STAT1005 Introduction to Probability and Data
Analysis
Dr Vasudevan Mangalam
Mathematics and Statistics
Curtin University
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Outline
Introduction to Probability and Statistics
Probability
Statistics
Basic Ideas of Probability
The Statistical Package
About STAT1005
Sample Spaces, Outcomes and Events
Naı̈ve Definition of Probability
Counting Methods
Permutations
Combinations
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Why Probability?
I If mathematics is the logic of certainty, probability is the logic
of uncertainty and chance.
I Everyone has a some intuitive notion of using probability to
express uncertainty:
I probability of rain tomorrow
I chance that your team will win next weekend
I likelihood that someone is guilty (Sally Clarke, OJ Simpson)
I What are your intuitive notions of probability?
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Probability and Quantum Mechanics
I Electron orbits are representations
of the space in which an electron is
likely to be found
I The result of wave functions,
where the amplitude of the wave
function is interpreted as the
probability of finding an electron at
any point in space
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Probability in Genetics
I Fundamental laws of genetics are
stated as probability laws, i.e.,
what is the likelihood of a
particular trait – eye colour, for
example – being passed on from
one generation to the next?
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Probability in Weather and Climate
I Forecasts are computed and
expressed in terms of probability.
I What is the probability of rain
tomorrow?
I Will extreme rainfall events become
more likely in the future, and how
extreme will they be?
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The Birthday Problem
I There are k people in a room. Assume each person’s birthday
is equally likely to be any of the 365 days of the year (we
exclude February 29).
I Assume that people’s birthdays are independent.
I Also assume there are no twins in the room.
I What is the probability that two or more people in the group
have the same birthday?
I On a different note, are birthdays equiprobable? Read the
following article:
https://www.abc.net.au/news/2017-12-13/australias-mostand-least-popular-birthdays-revealed/9241978
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Are Birthdays Equiprobable?
ABC News Report
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3.1 A Brief History of Probability
I A gambler’s dispute in 1654 led to
the creation of a mathematical
theory of probability by two French
mathematicians, Blaise Pascal and
Pierre de Fermat.
I Antoine Gombaud, (AKA Chevalier
de Méré) a French noble man and
a writer with an interest in gaming
and gambling questions, called
Pascal’s attention to an apparent
misconception concerning a
popular dice game.
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A Game of Dice
I The game consisted of throwing a pair of dice 24 times. The
problem was to decide whether or not to bet even money on
the occurrence of at least one double six during the 24
throws. A seemingly well-established gambling rule suggests
that betting even money on a double six in 24 throws would
be profitable. But de Méré’s own calculations showed that
this is not so.
I This problem led to an exchange of letters between Pascal
and Fermat in which the fundamental principles of probability
were formulated for the first time.
I Questions:
I What is the probability of getting at least one six in 4 throws
of a die?
I What is the probability of getting at least one double-six in 24
throws of a pair of dice?
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What is Probability Used for?
I Epidemiology: The spread of virus among humans
I Agriculture: Crop trials, plant genetics
I Meteorology: Weather prediction
I Sports strategies: Probabilities are used for planning and
gaining competitive advantage.
I Insurance options: Risk assessment
I Games and recreational activities: What are the chances of
getting a full house in a poker hand?
I Economics and finance: Performance of stock market
And many more. . .
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Probability and Statistics
I Probability focuses on a systematic study of randomness and
uncertainty. In any situation where one of several outcomes
may occur, probability provides a method for quantifying the
chance or likelihood associated with different outcomes.
I In inferential statistics, we often ask the question “could this
have occurred by chance alone, or is there something really
going on here?”
I Our intuition is often not very helpful in assessing randomness
and uncertainty, so probability theory provides us with a
principled framework for answering questions such as the one
above.
I Probability theory provides us with probability models for
describing different ‘types of randomness’.
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What is Statistics?
I Statistics is a powerful tool for researchers to
I discover new things
I prevent mistakes
I Example: 70% of graduates wish they had learned more
statistics.
I Scientists say they have discovered the 25 genes responsible
for 7 common serious diseases. How did they do it?
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What is Statistics?
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What is Statistics?
I From data we draw conclusions.
I We need statistics to draw appropriate conclusions from data.
I Statistics is the science of drawing conclusions from data.
