6
 
Favourable  3 
4
Probability =


Total
11 33
 
3
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A bag contains five blue and six red discs. Three discs are drawn at random
from the bag. Find the probability that exactly two of the discs are red.
Note: Since 3 discs are
drawn the third one must
be considered and is blue
Favourable Outcomes :
From 6 red choose 2 and from 5 blue choose 1
 6   5
  
Favourable  2  1  5
Probability =


Total
22
11
 
3
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A bag contains five blue and six red discs. Three discs are drawn at random
from the bag. Find the probability that at least one red disc is drawn.
Note: This is the negative of
drawing three blue discs.
Unfavourable Outcomes :
Total Outcomes :
Favourable Outcomes :
From 5 blue choose 3
From 11 discs choose 3
  


 5
3
 11
3
 11  5
3
3
11  5 
  
Favourable  3   3  31
Probability (at least one red) =


11
Total
33
 
 
3 
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Another method
A bag contains five blue and six red discs. Three discs are drawn at random
from the bag. Find the probability that at least one red disc is drawn.
Note: We can also do this by
counting all of the cases that
will yield a favourable
outcome.
Favourable Outcomes :
From 6 red choose 3
or
From 6 red choose 2 and from 5 blue choose 1
or
From 6 red choose 1 and from 5 blue choose 2
Probability (at least one red)=

  
  
6  20
3
6  5  75
2 1
6  5  60
1 2
Favourable 20  75  60 31


Total
33
11
 
3 
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RELATIVE FREQUENCY
number of successful trials
RELATVE FREQUENCY=
total number of trials
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Example 1.
A drawing-pin can land “point up” or “point down” when dropped.
Pat drops a drawing-pin 100 times and it lands “point up”35 times. Estimate
the probability of the drawing pin landing “point up”.
Solution:
Probability =
number of successful trials
total number of trials
35

100
7

20
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Example 2
A coin was tossed 1000 times and the number of heads were recorded.
It was found that a head occurred 495 times. Estimate the probability of
getting a head.
Solution:
Probability =
number of successful trials
total number of trials
495

1000
 0  495
Note: As the number of trials increases the relative
frequency tends to stabilise about a constant value. This
constant value is called the probability of the event
occurring.
Therefore the probability of getting a head is 0.5 or 50%
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Example 3
A five sided spinner, with sides marked A, B, C, D and E, was spun 10,000 times.
The frequency of occurrence of each side is shown in the following frequency
table.
Side
A
B
C
D
E
Frequency 1988 2023 2009 1981
Calculate,
correct to three decimal places, the relative frequency of
1999
occurrence of each side and hence estimate the probability of
occurrence of each side.
SOLUTION:
Side
A
Frequency 1988
Relative
0.20
Frequency
B
2023
0.202
C
D
E
2009 1981 1999
0.201 0.198 0.20
Probability of occurrence of each side = 20%
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PROBABILITY
1. P(event ) 
Number of favourable outcomes
Total number of outcomes
Number of outcomes of interest

Number of possible outcomes
2. 0  P(event )  1,
0
1
3. P(event not occuring)1P(event occuring)
This can be useful in certain questions but is not required
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SINGLE EVENT PROBLEMS
Ex . 1 A card is picked at random from a pack of 52 cards.
What is the probability that it is (i) an ace (ii) a spade.
Favourable 4
1
(i) P(ace) 


Total
52 13
(ii) P(spade) 
Favourable 13 1


Total
52 4
Ex . 2 A bag contains 8 blue marbles and 6 white marbles
A marble is drawn at random from the bag. What is the
the probability that the marble is white.
P(white) 
Favourable 6
3


