report review chapter 4 5 6 1 basic probability concepts

CHAPTER 4: BASIC PROBABILITY4.1 Basic Probability Conceptsa.Events and sample spaces:- Events are subsets of the sample space, the set of all outcomes that produce a specificresult.- The

REPORT REVIEW CHAPTER 4,5,6

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CHAPTER 4: BASIC PROBABILITY

4.1 Basic Probability Concepts

a.Events and sample spaces:

- Events are subsets of the sample space, the set of all outcomes that produce a specific result

- The complement of an event A, noted by the symbol A', is the subset of outcomes that are not part of the event

- A set of events are mutually exclusive if they cannot occur at the same

- A set of events are collectively exhaustive if one of the events must occur

- A certain event is an event that is sure to occur such as “roll a value greater than 0” for rolling one fair die

- An impossible event is an event that has no chance of occurring, such as “roll a value greater than 6” for rolling one fair die

b.Types of Probability

-In apriori probability, the probability of an occurrence is based on having prior knowledge

of the outcomes that can occur

- In the empirical probability approach, the probabilities are based on observed data, not

on prior knowledge of how the outcomes can occur Surveys are often used to generate empirical probabilities

- Subjective probability, differs from the other two approaches because subjective probability differs from person to person

c.Summarizing Sample Spaces

- Sample space is all possible outcomes of an experiment

- The symbol (S) is used to for the sample space

d.Simple Probability

- Simple probability is the probability of occurrence of a simple event A, P(A) in which each outcome is equally likely to occur

PROBABILITY OF OCCURRENCE

Probability of occurrence = X/T

where X = number of outcomes in which the event occurs

T = total number of possible outcomes

EX: How can you determine the probability of selecting a household that planned to purchase

a large TV? Using the Table summary table, you determine the value of X as 250, the total

of the Planned-to-Purchase Yes row and determine the value of T as 1,000, the overall total of respondents located in the lower right corner cell of the table

Probability of occurrence =X/T

P(Planned to purchase) =Number who planned to purchase:Total number of households

=250/1,000 = 0.25

Thus, there is a 0.25 (or 25%) chance that a household planned to purchase a large TV e.Joint Probability

-Whereas simple probability refers to the probability of occurrence of simple events, joint probability refers to the probability of an occurrence involving two or more events f.Marginal Probability

- The marginal probability of an event consists of a set of joint probabilities You can determine the marginal probability of a particular event by using the concept of joint probability just discussed

MARGINAL PROBABILITY

P(A) = P(A and B1) + P(A and B2) + + P(A and Bk)

where B1, B2, , Bk are k mutually exclusive and collectively exhaustive events

EX:You can use Equation (4.2) to compute the marginal probability of “planned to purchase”

a large TV:

P(Planned to purchase) = (Planned to purchase and purchased)+ PP (Planned to purchaseand did not purchase)

=200/1,000 + 50/1,000 = 250/1,000= 0.25

You get the same result if you add the number of outcomes that make up the simple event

“planned to purchase.”

g.General Addition Rule

-The probability of A or B is equal to the probability of A plus the probability of B minus the probability of A and B

P(A or B) = P(A) + P(B) - P(A and B)

-The general addition rule consists of taking the probability of A and adding it to the probability of B and then subtracting the probability of the joint event A and B from this total because the joint event has already been included in computing both the probability of A and the probability of B

EX:One way to calculate the probability of “planned to purchase or actually purchased” is

P(Planned to purchase or actually purchased) = (Planned to purchase)+ (Actually purchased)P P

- P(Planned to purchase and actually purchased)

=250/1,000 + 300/1000-200/1000=350/1,000 = 0.35

4.2 Conditional Probability

A conditional probability is the probability of one event, given that another event has occurred

P(A|B) = P(A and B) / P(B)

P(B|A) = P(A and B) / P(A)

Where P(A and B) = joint probability of A and B

P(A) = marginal or simple probability of A

P(B) = marginal or simple probability of B

For example Suppose you have two decks of playing cards, one with red backs and one with blue backs The red deck contains 26 cards, including 13 spades and 13 hearts The blue deck contains 26 cards, including 13 clubs and 13 diamonds You draw a card at random from the red deck and see that it is a spade What is the probability that the next card you draw from the blue deck is a diamond?