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Hubble’s Law
I Edwin Hubble discovered the relation between distances to
the galaxies and the speed at which they move away from us
(and from each other).
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Melanoma and Latitude: Sun Exposure Increases Cancer
I Henry Oliver Lancaster was the Foundation Chair of
Mathematical Statistics at the University of Sydney.
I Lancaster was the first person to quantify the relation
between melanoma and latitude.
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Henry Oliver Lancaster: Rubella Causes Infant Deafness
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What is Statistics Used for?
I Medical research: Drug trials, public health, genetics
I Agriculture: Crop trials, plant genetics
I Forensic science: DNA evidence
I Astronomy: Galaxy distribution, exoplanets
I Geology
I Internet security
I Environmental science: Ecology, Climate
I Many articles in scientific journals use statistics to analyze
data and draw conclusions.
Climate is a statistical description of weather conditions and
their variations, including both averages and extremes Australian Academy of Science The Science of Climate
Change, August 2010
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Statistics in Science
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Statistics in Science
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The Sally Clarke Case
I Applying probabilistic ideas with a poor understanding of the
underlying concepts can result in tragic errors.
I Sally Clarke was a young British solicitor who became the
victim of a famous case of incorrect statistics. She was
wrongly convicted of the murder of two of her children.
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The Sally Clarke Case
I Her first son died suddenly within a few weeks of birth in 1996.
I When her second son also died suddenly in 1998, she was
arrested for double murder.
I A key witness in her trial was paediatrician Professor Sir Roy
Meadow, who testified that the chance of two children from
an affluent family suffering sudden infant death syndrome
(SIDS) was 1 in 73 million.
I This probability calculation was based on the statistics of
previous SIDS deaths and the assumption that the chances of
a single cot death is 1 in 8543 and that such deaths in a
family are independent.
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The Sally Clarke Case
I You can read about the multiple errors committed in the
probability calculation and the conclusion drawn from it in the
‘Statistical evidence’ section of
https://en.wikipedia.org/wiki/Sally_Clark
I In 2001, the Royal Statistical Society issued a public statement
expressing its concern on the misuse of statistics in court.
I They calculated the odds of a second cot death in a family as
about 200 to 1.
I Sally Clark was convicted in November 1999.
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The Sally Clarke Case
I However, in an appeal in January 2003 it was revealed that
the prosecutor’s pathologist had failed to disclose
microbiological reports that suggested one of her sons had
died of natural causes.
I Clarke’s conviction was overturned and she was released from
prison having served more than three years of her sentence,
but she never recovered from the psychological damage and
the mistreatment she received in prison.
I Sally Clarke died of alcohol poisoning in 2007.
I The case was widely considered as one of the great
miscarriages of justice in modern British legal history.
I Hundreds of other cases were reviewed subsequently, and two
other women convicted of murdering their children had their
convictions overturned.
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Statistics to the Rescue
I Analysing data is not a mindless process.
I Great scientific discoveries have been made by analysing data.
I Terrible mistakes have been committed by wrongly
interpreting data.
I Statistics is the guiding light!
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Basic Ideas of Probability
I Chance behaviour is unpredictable in the short run, but has a
regular and predictable pattern in the long run.
I We call a phenomenon random if individual outcomes are
uncertain but there is nonetheless a regular distribution of
outcomes in a large number of repetitions.
I The probability of any outcome of a chance process is the
proportion of times the outcome would occur in a very long
series of repetitions.
I Mathematical probability is an idealization based on imagining
what would happen in an indefinitely long series of trials.
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Law of Large Numbers in a Coin Toss Experiment
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Probability Models
I Descriptions of chance behaviour contain two parts: a list of
possible outcomes and a probability for each outcome.
I The sample space S of a random phenomenon is the set of all
possible outcomes.
I An event is an outcome or a set of outcomes of a random
phenomenon. That is, an event is a subset of the sample
space.
I A probability model is a mathematical description of a random
phenomenon consisting of two parts: a sample space S and a
way of assigning probabilities to events.
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Probability Rules
I Any probability is a number between 0 and 1.
I All possible outcomes together must have probability 1.
I If two events have no outcomes in common, the probability
that one or the other occurs is the sum of their individual
probabilities.
I The probability that an event does not occur is 1 minus the
probability that the event does occur.
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Probability Rules
Mathematically,
I Rule 1. The probability P (A) of any event A satisfies
0 ≤ P (A) ≤ 1.
