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Nguyen An Khuong, Huynh Tuong Nguyen Contents Randomness Sampling with R Probability Probability Rules Probability with R Discrete RVs Some Discrete Probability Models Geometric Model Bi

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Nguyen An Khuong, Huynh Tuong Nguyen

Contents Randomness Sampling with R Probability Probability Rules Probability with R Discrete RVs Some Discrete Probability Models Geometric Model Binomial Model The built-in distributions in R Densities Cdf Quantiles Random numbers References

Chapter 7

Discrete Probability with R

Discrete Structures for Computer Science (CO1007) on

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5 Probability calculations and combinatorics with R

6 Discrete Random variables

7 Some Discrete Probability Models

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Motivations

• Gambling

• Real life problems

• Computer Science: cryptology, coding theory, algorithmic

complexity,

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Randomness

Which of these arerandom phenomena?

• The number you receive when rolling afair dice

• The sequence for lottery special prize (by law!)

• Your blood type (No!)

• You met the red light on the way to school

• The traffic light isnotrandom It has timer

• The pattern ofyour ridingis random

So what is special about randomness?

In thelong run, they are predictable and haverelative frequency

(fraction of times that the event occurs over and over and over)

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Randomness in Statistics

• Randomness and probability: central to statistics

• Empirical fact: Most experiments and investigations are not

perfectly reproducible

• The degree of irreproducibility may vary:

• Some experiments in physics may yield data that are accurate

to many decimal places,

• whereas data on biological systems are typically much less

reliable

• View of data as something coming from a statistical

distribution: vital to understanding statistical methods

• We outline the basic ideas of probability and the functions

that R has for random sampling and handling of theoretical

distributions

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Random Numbers with R

• Much of the earliest work in probability theory was about

games and gambling issues, based on symmetry

considerations

• The basic notion then is that of a random sample: dealing

from a well-shuffled pack of cards or picking numbered balls

from a well-stirred urn

• In R, we can simulate these situations with the sample

function

• If we want to pick five numbers at random from the set

1 : 40, then you can write

> sample(1:40,5)

[1] 4 30 28 40 13

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Sample function

• The first argument(x)is a vector of values to be sampled

• The second (size)is the sample size

• Actually, sample(40, 5)would suffice since a single number is

interpreted to represent the length of a sequence of integers

• Notice that the default behavior ofsampleis sampling

without replacement

• That is, the samples will not contain the same number twice,

and size obviously cannot be bigger than the length of the

vector to be sampled

• If we want sampling with replacement, then we need to add

the argument replace = TRUE

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Sampling with replacement

• Sampling with replacement is suitable for modelling coin

tosses or throws of a die

• So, for instance, to simulate 10 coin tosses we could write

> sample(c("H","T"), 10, replace=T)

[1] "T" "T" "T" "T" "T" "H" "H" "T" "H" "T"

• In fair coin-tossing, the probability of heads should equal the

probability of tails, but the idea of a random event is not

restricted to symmetric cases

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Data with nonequal probabilities

• You can simulate data with nonequal probabilities for the

outcomes (say, a90% chance of success) by using theprob

argument to sample, as in

> sample(c("succ", "fail"), 10, replace=T,

prob=c(0.9, 0.1))

[1] "succ" "succ" "succ" "succ" "succ"

"fail" "succ" "succ" "succ" "fail"

• This may not be the best way to generate such a sample,

though See the later discussion of the binomial distribution

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Terminology

• Experiment/trial(thí

nghiệm (ngẫu nhiên)/phép

thử ): a procedure that yields

one of a given set of possible

• Event(sự kiện): a subset of sample space

• You see Head after an experiment {Head} is an event

• {1, 3, 5}

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Example

Experiment: Rolling two dice What is the sample space?

Answer:It depends on what we’re going to ask!

• The total number?

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The Law of Large Numbers (LLN)

Definition

The Law of Large Numbers (Luật số lớn) states that thelong-run

relative frequencyof repeated independent events gets closer and

closer to thetruerelative frequency as the number of trials

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Be Careful!

Don’t misunderstand the Law of Large Numbers (LLN) It can

lead to money lost and poor business decisions

Example

I had 8 children, all of them are girls Thanks to LLN (!?), there

are high possibility that the next one will be a boy

(Overpopulation!!!)

Example

I’m playing Bầu cua tôm cá, the fish has not appeared in recent 5

games, it will be more likely to be fish next game Thus, I bet all

my money in fish (Sorry, you lose!)

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Probability

Definition

Theprobability(xác suất) of an eventE of a finite nonempty

sample space ofequally likely outcomesΩis:

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Examples

Example (1)

What is the probability of getting a Head when tossing a coin?

Answer:

• There are|Ω| = 2possible outcomes

• Getting a Head is|E| = 1outcome, so

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Examples

Example (3)

We toss a coin 6 times What is probability of H in 6th toss, if all

the previous 5 are T?

Answer:

Don’t be silly! Still 1/2

Example (4)

Which is more likely:

• Rolling an 8 when 2 dice are rolled?

• Rolling an 8 when 3 dice are rolled?

Answer:

Two dice:5/36 ≈ 0.139

Three dice:21/216 ≈ 0.097

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Formal Probability

Rule 1

A probability is a numberbetween 0 and 1

0 ≤ p(E) ≤ 1

Rule 2: Something has to happen rule

The probability of the set of all possible outcomes of a trialmust

be 1

p(Ω) = 1

Rule 3: Complement Rule

The probability of an event occurring is 1 minus the probability

that it doesn’t occur

p(E) = 1 − p(E)

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Example (Birthday Problem)

Given a group of n < 365students We’ll ignore leap years and

assume that all birthdays are equally likely

i) If we pick a specific day (say December 7th), then what is the

chance that at least one student was born on that day?

ii) What is the probability that at least one student has the

same birthday as any other student?