Total
14 7
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MULTIPLE EVENTS PROBLEMS. F.P.C
Ex.1 A die is thrown and a coin is tossed. What is the probability of
getting a head and a 5?
Method 1. Sample Space
H
1
1,H
2
2,H
3
4
3,H 4,H
5
5,H
6
6,H
T
1,T
2,T
3,T
5,T
6,T
4,T
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MULTIPLE EVENT PROBLEMS.
Ex. 1 An unbiased die is thrown twice. Find the probability
of getting two equal scores or a total of 10
Method 1. Grid method F.P.C
1
1
1,1
2
2,1
3
3,1
4
5
4,1 5,1
6
6,1
2
3
1,2
1,3
2,2
2,3
3,2
3,3
4,2 5,2
4,3 5,3
6,2
6,3
4
1,4
2,4
3,4
4,4 5,4
6,4
5
1,5
2,5
3,5
4,5 5,5
6,5
6
1,6
2,6
3,6
4,6 5,6
6,6
We have 36 outcomes and 8
favourable outcomes
8
P ( equal or total 10) 
36
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1999 Q. 9 (a)
An unbiased die is thrown twice.
Find the probability of getting a total less than four.
1
1
1,1
2
2,1
3
4 5
3,1 4,1 5,1
6
6,1
2
3
1,2
1,3
2,2
2,3
3,2 4,2 5,2
3,3 4,3 5,3
6,2
6,3
4
1,4
2,4
3,4 4,4 5,4
6,4
5
1,5
2,5
3,5 4,5 5,5
6,5
6
1,6
2,6
3,6 4,6 5,6
6,6
P(event) 
Favourable 3
1


Total
36 12
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REPLACEMENT AND NO REPLACEMENT
Ex. 1 A card is picked at random from a pack of 52 cards.
The card is replaced and again a card is then picked.
What is the probability that the two cards picked are clubs.
Note: Replacement does not occur unless explicitly stated
13  13 
  
Favourable  1   1  13 13 1



Total
 52   52  52  52 16
   
1 1
Ex. 2 Two cards are picked at random from a pack of 52 cards.
What is the probability that the two cards picked are clubs.
13 
 
Favourable  2  13 12 12
1




Total
 52  52  51 204 17
 
2
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ORDER DOES NOT MATTER
Ex.1 Three cards are drawn at random, without replacement, from
from a pack of 52.
Find the probability that the three cards drawn are,
the Jack of hearts, the Queen of diamonds
and the King of clubs.
No. of favourable outcomes =
Total no. of outcomes =
P(
)
 52 
 
3
33

3
3
1


52
  22100
 
3
47
© MSS 2007
2005
Discrete maths questions
48
© MSS 2007
2005
1
2
Outcomes of Interest:
(Favourable Outcomes) :
Total Outcomes :
3
4
5
6
7
8
9
From remaining cards choose 3.
Therefore we have 8 cards left to choose 3 =
We have 9 cards to choose 3 =
8
 
Favourable  3  56 2
Probability =



Total
 9  84 3
 
3
8
   56
 3
9
   84
 3
49
© MSS 2007
1
2
Outcomes of Interest:
(Favourable Outcomes)
3
4
From remaining
cards choose 3
Probability =
5
6
7
8
9
Therefore we have 5 cards left to choose 3 =
5
   10
 3
Favourable 10 5


Total
84 42
50
© MSS 2007
1
7
+
8
2
+ 9
3
24
4
>
5
1
6
+ 2
7
+
8
3
+ 4
9
+ 5
+ 6
21
6
+
8
+ 9
23
>
1
+ 2
+
3
+ 4
+ 5
+ 7
22
5
+
8
+ 9
22
<
1
+ 2
+
3
+ 4
+ 6
+ 7
23
Only Two Favourable Outcomes
Probability =
Favourable 2
1


Total
84 42
51
© MSS 2007
What the Chief Examiner had to
say about Q6 -2005
• (c) Parts (i) and (ii) involved relatively standard counting
techniques in a familiar context. Nonetheless they
required some clarity of thought,
• Part (iii) was poorly answered. Candidates seemed
unable to identify favourable outcomes and many did not
even attempt the question. It would appear that, other
than with certain well-rehearsed question types,
candidates lack the capacity to systematically list and/or
systematically count outcomes satisfying particular
criteria. Even the basic generic skills for this topic, such
as exploring the situation by looking at examples of
outcomes that satisfy or do not satisfy relevant criteria,
were lacking. –
52
© MSS 2007
2005
Outcomes of Interest: Any 4 blue discs
Favourable Outcomes :
Total Outcomes :
We have 16 discs to choose 4 =
Probability =
5
 5
 4
Therefore we have 5 blue to choose 4 =
16 
   1820
4
Favourable
5
1


Total
1820 364
53
© MSS 2007
Outcomes of Interest:
Favourable Outcomes :
5 blue choose 4 or
6 red choose 4:
5
 