I Rule 2. If S is the sample space in a probability model, then
P (S) = 1.
I Rule 3. If A and B are disjoint, P (A ∪ B) = P (A) + P (B).
This is the addition rule for disjoint events.
I Rule 4. For any event A, P (A does not occur) = 1–P (A).
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Finite and Discrete Probability Models
I One way to assign probabilities to events is to assign a
probability to every individual outcome, then add these
probabilities to find the probability of any event. This idea
works well when there are only a countable (finite or
countably infinite) number of outcomes.
I A finite probability model is one with a finite sample space.
I Discrete probability models include finite models as well as
sample spaces that are countably infinite. Countably infinite
sets are ones that are equivalent to the set of all positive
integers.
I To assign probabilities in a finite model, list the probabilities
of all the individual outcomes. These probabilities must be
numbers between 0 and 1 that add to exactly 1. The
probability of any event is the sum of the probabilities of the
outcomes making up the event.
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Continuous Probability Models
I Suppose we want to choose a number at random between 0
and 1, allowing any number between 0 and 1 as the outcome.
I We cannot assign probabilities to each individual value
because there is an infinite interval of possible values.
I A continuous probability model assigns probabilities as areas
under a density curve. The area under the curve and above
any range of values is the probability of an outcome in that
range.
I Normal distributions are examples of continuous probability
models.
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Random Variables
I A random variable is a numerical outcome of a random
phenomenon (such as random sampling).
I The probability distribution of a random variable X tells us
what values X can take and how to assign probabilities to
those values.
I Random variables that have a finite or countably infinite list
of possible outcomes are called discrete random variables.
I Random variables that can take on any value in an interval
are continuous random variables.
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The Statistical Package R
R is the software package we shall use
for probabilistic and statistical
computations.
> (x <- rnorm(9,3.2,1.6))
[1] 3.623632 4.100666 3.925909
[4] 4.342729 2.045807 2.883580
[7] 2.566033 3.890913 3.566257
> mean(x)
[1] 3.438392
> sd(x)
[1] 0.7711733
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The Statistical Package R
Why Use R?
I Free, open-source software
I Works on Windows, Mac, Linux. . .
I Standard platform for data analysis
I Relatively easy to learn
I Saves us a lot of work
I Many online resources available
I Thousands of extension packages
for specific tasks
I Used in many other units
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Example
> QuizMarks
[1] 13 14 10 14 13
[212] 4 7 7 11 0
> mean(QuizMarks)
[1] 13.48444
> sd(QuizMarks)
[1] 5.067489
> median(QuizMarks)
[1] 14
9 14 14 ...
6 8 0 5 ...
> hist(QuizMarks, xlab = "Marks",
main = "Histogram of AiM Quiz Marks",
col = "darkseagreen")
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STAT1005 Unit Objectives
I Introduce the theory of probability, random variables, discrete
and continuous distributions
I Introduce you to statistics as a way of thinking
I Provide you with some simple tools for analysing data
I Gain understanding of the role of statistical inference in
drawing conclusions about the world
I By the end of the unit, you should
I Have a good grounding that will allow you to take more
advanced units;
I Have a firm grasp of the fundamental principles of probability;
I Be on your way to becoming a statistically literate citizen who
can cast a critical eye over claims made in the media and
elsewhere.
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UniPass
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Blackboard
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Sample Spaces, Outcomes and Events
I The mathematical framework for probability is built around
sets.
I Imagine that an experiment is performed, resulting in one out
of a set of possible outcomes.
I An experiment is any action or process that generates
observations. Before the experiment is performed, it is
unknown which outcome will be the result; afterwards, the
result ‘crystallises’ into the actual outcome.
Definition 1
The sample space S of an experiment is the set of all possible
results of the experiment. Elements of the sample space are called
outcomes. An event A is a subset of the sample space S, and we
say that A occurred if the actual outcome is in A.
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Sample Spaces, Outcomes and Events
Example 1
I Toss a coin once: S = {H, T} = {Head, Tail}
I The number of customers to a bank on a randomly selected day:
S = {0, 1, 2, . . . }
I Measure someone’s systolic blood pressure: S = {x : x ≥ 0} = [0, ∞)
Remark 1
I In the first example, the sample space is finite.
I In the second example, the sample space is countably infinite.
