Download Maths Leaving Cert Probability and Statistics Question 6,7 Paper 2

Maths
Leaving Cert
Probability and Statistics
Question 6,7 Paper 2
By Cillian Fahy
and Darron Higgins
Pape
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Page 1
r II Q.6 and 7
Discrete Mathematics, or (Probability & Statistics)
Most students tackle at least one of these questions if not both. It is very difficult to avoid this
area as it leaves you with no choice at all in paper II. Probability causes some students a great
amount of difficulty and it is important to thoroughly think through the question before
committing to an answer. It is very easy to muddled and to jump to incorrect conclusions.
This area is broken up into the following areas
1.
2.
3.
4.
Permutations and Combinations
Probability
Difference Equations
Statistics
These questions have followed a similar pattern over the past years. Question 6(b) traditionally
deals with Difference equations and Question 6(c) and 7(b) usually deals with Probability.
Question 7(c) involves Statistics and the part (a)’s are made up of Permutations and
Combinations. However, although this has been the pattern for the last number of years there is
no guarantee that it will be the same this year. You should know something about all of these
topics instead of avoiding one.
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Page 2
Paper II Q.6 and 7 ......................................................................................................................................... 1
Discrete Mathematics, or (Probability & Statistics) ..................................................................................... 2
1.
Permutations and Combinations ....................................................................................................... 4
1.1 ......................................................................................................................................................... 4
1.2
Factorial Notation: .................................................................................................................... 4
1.3 ......................................................................................................................................................... 4
1.4 ......................................................................................................................................................... 5
2
1.5
Combinations ............................................................................................................................ 6
1.6
Combinations from two different sets ....................................................................................... 8
Probability ....................................................................................................................................... 10
2.1 ....................................................................................................................................................... 10
2.2. ...................................................................................................................................................... 10
3.
2.3
Probability of an event not happening .................................................................................... 10
2.4
Sample Spaces......................................................................................................................... 11
2.5
Rules of Probability ................................................................................................................ 11
2.6
Probability Involving Permutations and Combinations .......................................................... 13
Difference Equations....................................................................................................................... 14
3.1
Introduction ............................................................................................................................. 14
3.2
Solving Difference Equations ................................................................................................. 15
3.3 ....................................................................................................................................................... 16
4.
Statistics .......................................................................................................................................... 18
4.1
Standard question. ................................................................................................................... 18
4.2 ....................................................................................................................................................... 19
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1.
Permutations and Combinations
Definition
Permutation:
Combination:
1.1
Must be in a set order (permutation locks, eg brief cases, bike locks)
Can be in any order.
Permutations deal with the number of ways you can arrange people or things.
E.g.
How many ways can 3 people sit in 3 chairs?
Ans.
Lets call the people A B and C. Therefore, we can have the following ways
ABC
ACB
BAC
BCA
CAB
CBA
Therefore there are 6 ways for 3 people to sit on 3 chairs.
However, it is easier to look at permutations with boxes.
Giving
1.2
3
2
1
options
options
option
3  2 1  6 ways
Factorial Notation:
3  2 1  3!
5  4  3 2 1  5!
 n!  n   n  1   n  2   ...... 3  2 1
1.3
After this all the questions are basically the same, although constraints can be added. Always start
with the constraint.
E.g.
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How many ways can the letters PROUD be arranged if a vowel must go first?
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Ans.
2  4  3  2 1  48 ways
Or identical values might be included.
E.g.
How many ways can we arrange the letters BOOK?
Ans. Note that BOOK and BOOK are the same, even though the O’s have moved.
4!
 12 ways
2! If we arrange n objects of which p of one type are alike and q of
Note:
another are alike is given by
n!
p ! q !
E.g.
How many ways can the letters MINIMUM be arranged?
Ans.
7!
 420 ways
2! 3!
E.g.
In how many arrangements do the 3 M’s come together?
Ans.
(ie MMM INIU) Take the 3 M’s as one letter. Therefore only 5 letters now, but 2 i’s.
5!
 60 ways
2!
1.4
Permutations of n objects taking r at a time
E.g.
In how many ways can 5 books be arranged taking 3 at a time?
Ans. Two methods to do this
(i)
Boxes
(ii)
P notation
5  4  3  60
5
P3 
{Note: P notation
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5!
 60
(5  3)!
n
Pr 
n!
}
(n  r )!
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Page 5
It is normally easier to use the box method unless directly asked a P question.
E.g.
Ans.
Find n if 7[n P3 ]  6[n1 P3 ]
7[ n P3 ]  6[ n 1 P3 ]
7  n   n  1   n  2    6  n  1  n   n  1 
7  n  2  6  n  1
7n  14  6n  6
n  20
The other main area of confusion is when asked to arrange people around a round table.
E.g.
In how many ways can 5 people be arranged around a round table?
1
Ans.
5
2
5
4
1
is the same as
4
3
3
2
To overcome this repetition we must fix one person in place and arrange the other 4 around the 5th
Therefore, 4! ways  24 ways
In general, n people can be arranged around a round table in  n  1! ways.
1.5
Combinations
In combinations order is not important. If asked to arrange the letters ABC in a permutation we
would have 3! Ways however in a combination question we would have only 1 way. ABC is the
same as BCA if order is not important.
A combination therefore deals with selecting or choosing a certain amount of objects from a
larger set and in how many ways this can be done. For example if there are 22 players in a squad
how many different teams of 11 can be selected for this pool?
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This is found using the n Cr button on your calculator.
22 C11  705432
n
n
Pr
n!
In general terms n choose r is Cr    
or
r!
 r  r !(n  r )!
n
Note:
1.
 7  7  6 5 4
 8  8 7
and   
 