 4
OR
6
 
 4
5
+
15
Probability =
= 20
Favourable
20
1


Total
1820 91
54
© MSS 2007
Outcomes of Interest: 5 Blue choose 1 and 3 Green choose 1 and 6 Red choose 1 and 2 Yellow choose 1
Favourable Outcomes :
5
 
1
AND
 3
 
1
AND
6
 
1
AND
 2
 
1
5
x
3
x
6
x
2
Probability =
= 180
Favourable 180
9


Total
1820 91
55
© MSS 2007
Outcomes of Interest:
Favourable Outcomes :
5 blue choose 2 and other discs, 11 choose 2
5
 
 2
AND
11
 
2
10
x
55
Probability =
= 550
Favourable 550
55


Total
1820 182
56
© MSS 2007
What the Chief Examiner had to
say about Q7 2005
• (b) Parts (i) and (ii) were very well answered by most
candidates. Parts (iii) and (iv), however, caused much more
difficulty than ought to be expected at this level. It is worth
noting that the majority of candidates approached these parts
using multiplicative laws for probability, which are not on the
core course. Such candidates (unless they have developed their
understanding substantially through studying the material on
the option) tend to favour a blind application of rules over
clear thinking, and generally suffer the consequences in all but
the most basic of situations. In this case, they made errors
related to ordering. As might be expected, the candidates who
stuck to the basic principle for dealing with all situations
involving equally likely outcomes (i.e.,
count the outcomes and put favourable over possible)
fared much better than those attempting to use the more
complex rules.
57
© MSS 2007
LC 2003 Paper 2 Question 6 (a) (i)
(a) Eight people, including Kieran and Anne, are available to
form a committee.
Five people must be chosen for the committee
(i) In how many ways can a committee be formed if both Kieran
and Anne must be on the chosen ?
How do you get pupils to decide between combinations and
permutations ?
58
© MSS 2007
LC 2003 Paper 2 Question 6 (a) (i)
(a) Eight people, including Kieran and Anne, are available to
form a committee.
Five people must be chosen for the committee
(i) In how many ways can a committee be formed if both Kieran
and Anne must be on the chosen ?
How do you get pupils to decide between combinations and
permutations ?
59
© MSS 2007
LC 2003 Paper 2 Question 6 (a) (i)
(a) Eight people, including Kieran and Anne, are available to
form a committee.
Five people must be chosen for the committee
(i) In how many ways can a committee be formed if both Kieran
and Anne must be on the chosen ?
How do you get pupils to decide between combinations and
permutations ?
If combinations is being used what are the two important
questions ?
1) How many are available ?
n
2) How many am I choosing/picking/selecting ? r
n
r 
 
60
© MSS 2007
(i) In how many ways can a committee be formed
if both Kieran and Anne must be on the chosen ?
Kieran Anne
Committee (of 5)
? ? ?
1) How many are available ?
n
2) How many am I choosing/picking/selecting ? r
61
6
3
 
© MSS 2007
(ii)
In how many ways can a committee be formed
if neither Kieran nor Anne can be chosen
Kieran Anne
Committee (of 5)
? ? ? ? ?
1) How many are available ?
n
2) How many am I choosing/picking/selecting ? r
62
6
5
 
© MSS 2007
LC 2003 Paper 2 Question 6 (c)
Ten discs, each marked with a different whole number from 1 to 10,
are placed in a box. Three of the discs are drawn at random
(without replacement) from the box.
(i) What is the probability that the disc with the number 7 is drawn ?
(ii) What is the probability that the three numbers on the discs drawn
are odd ?
(iii) What is the probability that the product of the three numbers on
the discs drawn is even ?
(iv) What is the probability that the smallest number on the discs
drawn is 4 ?
63
© MSS 2007
(i) What is the probability that the disc with the number 7 is drawn ?
Method F.P.C
The number of things we want (favourable outcomes)
P=
The total number of things available (total outcomes)
1
2
3
4
5
6
Favourable Outcomes :
Total Outcomes :
7
8
From the 10 discs choose any 3
9
10
P
A seven and two other discs
A seven and from 9 choose any 2 discs
From 10 choose any 3 discs
 1  9 
 1   2 
3