I In the third example, the sample space is uncountably infinite.
I A set that has either a finite or a countably infinite number of elements
will be called countable.
I In the first two examples the sample space is said to be discrete
(countable), while in the last example S is said to be continuous
(interval, uncountably infinite, uncountable).
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Union, Intersection and Complement
An event is just a set of outcomes, so that relationships and results
from elementary set theory can be used to study events.
Definition 2
I The union of two events A and B, denoted by A ∪ B and
read “A or B”, is the event consisting of all outcomes that
are either in A or in B or in both events.
I The intersection of two events A and B, denoted by A ∩ B
and read “A and B”, is the event consisting of all outcomes
that are in both A and B.
I The complement of an event A, denoted by Ac or Ā, is the
set of all outcomes in S that are not contained in A.
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Union and Intersection
Intersection: A ∩ B is the set of
points that are both in A and in
B.
Union: A ∪ B is the set of points
that are in A or B or both.
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Complementary and Disjoint Events
The complement of an event A,
denoted A0 , A or Ac , is the set of
points which are not in A.
Definition 3
When A and B have no
outcomes in common, they are
said to be mutually exclusive or
disjoint events. (A ∩ B = ∅)
Exercises
1. What is the complement of Ac ?
2. What is the intersection of A and Ac ?
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De Morgan’s Laws
Example 2
I Two coins are tossed.
I Let A denote the event that the first toss yields a H.
(a) What is the complement of A?
S = {HH, HT, T H, T T } ,
A = {HH, HT } ,
Ac = {T H, T T }
(b) Write down an event that is disjoint with A.
Ac is disjoint with A. {T H} and {T T } are also disjoint with
A.
Remark 2
De Morgan’s laws:
For two events A, B ⊆ S:
(A ∪ B)c = Ac ∩ B c
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Example 3 (Coin flips)
A coin is flipped 10 times. Writing Heads as H and Tails as T , a possible
outcome is HHHT HHT T HT , and the sample space is the set of all
possible strings of length 10 of Hs and T s. We can encode H as 1 and T
as 0, so that an outcome is a sequence (s1 , . . . , s10 ) with sj ∈ {0, 1},
and the sample space is the set of all such sequences.
Now let’s look at some events:
1. Let A1 be the event that the first flip is Heads.
As a set, A1 = {(1, s2 , . . . , s10 ) : sj ∈ {0, 1} for 2 ≤ j ≤ 10}.
This is a subset of the sample space, so it is indeed an event;
saying that A1 occurs is the same as saying that the first flip
is Heads.
Similarly, let Aj be the event that the j th flip is Heads for
j = 2, 3, . . . , 10.
2. Let B be the event
that at least one flip was Heads.
S
As a set B = 10
A
j=1 j .
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3. Let C be the event
that all flips were Heads.
T
As a set C = 10
A
j=1 j .
4. Let D be the event that there were at least two consecutive
Heads.
S
T
As a set, D = 9j=1 (Aj Aj+1 )
Events can be described using words or in set notation.
Events and occurrences
sample space
s is a possible outcome
A is an event
A occured
something must happen
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S
s∈S
A⊆S
sactual ∈ A
sactual ∈ S
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New events from old events
A or B (inclusive)
A and B
not A
A or B, but not both
at least one of A1 , . . . , An
all of A1 , . . . , An
A∪B
A∩B
Ac
(A ∩ B c ) ∪ (Ac ∩ B)
A1 ∪ · · · ∪ An
A1 ∩ · · · ∩ An
Relationships between events
A implies B
A and B are mutually exclusive
A1 , . . . , An are a partition of S
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A⊆B
A∩B =∅
A1 ∪ · · · ∪ An = S,
Ai ∩ Aj = ∅ for i 6= j
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Naı̈ve Definition of Probability
The naı̈ve definition applies when every outcome in S is equally
likely to occur.
Definition 4 (Naı̈ve definition of probability)
Let A be an event for an experiment with a finite sample space S.
The naı̈ve probability of A is
P (A) =
number of outcomes favourable to A
|A|
=
|S|
total number of outcomes in S
Remark 3
Obviously, P (A) can only be defined in this way only if S is a finite
set.
Experiments in which each outcome is equally likely to occur are
also called Laplace experiments.
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Naı̈ve Definition of Probability
There are several important types of problems where the naı̈ve
definition is applicable:
I when there is symmetry in the problem that makes outcomes
equally likely. . .