 4  1 2  3  4
 2  1 2
2.
n  n 
 

r  nr
3.
n n
     1
n 0
6 6
 2  4
e.g.     
We will be able to use the calculator for most questions but should also know the above formula
and the notes.
E.g.
Ans.
E.g.
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How many different selections of 5 letters can be made from the letters of the word
CHEMISTRY?
(i)
How many if C must be in each?
(ii)
How many if C must be included and Y excluded?
9 9
   C5  126
5
(i)
8
   70 ways
 4
(ii)
7
   35 ways
 4
n
Solve for n 5  n P3   24   .
4
 
Note: can’t use our calculator for this.
Need to use above formula
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Ans.
5  n    n  1   n  2    24
 n    n  1   n  2    n  3
4!
5  n 3
n  8.
1.6
Combinations from two different sets
When combining 2 sets just work out the number of ways for the 1st set and multiply it by the
number of ways for the 2nd set.
E.g.
Ans.
In how many ways can a committee of 4 men and 3 women be chosen from 7 men and 5
women?
 7   5
      35 10  350 ways.
 4   3
Constraints can be added, but if thought out fully they should not cause too much difficulty.
Note: Remember
AND = Multiply
OR = Add
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E.g.
From a group of 5 boys and 8 girls. In how many ways can a team of 5 be chosen if it is
to contain;
(i)
No girls.
(ii)
No boys.
(iii)
At least 3 boys.
Ans.
(i)
No girls
(ii)
No boys 
(iii)
At least 3 boys

5
   1 ways
5
8
   56 ways
 5

BBB
And
GG
Or
BBBB And
G
Or
BBBBB
 5  8 
      10  28  280
 3  2 

 5  8
      5  8  40
 4 1

5
  1
5
Therefore, 280  40  1  321 ways.
Remember think through the question and don’t rush. These questions will not take as
long as an Algebra or Trigonometry question but the time saved writing should be spent
thinking. It is very easy to get confused and to make errors.
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2
Probability
2.1
The probability of an event happening P(E) 
E.g.
Number of favourable outcomes
Number of possible outcomes
1
6
(i)
P(getting a six) 
(ii)
P(getting an Ace) 
1
13
1
P(getting a Spade) 
4
(iii)
Note: You need to know that there are 52 cards in a deck, 4 different suits (Spades, Clubs,
Diamonds and Hearts), 4 of each type of card (i.e. 4 Aces etc…)
2.2.
Results will always be a fraction between 0 and 1. Where 0 can’t happen and 1 must happen.
2.3
Probability of an event not happening
P(E not happening) = 1 – P(E)
E.g.
What is the probability of rolling a dice and not getting a 6?
Ans.
P(not a 6) = 1 – P(getting a 6)
 1

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Note: From 2.2 we know
that all events must add up
to give 1.
1
6
5
6
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2.4
Sample Spaces
Sometimes a sample space can be helpful, especially if there are a lot of results to choose from.
It might seem like a lot of extra work but it will save you time in the end.
E.g.
If 2 dice are rolled, what is the probability that the two numbers add up to 7?
Ans.
2 dice gives a sample space of
1,1
2,1
3,1
4,1
5,1
6,1
1,2
2,2
3,2
4,2
5,2
6,2
P(total of 7) =
2.5
1,3
2,3
3,3
4,3
5,3
6,3
1,4
2,4
3,4
4,4
5,4
6,4
1,5
2,5
3,5
4,5
5,5
6,5
1,6
2,6
3,6
4,6
5,6
6,6
6 1