P

10
10 
3 
64
 
© MSS 2007
(ii) What is the probability that the three numbers on the discs are odd ?
Method F.P.C
The number of things we want (favourable outcomes)
P=
The total number of things available (total outcomes)
Favourable Outcomes :
Total Outcomes :
1
2
Three odd discs
From the 10 discs choose any 3
3
P=
4
7
10
5
8
6
9
From the 5 odd discs choose any 3
From 10 choose any 3 discs
5
3
1


P

10  12
3 
 
65
© MSS 2007
(iii) What is the probability that the product of the three numbers on the
discs is even ?
Note: If any or all the numbers are even then the product will be even.
Unfavourable Outcomes : Three odd discs
Total Outcomes :
From the 10 discs choose any 3
Favourable Outcomes = Total – unfavourable =
1
2
3
4
5
6
7
8
9
10
10   5 
    
 3   3
10   5 
 3   3
11




P

10
12
 
3 
 
66
© MSS 2007
(iv) What is the probability that the smallest number on the discs drawn is 4 ?
P=
The number of things we want (favourable outcomes)
The total number of things available (total outcomes)
Favourable Outcomes :
Total Outcomes :
1
2
From the 10 discs choose any 3
3
4
5
6
7
8
9
10
A four and two other discs larger than 4
A four and from (5, 6, 7, 8, 9, 10) any 2 discs
P=
From 10 any 3 discs
1  6 
1   2 
1




P

8
10 
3 
 
67
© MSS 2007
LCH 2003 Paper 2 Question 7 (a) (i)
(a) Five cars enter a car park. There are exactly five vacant spaces
in the car park.
(i) In how many different ways can the five cars park in the vacant
spaces ?
5
4
3
2
1
5.4.3.2.1 = 120 ways
68
© MSS 2007
LCH 2003 Paper 2 Question 7 (a) (ii)
Two of the cars leave the car park without parking. In how many
different ways can the remaining three cars park in the five vacant
spaces ?
There are 5 different spaces this car could choose.
However once a space is picked there is only 4 spaces left for
the next car.
Similarly, only 3 spaces are left for
the final car.
No. of ways 5 . 4 . 3 = 60
69
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (i)
L and k are distinct parallel lines. a, b,c and d are points on L such that
|ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm.
(i)How many different triangles can be constructed using three of the
named points as vertices?
L
. . . .
b
a
K
c
d
. . .
x
y
z
Two identified methods here.
(1) Traditional C Notation
(2) Inspection
70
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (i)
L and k are distinct parallel lines. a, b,c and d are points on L such that
|ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm.
(i)How many different triangles can be constructed using three of the
named points as vertices.
Choose 1 point from line L and
b
c
d
a
L
two points from line K
. . . .
K
. . .
x
y
 4  3 
1  2 
  