Answer i)

• The sample space is the set of all365n possible choices of

birthdays fornindividuals

• p1(n) = P (At least one student was born on December 7th)

= 1 − P (No students were born on December 7th)

= 1 −364365nn

• We havep1(30) ≈ 7.9%, and p2(91) ≈ 21.8%

• In order for the probability of at least one other person to

share your birthday to exceed50%,we neednlarge enough

that

p1(n) = 1 −364n > 0.5,orn > 253

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Birthday Problem (cont’d)

Answer ii)

• p2(n) = P (At least 1 same birthday)

= 1 − P (No same birthdays)

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Generalization and Variation of Birthday Problem

• More generally, suppose we haveN objects, whereN is large

There arerpeople, and each chooses an object

• Then, similarly to above approximation,

• If there areN possibilities and we have a list of length√N,

then there is a good chance of a match:≈ 40%

• If we want to increase the chance of a match, we can make a

list of length of a constant times√N

• As a variation, suppose there areN objects and there are

two groups ofrpeople Each person from each group selects

an object What is the probability that someone from the first

group choose the same object as someone from the second

group?

• P (there is a match between two groups) = 1 − e−r2N.(Rather

difficult!)

• Eg If we takeN = 365andr = 30, then

P (there is a match between two groups) = 1 − e−302/365=

0.915

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A birthday attack on discrete logarithm

• We want to solveαx≡ β (mod p)

• Make two lists, both of length around√p:

• 1st list:αk (mod p)for randomk

• 2nd list:βα−h(mod p)for randomh

• There is a good chance that there is a match:

αk≡ βα−h (mod p)

• Hence, x = h + k

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Formal Probability

General Addition Rule

p(E1∪ E2) = p(E1) + p(E2) − p(E1∩ E2)

• IfE1∩ E2= ∅: They aredisjoint, which means they can’t

occur together

• then,p(E1∪ E2) = p(E1) + p(E2)

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Example

Example (1)

If you choose a number between 1 and 100, what is the probability

that it is divisible by either 2 or 5?

There are a survey that about 45%of VN population hasType O

blood,40% type A,11% type Band the resttype AB What is the

probability that a blood donor has Type A or Type B?

Short Answer:

40% + 11% = 51%

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Contents Randomness Sampling with R Probability Probability Rules Probability with R Discrete RVs Some Discrete Probability Models Geometric Model Binomial Model The built-in distributions in R Densities Cdf Quantiles Random numbers ReferencesConditional Probability (Xác suất có điều kiện)

• “Knowledge” changes probabilities

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Example

What is the probability of drawing a red card and then another red

cardwithout replacement(không hoàn lại )?

Solution

E: the event of drawing the first red card

F: the event of drawing the second red card

p(E) = 26/52 = 1/2

p(F | E) = 25/51

So the event of drawing a red card and then another red card is

p(E ∩ F ) = p(E) × p(F | E) = 1/2 × 25/51 = 25/102

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• Example:p(“Head”|“It’s raining outside”) =p(“Head”)

• IfE andF are independent

p(E ∩ F ) = p(E) × p(F )

Disjoint 6=Independence

Disjoint events cannot be independent They have no outcomes in

common, so knowing that one occurred means the other did not

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Probability of sampling without replacement

• Let us return to the case of sampling without replacement,

specificallysample(1 : 40, 5)

• The probability of obtaining a given number as the first one of

the sample should be1/40,the next one1/39,and so forth

• The probability of a given sample should then be

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Probability of a set

• The above probability is the probability of getting given

numbers in a given order

• If this were a Lotto-like game, then you would rather be

interested in the probability of guessing a givensetof five

numbers correctly

• Thus you need also to include the cases that give the same

numbers in a different order

• Since obviously the probability of each such case is going to

be the same, all we need to do is to figure out how many

such cases there are and multiply by that

• There are five possibilities for the first number, and for each

of these there are four possibilities for the second, and so

forth;

• that is, the number is5 × 4 × 3 × 2 × 1

• This number is also written as 5!(5factorial)

• So the probability of a “winning Lotto coupon” would be

> prod(5:1)/prod(40:36)

[1] 1.519738e-06

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The choose function

• There is another way of arriving at the same result

• Notice that since the actual set of numbers is immaterial, all

sets of five numbers must have the same probability

• So all we need to do is to calculate the number of ways to

choose 5 numbers out of40

• This is denoted by 405 = 40!

5! · 35! = 658008.

• In R, the choosefunction can be used to calculate this

number, and the probability is thus

> 1/choose(40,5)

[1] 1.519738e-06

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Random Variables

• When looking at independent replications of a binary

experiment, we would not usually be interested in whether

each case is a success or a failure but rather in the total

number of successes (or failures)

• Obviously, this number is random since it depends on the

individual random outcomes,

• and it is consequently called a random variable

• In this case it is a discrete-valued random variable that can

take values0, 1, , n,where nis the number of replications

• A random variable X has a probability distribution that can

be described using point probabilities fX(x) = p(X = x),

• or the cumulative distribution functionF (x) = p(X ≤ x)

• Expected value (giá trị kỳ vọng):

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Expected Value: An Example

An insurance company charges $50 a year Can company make a

profit? Assuming that it made a research on 1000 people and have

X: amount of payment, is adiscrete random variable(biến ngẫu

nhiên rời rạc) The companyexpectsthat they have to pay each

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