I when the outcomes are equally likely by design. For example,
consider conducting a survey of n people in a population of
N people. . .
I when the naı̈ve definition serves as a useful null model. In this
setting, we assume that the naı̈ve definition applies just to see
what predictions it would yield, and then we can compare
observed data with predicted values to assess whether the
hypothesis of equally likely outcomes is tenable.
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Counting Methods
I Calculating the naı̈ve probability of an event A involves
counting the number of outcomes in A and in the sample
space S.
I Enumerating all outcomes is straightforward when the list is
small, but in many real instances, such lists are prohibitively
long. Exploiting general counting rules we can calculate
probabilities without listing all outcomes.
I In many instances, we can count the number of possibilities
using the multiplication rule.
I Consider a compound experiment consisting of two
sub-experiments such that the first has n1 outcomes and for
each possible outcome for the first, the second has n2
outcomes. Then the compound experiment has n1 × n2
possible outcomes.
I This rule is known as the fundamental principle of counting.
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An Example of the Multiplication Rule
I At your local cinema, you can buy ice
cream with a choice of Bacio, Hazelnut,
and Tiramisu flavours, and with wafer or
waffle cones. By the multiplication rule,
there are 2 × 3 = 6 choices, and you can
visualize the decision process with a tree
diagram.
I It doesn’t matter whether you choose the
cone first or the flavour first.
I As long as there is the same number of
choices of the second element for each
first element, the product rule is valid
even when the set of possible second
elements depends on the first, i.e. the
option of flavours are different for
different types of cones.
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General Multiplication Rule
I If a compound experiment consists of k sub-experiments such
that the ith sub-experiment has ni outcomes and for each
combination of outcomes
for the others, then the compound
Qk
experiment has i=1 ni = n1 × n2 × · · · nk possible outcomes.
I This can also be stated another way: Suppose that a set
consists of ordered collections of k-elements (k-tuples) such
that there are n1 possible choices for the first element, for
each choice of the first element, there are n2 possible choices
of the second element, etc., and for each possible choice of
the first k − 1 elements, there are nk choices of the kth
element. Then there are n1 × n2 × · · · nk possible k-tuples.
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Examples
I Suppose a set has n elements. How many possible subsets of
S exist?
I For each element of S, we can choose whether to include it or
exclude it in the construction of a subset. This gives us 2
possibilities for each of the n elements. So from the product
rule, it follows that there are 2n subsets.
I If we perform sampling with replacement of size k from a set
of n objects, there are nk possible outcomes.
I Using the 26 letters in the English alphabet, a 3-letter ‘word’
is generated at random. What is the probability that the word
contains neither of the letters ‘a’ and ‘e’ ?
I Here |S| = 263 . If A is the set of words that have no ‘a’ or ‘e’,
then |A| = 243 . Hence by the naı̈ve definition of probability,
3
12
243
.
P (A) = 3 =
26
13
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Examples
I If we perform sampling with replacement of size k from a set
of n objects, there are n × (n − 1) × · · · (n − k + 1) possible
outcomes if k ≤ n (and zero possibilities if k > n).
I In the ‘words’ example, what if we now specify that the words
cannot contain the same letter more than once?
I Now |S| = 26 × 25 × 24 and |A| = 24 × 23 × 22. Thus
P (A) =
24 × 23 × 22
26 × 25 × 24
I This leads us to the definition of permutations. Any ordered
sequence of k objects taken from a set of n distinct objects is
called a permutation of size k of the objects.
I We first introduce the factorial notation: n!, pronounced as ‘n
factorial’, is defined as n! = n × (n − 1) × · · · 2 × 1.
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Permutations
I The number of permutations of n objects in k places is
denoted by Pkn (or Pn,k ) and is given by
Pkn = n × (n − 1) × · · · (n − k + 1) =
n!
.
(n − k)!
I Ten markers are available for grading a test. The test has four
questions, and the professor wishes to select a different marker
to grade each question with only one marker per question. In
how many ways can markers be chosen to grade the test?
I Here n = 10 and k = 4, so the number of different ways
markers can be assigned is
P410 =
Dr Vasudevan Mangalam
10!
= 5040.
6!
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Permutations
I A deck of 52 cards is shuffled and 5 cards are dealt face up in
a row. Find the probability that four aces present in adjacent
position.