OR Choose 1 point from line K
and two points from line L
z
 4  3 
 2 1 
  
Two identified methods here.
(1) Traditional C Notation
4.3 + 6.3 = 12+18 = 30 triangles
71
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (i)
Land k are distinct parallel lines. A, b,c and d are points on L such that
|ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm.
(i)How many different triangles can be constructed using three of the
named points as vertices.
Using just point a
b
c
d
a
L
. . . .
1
K
. . .
x
y
z
Two identified methods here.
(2) Inspection
72
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (i)
Land k are distinct parallel lines. A, b,c and d are points on L such that
|ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm.
(i)How many different triangles can be constructed using three of the
named points as vertices.
Using just point a
b
c
d
a
L
. . . .
2
K
. . .
x
y
z
Two identified methods here.
(2) Inspection
73
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (i)
Land k are distinct parallel lines. A, b,c and d are points on L such that
|ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm.
(i)How many different triangles can be constructed using three of the
named points as vertices.
Using just point a
b
c
d
a
L
3 triangles from a
. . . .
K
. . .
3
x
y
z
4 points (a,b,c,d)
4 x 3 = 12 triangles
Two identified methods here.
(2) Inspection
74
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (i)
Land k are distinct parallel lines. A, b,c and d are points on L such that
|ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm.
(i)How many different triangles can be constructed using three of the
named points as vertices.
Using just point x
b
c
d
a
L
1
. . . .
K
. . .
3
x
y
z
Two identified methods here.
(2) Inspection
75
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (i)
Land k are distinct parallel lines. A, b,c and d are points on L such that
|ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm.
(i)How many different triangles can be constructed using three of the
named points as vertices.
Using just point x
b
c
d
a
L
2
. . . .
K
. . .
x
y
z
Two identified methods here.
(2) Inspection
76
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (i)
Land k are distinct parallel lines. A, b,c and d are points on L such that
|ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm.
(i)How many different triangles can be constructed using three of the
named points as vertices.
Using just point x
b
c
d
a
L
3
. . . .
K
. . .
x
y
z
Two identified methods here.
(2) Inspection
77
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (i)
Land k are distinct parallel lines. A, b,c and d are points on L such that
|ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm.
(i)How many different triangles can be constructed using three of the
named points as vertices.
Using just point x
b
c
d
a
L
4
. . . .
K
. . .
x
y
z
Two identified methods here.
(2) Inspection
78
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (i)
Land k are distinct parallel lines. A, b,c and d are points on L such that
|ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm.
(i)How many different triangles can be constructed using three of the
named points as vertices.
Using just point x
b
c
d
a
L
5
. . . .
K
. . .
x
y
z
Two identified methods here.
(2) Inspection
79
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (i)
Land k are distinct parallel lines. A, b,c and d are points on L such that
|ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm.
(i)How many different triangles can be constructed using three of the
named points as vertices.
Using just point x
b
c
d
a
L
6 triangles from x
6
3 points (x,y,z)
. . . .
K
. . .
x
y
z
Two identified methods here.
3 x 6 = 18 triangles
12 red + 18 blue = 30 triangles
(2) Inspection
80
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (ii)
(ii) How many different quadrilaterals can be constructed ?
Again two methods could be used here.
(1) Traditional C Notation
(2) Inspection
a
L
. . . .
K
b
c
d
. . .
x
y
z
81
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (ii)
(ii) How many different quadrilaterals can be constructed ?
Again two methods could be used here.
(1) Traditional C Notation
L
K
Quadrilateral:
. . . .
b
a
c
d
. . .
x
y
z
Choosing 2 points from line L
 4
 2
 
Choosing 2 points from line K
3
 2
 
6 x 3 = 18 different quadrilaterals
82
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (ii)
(ii) How many different quadrilaterals can be constructed ?
Again two methods could be used here.
(2) Inspection
L
Using just [ab]
. . . .
b
a
c
d
1
K
. . .
x
y
z
83
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (ii)
(ii) How many different quadrilaterals can be constructed ?
Again two methods could be used here.
(2) Inspection
L
Using just [ab]
. . . .
b
a
c
d
2
K
. . .
x
y
z
84
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (ii)
(ii) How many different quadrilaterals can be constructed ?
Again two methods could be used here.
(2) Inspection
L
. . . .
b
a
c
d
There are 6 possible line segments
3
K
. . .
x
Using just [ab]
There are just 3 quadrilaterals
which can be formed using [ab]
y
z
([ab],[ac],[ad],[bc],[bd],[cd])
6 x 3 = 18 different quadrilaterals
85
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (iii)
(iii)how many different parallelograms can be constructed using
four of the named points as vertices ?
. . . .
b
a
c
d
[ab]
1
. . .
x
y
z
86
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (iii)
(iii)how many different parallelograms can be constructed using
four of the named points as vertices ?
. . . .
b
a
c
d
[ab]
2
. . .
x
y
z
87
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (iii)
(iii)how many different parallelograms can be constructed using
four of the named points as vertices ?
. . . .
b
a
c
d
[ac]
3
. . .
x
y
z
88
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (iii)
(iii)how many different parallelograms can be constructed using
four of the named points as vertices ?
. . . .
b
a
c
d
[bc]
4
. . .
x
y
z
89
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (iii)
(iii)how many different parallelograms can be constructed using
four of the named points as vertices ?
. . . .
b
a
c
d
[bc]
5
. . .
x
y
z
90
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (iii)
(iii)how many different parallelograms can be constructed using
four of the named points as vertices ?
. . . .
b
a
c
d
[bd]
6
. . .
x
y
z
91
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (iii)
(iii)how many different parallelograms can be constructed using
four of the named points as vertices ?
. . . .
b
a
c
d
[cd]
7
. . .
x
y
z
92
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (iii)
(iii)how many different parallelograms can be constructed using
four of the named points as vertices ?
. . . .
b
a
c
d
[cd]
8
. . .
x
y
8 parallelograms
z
93
© MSS 2007
LCH 2003 Paper 2 Question 7 (b) (iv)
(iv) If one quadrilateral is constructed at random, what is the probability that it is not a
parallelogram ?
Again, keep pupils on alert for total – the opposite of what we are looking for.
Quadrilateral which is not a parallelogram = 18-8
P( Quadrilateral which is not a parallelogram) =
or
18  8 5