36 6
Rules of Probability
(i)
The Addition Rule {Independent events}
P(E or F) = P(E) + P(F)
E.g.
What is the probability of getting a 5 or a 6 when a dice is rolled?
Ans.
P(5 or 6) = P(5) + P(6)
1 1

6 6
1

3

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(ii)
The Addition Rule {Dependent events}
P(E or F) = P(E) + P(F) – P(E with F)
E.g.
What is the probability of picking a card from a deck and getting a Queen
or a Club?
Ans.
P(Q or Club) = P(Q) + P(Club) – P(Queen of Clubs)
4 13 1
 
52 52 52
4

13

(iii)
Note: The Queen of Clubs was
counted twice. Once as a Queen
and then again as a Club.
Therefore, we must take away one
of these repetitions.
The Multiplication Rule
P(E and F) = P(E)  P(F)
E.g.
Independent events
What is the probability of rolling a 6 and a 2?
Ans.
P(6 and 2)  P(getting a 6)  P(getting a 2)
1 1
 
6 6
1

36
Note you must be careful if the events are dependent on each other, for example when
cards are not replaced.
E.g.
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What is the probability of picking two cards from a deck and getting an Ace and a King?
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Page 12
Ans.
P(Ace and King)=
4 4
4
 
52 51 663
Note: Only 51 cards left in the deck
after the first was taken out and not
replaced.
(iv)
At least once
P(E occurring at least once) = 1 – P(E not happening at all)
If there are not too many events you might find it easier to simply count each one.
However, if there are a lot of events it is easier to use the above rule.
2.6
Probability Involving Permutations and Combinations
The more difficult questions will combine Permutations and Combinations and Probability.
Remember that probability looks for the number of favourable outcomes divided by the number
of possible outcomes. Use Permutations & Combinations to find the number of each outcome and
then use these with the Probability rules.
E.g.
In an examination a candidate is required to select any seven questions from ten.
(i)
In how many ways can this be done?
(ii)
How many of the selections contain the first and last questions?
Now calculate the probability that the candidate selects
(iii)
both the first and second questions
(iv)
at least one of the first two questions.
Ans.
(i)
(ii)
10 
   120
7
8
   56
 5
Note: Parts (i) and (ii) are not asking
for the probability of an event.
Remember don’t rush into the
question.
56
7

120 15
(iii)
P(both 1st and 2nd) =
(iv)
P(at least one of first two) = 1 – P(neither of first two)
8
7
Neither 1st or 2nd     8
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Therefore, P(neither 1st or 2nd) 
8
1

120 15
Then, P(at least one of first two) = 1 – P(neither of first two)
Gives
 1

3.
Difference Equations
3.1
Introduction
1
15
14
15
A common way to define a sequence of numbers is by a set of instructions which explains how to
carry on after a given start.
For example, consider the sequence given by a1  2 , an1  2an  1 .
The equation an1  2an  1 is called a difference equation and can be used to calculate any term
if we know the previous term. The statement a1  2 is called the initial condition. Without it we
could not generate any of the terms.
Thus, for the given sequence
a1  2
a2  2a1  1  2  2   1  3
a3  2a2  1  2  3  1  5
a4  2a3  1  2  5  1  9 etc…
Definition:
E.g. I
A difference equation is an equation which allows us to calculate a term of
a sequence from the preceding term or terms.
A sequence a1 , a2 , a3 ,...... is defined by a1  1 , and an 1  2 
1
, for n  1.
an
Find a5 .
Ans:
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a1  1
a2  2 
1
a1
1
 2  3
1
a3  2 
1
a2
1 7
 2 
3 3
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Page 14
a4  2 
1
a3
 2
3 17