I The total number of arrangements is P552 = 311875200.
I The number of arrangements where the four aces are present
and are in adjacent position is 48 × 4! × 2! = 2304.
I Thus the required probability is
2304
= 0.000007388.
311875200
I How many permutations are there of the word STATISTICS?
I First we think of the repeated letters as distinct to get 10!.
I Then we divide it by the factors by which the repetitions
occur to get the number of permutations to be
10!
= 50400.
3!3!2!
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The Birthday Problem
I There are k people in a room. Assume each person’s birthday
is equally likely to be any of the 365 days of the year. We
assume that people’s birthdays are independent of each other.
What is the probability that two or more people in the group
have the same birthday?
I We can calculate the probability of the complement, that of
P 365
k
all birthdays being distinct. This is given by 365
k . Then the
probability that there is a clash is given by
k
P (k)
k
P (k)
1
0
22
.476
P (k) = 1 −
Pk365
365k
2
.003
23
.507
10
.117
40
.891
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3
.008
30
.706
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.411
50
.9704
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21
.444
80
.9999
The Birthday Problem
Thus we need only 23 people in the group to have a higher than
50% chance of there being a clash.
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Simulating the Birthday Problem
I One way to simulate the birthday problem using R is as
follows.
I Randomly generate birthdays for k = 50 people by sampling
from the numbers 1 to 365 with replacement.
I Find the frequency each day and record the maximum
frequency r.
I Program R to do this N = 10000 times, and calculate the
fraction of time that r ≥ 2.
I You will probably find that this fraction is close to
P (50) = 0.9704.
Dr Vasudevan Mangalam
STAT1005
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Combinations
I In permutations, order matters. In the ‘words’ example, ‘pqr’
is different from ‘rpq’.
I In many situations, counting the number of unordered subsets
of a specific size is required.
I Given a set of n distinct objects, any unordered subset of size
k of the objects is called a combination.
I The number of combinations of size k that can be formed
from n distinct objects will be written by
n
= Ckn = Cn,k .
k
I For given n and k, the number of combinations is smaller
than the number of permutations (unless k = 1, in which case
they are equal), because when order is disregarded, some of
the permutations correspond to the same combination.
Dr Vasudevan Mangalam
STAT1005
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Combinations
I In the set {A, B, C, D, E}, the number of permutations of
5!
= 60.
size 3 is (5−3)!
I Consider a single subset {A, B, C}. There are 3! = 6
permutations that corresponding to a single combination.
I Similarly, for any other combination of size 3 there are 6
permutations, each obtained by ordering the three objects.
I Thus, we’re overcounting by a factor of 6, and hence the
5!
× 3!1 = 10, i.e.,
number of combinations is (5−3)!
{A, B, C}
{A, B, D}
{A, B, E}
{A, C, D}
{A, C, E}
{A, D, E}
{B, C, D}
{B, C, E}
{B, D, E}
{C, D, E}
Dr Vasudevan Mangalam
STAT1005
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Combinations
I More generally, when there are n distinct objects, the number
of combinations of size k is given by
Pn
n!
n
= k =
k!
k!(n − k)!
k
I n0 = nn = 1 because there is only one way to choose a set
of all n elements or of no elements.
I n1 = n since there are n subsets of size 1.
I In how many different ways can 6 tosses of a coin yield 2
heads and 4tails?
Answer: 62 = 15.
I How many committees of two actuarial science students and
one data science student be formed from four actuarial
science students
and
three data science students?
Answer: 42 × 31 = 6 × 3 = 18.
Dr Vasudevan Mangalam
STAT1005
64/ 65
Combinations
I Eight politicians meet at a fund-raising dinner. How many
greetings can be exchanged if each politician shakes hands
with every other politician exactly once?
I Think of the politicians as being numbered from 1 to 8. A
handshake corresponds to an unordered sample
of size 2
8
chosen from those 8. So the answer is 2 = 28.
I If there are 20 points on a paper such that no three are
collinear,
1. how many line segments can one create with two of these
points as endpoints?
2. how many triangle can one form with three of these points as
vertices?
I The answers are
I Note that nk =
20
20
2 = 190 and 3 = 1140 respectively.
n
20
20
n−k . For instance, 18 = 2 = 190.
Dr Vasudevan Mangalam
STAT1005
65/ 65