18 9
P(Success) = 1- P(Failure)
Quadrilateral which is
not a parallelogram
8
P=1 
18
Parallelogram
10 5
P=
=
18 9
94
© MSS 2007
2004 Question 6 (c)
1
2
3
4
5
6
77
78
95
© MSS 2007
1
2
3
4
5
6
77
78
We want to pick 4 cards all different in colour
7
Number of Favourable Outcomes
Probability =
=
Favourable Outcomes =
Total Outcomes
Total Number Outcomes
=
Number of Favourable Outcomes
Probability =
Total Number Outcomes
 2   2   2   2  = 16
        
1 1 1 1
8
= 70
 
4
16 3


70 35
96
© MSS 2007
1
2
3
4
5
77
6
78
We want to pick 4 cards 2 odd and 2 even
Colour does not matter here.
O
O
E
Favourable Outcomes = From 4 odd pick 2 and
from 4 even pick 2
Number of Favourable Outcomes
Probability =
Total Number Outcomes
E
4 4
  
2 2
36 18


70 35
= 36
97
© MSS 2007
1
Example
2
O
3
4
O
5
E
O
O O O
E
E
E
6
77
78
E
E
4

If we pick any the odd cards first there are   ways of doing this.
2
From the cards that are left we have one option: ie. to pick the even blue and the even yellow
4
 
Number of Favourable Outcomes
6
3
Probability =
  82  

98 © MSS 2007
Total Number Outcomes
 
70 35
 
4
2004 Question 7
(i)
The first runner can go in any one of 8 lanes.
The second runner can now occupy any of the 7 remaining lanes.
And so on until the eight runner has only one lane left
99
Possibilities
= 8  7  6  5  4  3  2  1 = 8!
© MSS 2007
2004 Question 7
(ii)
The first runner can go in any one of 8 lanes.
The second runner can now occupy any of the 7 remaining lanes.
And so on until the fifth runner has 4 lanes left
Possibilities
=87654=
8
100
© MSS 2007
P5  8.7.6.5.4
Q.7
These students do not study Biology = 21
Total number of students = 56
P (Not Biology) = 21
56
101
© MSS 2007
This gives a total of 26 who study
at least two subjects
These 4 do not study Biology
P=
4
26
102
© MSS 2007
Total of 28 study Physics
Number of Favourable Outcomes
Probability =
Favourable =
Total =
Total Number Outcomes
28 Choose 2
 
 28
2
56 Choose 2 
562
Probability =
 282  378  27
562  1540 110
103
© MSS 2007
104
© MSS 2007
2006 Question 6 (a)
(i)
n  11
 3   165
2) How many am I choosing/picking/selecting ? r  
1) How many are available ?
(ii)
Number of Favourable Outcomes 
Probability =
Total Number Outcomes
5
 
 3
 11
 
3
105

10
2

165 33
© MSS 2007
2006 Question 6 (c)
No. of favourble outcomes =
       
No. of possible outcomes =
301 301 301 301 301 301 301 
(i)
P
30  1  1  1  1  1  1
1
1 1 1 1 1 1
No. of favourble outcomes
30  1  1  1  1  1  1
30
1


7 
6
No. of possible outcomes
30  30  30  30  30  30  30  30   30 
106
© MSS 2007
2006 Question 6 (c)
No. of favourble outcomes =
       
No. of possible outcomes =
       
(ii)
P
30  29  28  27  26  25  24
1
1
1
1
1
1
1
30  30  30  30  30  30  30
1
1
1
1
1
1
1
No. of favourble outcomes 30  29  28  27  26  25  24 2639


 0  469
No. of possible outcomes
30  30  30  30  30  30  30 5625
107
© MSS 2007
2006 Question 6 (c)
No. of unfavourble outcomes =
No. of possible outcomes =
No. of favourble outcomes =
       