7 7
a5  2 
1
a4
 2
7 41

17 17
One obvious drawback is that to find a particular term we must first find all of the preceding
terms. Not too bad if we want a5 but what if we were asked for a50 or even a500 .
3.2
Solving Difference Equations
You must be able to prove the Difference Equation theorem.
Prove: If  and  are the roots of the quadratic equation
px 2  qx  r  0 and
un  l n  m n for all n, then:
pun2  qun1  run  0 for all n.
Proof:
un  l n  m n
 un 1  l n 1  m n 1   l n   m n
 un  2  l n  2  m n  2   2l n   2 m n
If  and  are the roots of px 2  qx  r  0 , then
p 2  q  r  0 and p 2  q  r  0
Thus,
pun  2  qun 1  run  p  2l n   2l  n   q  l n   l  n   r  l n  l  n 
 p 2l n  p 2l  n  q l n  q  l  n  rl n  rl  n
 l n  p 2  q  r   m n  p  2  q   r 
 l n  0   m n  0 
0
 pun  2  qun 1  run  0
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Q.E.D.
3.3
However in practice there is a very straightforward method to solve difference equations.
To find the general solution of the difference equation
pun2  qun1  run  0
1.
Write down and solve the quadratic equation
px 2  qx  r  0
This is called the characteristic equation. Let the roots of this equation be a and b.
2.
Write down the general solution, un  la n  mbn .
l and m are found using the two initial conditions that are given in the question.
The best way to understand how all this ties together is to see an example.
E.g.
{2006, Paper II Q.6 (b)}
Solve the difference equation 6un 2  7un1  un  0 , where n  0 , given that u0  8
and u1  3 .
6un2  7un1  un  0
Ans.
6 x2  7 x  1  0
Gives
…the characteristic equation
 x 1 6 x 1  0
1
6
1
a  1 and b 
6
 x  1 or x 
Therefore,
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Putting into
un  la n  mbn
We get
n
1
un  l 1  m  
6
n
0
When u0  8
1
l 1  m    8
6
l m8
0
…(1)
1
When u1  3
1
1
l 1  m    3
6
6l  m  18
…(2)
Now we have two equation in terms of l and m. Solving simultaneously we get
l m8
…(1)
6l  m  18
…(2)
5l  10
l  2
m  6
Note: If we were then asked to find a
particular term, for example u10 , we
1
Now we have our general solution un  2 1  6  
6
n
simply substitute 10 in for the n in the
solution.
n
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4.
Statistics
Statistics is usually asked in Q.7 (c) and the most common question deals with finding the mean
and standard deviation of a list of abstract values.
4.1
Standard question.
E.g.
{2004, Paper II Q.7 (c)}
The mean of the real number p, q and r is x and the standard deviation is  .
(i)
Show that the mean of the four numbers p, q, r and x is also x
(ii)
The standard deviation of p, q, r and x is k.
Show that k :   3 : 2
Ans.
(i)
x
pqr
3
Note: Just do what you would
normally do if they were numbers., i.e.
for the mean add up all the terms and
divide by how many there are.
 p  q  r  3x
New mean is
Therefore,
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pqrx
4
3x  x
4
4x
x

4
For the standard deviation take your
time and don’t get lost among all of
the terms.
as asked.
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Page 18
 p  x  q  x  r  x
2

(ii)

2
3
 p  x  q  x  r  x
1
3
2
2
k
k
4.2
2
 p  x  q  x  r  x   x  x
2
And
2
2
2
2
4
1
2
 p  x  q  x  r  x
2
2
2
1 1
:
3 2
Therefore,
k : 
which gives
k :  3 : 2
as asked.
Even in trickier questions if you follow the mean and standard deviation formula you will be able
to answer any question that might be asked. However, don’t get lost in all of the terms. Stay
focused on the question and pause every few steps to make sure that you haven’t trayed from the
correct path.
E.g.
{2006, Paper II Q7 (c)}
The mean of the integers from n to n , inclusive is 0.
n  n  1
3
Show that the standard deviation is
Ans.
Note: It’s always a good idea to get an
expanded view of the numbers that
you have to deal with.
Dealing with the numbers
 n  ,  n  1 ,  n  2 ,.....,  2  ,  1 , 0  , 1 ,  2  ,.....,  n  2  ,  n 1 ,  n 
 n  x    n  1  x 
2
 
2

 
2
 

2

2
 n    n  1
 
2
Mocks.ie
2

2
2n  1
But told that x  0
2
 
 ...  1  x  0  x  1  x  ...  n  1  x  n  x
2
 ...   1   0   1  ...   n  1   n 
2n  1
2
2
2
Maths LC HL Probability and Series © mocks.ie
2
2
Note: It might be very
long but it is just the
same as any standard
deviation question.
Page 19
 
2
 2 
n 2   n 2  2n  1  ...  1   0   1  ...   n 2  2n  1  n 2
2n  1
2[12  22  32  ...  n 2 ]
2n  1
n

2   n  1 2n  1 
6

 2  
2n  1
2n  n  1
 2 
6
n  n  1
 2 
3
 
Mocks.ie
n  n  1
3
as asked.
Maths LC HL Probability and Series © mocks.ie
Note: It’s important to spot that the
sequence of numbers is repeated and
can be written on its own when
multiplied by 2.
Note: From Sequences and Series you
will spot that this is the sum of the
squares of the first n natural numbers.
Note: Even if you are not planning on
doing the Sequences and Series
question you must know the basics of
this topic as it can crop up anywhere.
Page 20