30  29  28  27  26  25  24
1
1
1
1
1
1
1
307
307 
       
30  29  28  27  26  25  24
1
1
1
1
1
1
1
No. of favourble outcomes  30   30  29  28  27  26  25  24 2986
P


 0  53
7
No. of possible outcomes
5625
 30 
7
(iii)
 0 5
or
Probability = 1- P( all seven have the same birthday)
1
30  29  28  27  26  25  24
2639

1

 0  5309  0  5
7
30
5625
108
© MSS 2007
2006 Question 7 (a)
(i)
(ii)
10
1
10
10
10
10
10
10
10
5
 105
 1 103  5
109
© MSS 2007
2006 Question 7 (b)
 35 
(i)    324632
5 
 5   30 
(ii)      150
 4  1 
(iii)  5   30 
      4350
 3  2 
(iv)
P(matching at least three numbers)
= P( match three or match four or match five)

43501501
4501

0138649  0014

423632
423632
110
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The
Odds
Real
Lotto
Source www.lotto.ie
Odds on winning
111
© MSS 2007
Before the Lotto Draw
From 45 numbers you choose 6 numbers
Lotto Draw
B
A
C
E
D
X
Bonus
F
38 balls remain
6 numbers
112
© MSS 2007
Jackpot – Match 6
B
A
C
E
D
X
Bonus
F
38 balls remain
All 6 match
N/A
N/A
6
No. of favourable =    1
6
total possible
 45 
    8145060
 6
6
 
No. of favourble  6 
1
P


45
total possible
  8145060
6 
 
113
© MSS 2007
Match 5 + Bonus
B
A
C
E
D
X
Bonus
F
38 balls remain
From 6 you match 5
Yes
N/A
 6  1
favourable =       6
 5  1
6
 5 1
favourble
6
1
P
 


total possible
 45  8145060 1357510
6 
 
114
© MSS 2007
Match 5
B
A
C
E
D
X
Bonus
F
38 balls remain
From 6 you match 5
N/A
From 38
choose 1
 38 
 1   38
 
 6
 5 = 6
 
 6   38 
favourable =       228
 5  1 
favourble
228
1
1
1
P




8145060
total possible 8145060
35723.94 35724
228
115
Trick
© MSS 2007
Match 4 + Bonus
B
A
C
E
D
X
Bonus
F
38 balls remain
From 6 you match 4
Yes
 6
 4   15
 
 1
   1
 1
From 38
choose 1
 38 
 1   38
 
favourable = 15  1  38  570
favourble
570
1
1
1
P




8145060
total possible 8145060
14289.57 14290
570
116
© MSS 2007
Match 4
B
A
C
E
D
X
Bonus
F
38 balls remain
From 6 you match 4
N/A
From 38
choose 2
 38 
 2   703
 
 6
 4   15
 
favourable = 15  703  10545
favourble
10545
1
1
1
P




8145060
total possible 8145060
772.409 772
10545
117
© MSS 2007
Match 3 + Bonus
B
A
C
E
D
X
Bonus
F
38 balls remain
From 6 you match 3
 6
 3   20
 
Yes
1
  1
1
From 38
choose 2
 38 
 2   703
 
favourable = 20  1  703  14060
favourble
14060
1
1
1
P




8145060
total possible 8145060
579.30 579
14060
118
© MSS 2007
And finally back to the Socks Problem
Four distinguishable pairs of socks are in a
drawer. If two socks are selected at random
find the probability that a pair is chosen.
No of favourable outcomes =
(There are 4 pairs in the drawer)
Total Number of Outcomes =
(From 8 objects choose 2)
Number of Favourable Outcomes
Probability =
Total Number Outcomes
 4
 4
1 
8 
   28
 2
4
1


28 7
119
© MSS 2007
1
1. 2004 Ordinary Level Q6 (c)
2
2. 2003 Higher Level Q6 (c)
120
© MSS 2007
3
3. 2006 Ordinary Level Q6 (c)
4
4. 2003 Foundation Level Q6 (c)
121
© MSS 2007
5
5. 2005 Higher Level Q6 (c)
6
6. 2006 Foundation Level Q6 (c)
122
© MSS 2007
123
© MSS 2007