To solve this problem, we need to use conditional probability Let A be the event that the first card drawn is a spade from the red deck, and let B be the event that the second card drawn

is a diamond from the blue deck Then we want to find P(B|A), the probability of B given that

A has occurred

Using the formula for conditional probability, we have:

P(B|A) = P(A and B) / P(A)

To find P(A and B), we can use the multiplication rule:

P(A and B) = P(A) * P(B|A)

We know that P(A) is the probability of drawing a spade from the red deck, which is 13/26 or 1/2 To find P(B|A), we need to calculate the probability of drawing a diamond from the blue deck, given that we have already drawn a spade from the red deck Since we have already drawn one card, there are 51 cards remaining in the two decks, of which 25 are diamonds Therefore:

P(B|A) = 25/51

Putting it all together, we have:

P(A and B) = P(A) * P(B|A) = (1/2) * (25/51) = 25/102

And:

P(B|A) = P(A and B) / P(A) = (25/102) / (1/2) = 25/51

So the probability of drawing a diamond from the blue deck, given that we have already drawn a spade from the red deck, is 25/51

4.3 Ethical Issues and Probability:

Ethical issues can arise when any statements related to probability are presented to the public

Unintended misinterpretations can occur with people who are not comfortable with numerical concepts

Advertising quoting probabilities can also be intentionally misleading

4.4 Bayes’ Theorem

Bayes’ theorem is an extension of what you previously learned about conditional probability Bayes’ theorem revises previously calculated probabilities using additional information and forms the basis for Bayesian analysis

- Bayes’ Theorem is used to revise previously calculated probabilities based on new information It is an extension of conditional probability

Ex:- One drilling company estimated a 40% chance of finding oil for their new well.

A detailed test has been scheduled for more information Previously, 60% of successful wells were examined in detail and 20% of failed wells were examined in detail Given that this well has been scheduled for a detailed inspection, what is the probability that

the well will be successful?

Set S = good success

U = failed well P(S) = 0.4 , P(U) = 0.6 (previous probability) Defining a detailed test event is EASY Conditional Probability:

P(D|S) = 0.6 P(D|U) = 0.2 The goal is to find P(S|D)

P(S|D) = 𝑃(𝐷|𝑆)𝑃(𝑆) =

𝑃(𝐷|𝑆)𝑃(𝑆)+𝑃(𝐷|𝑈)𝑃(𝑈)

(0.6)(0.4) (0.6)(0.4)+(0.2)(0.6) = 0.24+0.120.24 = 0 667

So, the modification success probability, provided that this well has been scheduled for

a detailed inspection, is 0.667

With detailed testing, the modified probability of a successful well has increased to 0.667 from the original estimate of 0.4

4.5 : Counting Rules

Rules for counting the number of possible outcomes

Counting Rule 1:

If any one of k different mutually exclusive and collectively exhaustive events can occur

on each of n trials, the number of possible outcomes is equal to

kn

Ex: If you roll a fair die 3 times then there are 63 = 216 possible outcomes

Counting Rule 2:

If there are k1 events on the first trial, k2 events on the second trial, … and kn events

on the nth trial, the number of possible outcomes is

(k )(k ) (k )1 2…

n

Ex: You want to go to a park, eat at a restaurant, and see a movie There are 3 parks, 4 restaurants, and 6 movie choices How many different possible combinations are there?

Answer: (3)(4)(6) = 72 different possibilities

Counting Rule 3:

The number of ways that n items can be arranged in order is

n! = (n)(n – 1) (1)…

Example:

You have five books to put on a bookshelf How many different ways can these books

be placed on the shelf?

Answer: 5! = (5)(4)(3)(2)(1) = 120 different possibilities

Counting Rule 4:

Permutations: The number of ways of arranging X objects selected from n objects in order is

You have five books and are going to put three on a bookshelf How many different ways can the books be ordered on the bookshelf?

Counting Rule 5:

Combinations: The number of ways of selecting X objects from n objects, irrespective of order, is

Example:

You have five books and are going to select three are to read How many different combinations are there, ignoring the order in which they are selected?

CHAP 5 :Discrete Probability Distributions

I The Probability Distribution for a Discrete Variable

- A probability distribution for a discrete variable is a mutually exclusive list of all possible numerical outcomes for that variable and a probability of occurrence associated with each outcome

1 Expected Value Of Discrete Variables

where:

E(X) = Expected value of the discrete variable X

xi = the ith outcome of X

P(X=xi) = Probability of the ith occurrence of X

i i

II Variance and Standard Deviation of a Discrete Variable

1 Variance of a discrete variable

2 Standard Deviation of a discrete variable

where:

E(X) = Expected value of the discrete variable X

xi = the ith outcome of X

P(X=xi) = Probability of the ith occurrence of X

Interruptions Per Day

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(x )i

Probability P(X = x )i

[xi– E(X)]2 [xi– E(X)]2P(X = x )i

1.96 (1.96)(0.35) = 0.686

0.16 (0.16)(0.25) = 0.040

0.36 (0.36)(0.20) = 0.072

3 0.10 (3 – 1.4) =

2.56 (2.56)(0.10) = 0.256

6.76 (6.76)(0.05) = 0.338

12.96 (12.96)(0.05) = 0.648

σ2= 2.04, σ = 1.4283 III Binomial Distribution

1 Properties of the Binomial Distribution

- A fixed number of observations, n

- e.g., 15 tosses of a coin; ten light bulbs taken from a warehouse

- Each observation is classified into one of two mutually exclusive & collectively exhaustive categories

- e.g., head or tail in each toss of a coin; defective or not defective light bulb

- The probability of being classified as the event of interest, π, is constant from observation to observation

- Probability of getting a tail is the same each time we toss the coin

- Since the two categories are mutually exclusive and collectively exhaustive, when the probability of the event of interest is π, the probability of the event of interest not occurring is 1 – π

- The value of any observation is independent of the value of any other observation

2 Combinations

where:

n! =(n)(n - 1)(n - 2) (2)(1)

x! = (X)(X - 1)(X - 2) (2)(1)

0! = 1 (by definition)

Ex: How many possible 3 scoop combinations could you create at an ice cream parlor if you have 31 flavors to select from and no flavor can be used more than once in the 3 scoops?

The total choices is n = 31, and we select X = 3

P(X=x|n,π) = probability that X = x events of interest, given n and π

x = number of “events of interest” in sample,

(x = 0, 1, 2, , n)

n = sample size (number of trials

or observations)

π = probability of “event of interest”

1 – π = probability of not having an

event of interest

Ex: Suppose in a casino a player bets $10 that there will be 6 heads out of 20 tosses, the probability of a heads-up is 0.5 That player wants to calculate the probability of this happening, so he uses the binomial distribution

The probability of getting exactly 6 heads in 20 tosses is 0.037 or 3.7% So this player made a bad bet

Binomial Distribution Characteristics

The Poisson distribution is a discrete probability function which means that the variable can only take specific values within a given list of numbers, possibly infinite

The Poisson distribution measures the number of times an event is likely to occur in the "x" period In other words, we can define it as the probability distribution that is the result of the Poisson experiment

The Poisson test is a statistical test that classifies testing into two categories, such as success or failure The Poisson distribution is a limiting process of the binomial distribution The Poisson random variable “x” determines the number of successes in the experiment This distribution occurs when there are events that do not occur as a result of certain outcomes The Poisson distribution is used under certain conditions They are:

● The number of trials “n” tends to be infinite

● The probability of success “p” tends to be zero

● np = 1 is finite

CHAPTER 6

The normal distribution

The normal distribution is a continuous probability distribution shaped like a symmetric bell curve It is widely used in statistics to model natural phenomena such as height, weight, IQ, test scores, and many others The normal distribution is characterized by two main parameters, the mean and the standard deviation

Any normal distribution (with any combination of mean and standard deviation) can be converted to a normal (Z) distribution

To calculate the normal probability, it is necessary to convert X units to Z units

The standard normal distribution (Z) has a mean of 0 and a standard deviation of 1

If X is normally distributed with a mean of 100 and a standard deviation of 50, the Z value for

X = 200 is:

This says that X = 200 is two standard deviations (2 increments of 50 units) above the mean

of 100

A normalized probability density function is a mathematical function that describes the probability distribution of a normally distributed random variable with a mean of 0 and a standard deviation of 1

The graph of the normal normalized probability density function is a bell curve that is symmetric around the mean (which is 0) and has a total area of one The curve peaks at the mean and decreases as the distance from the mean increases

The normalized probability density function is useful in many areas of statistics and probability theory because it allows for probability calculations involving normally distributed random variables It is also used in statistical inference, hypothesis testing, and calculation

of confidence intervals.x`

III TheUniformDistributions

- Evenly distributed (uniform distribution) isaprobabilitydistributionof equal probability (equal probabilities) for all possible outcomes of the randomvariable

-IV.TheExponentialDistributions(exponentialdistribution)

- Usedtomodelthetimeintervalbetweentwooccurrencesofanevent

- Forexample:timetimebetween2timeexportpresentlybelongtomoonreal

- Biến cố là tập con của không gian mẫu, tập hợp tất cả các kết cục tạo ra một kết quả xác định.vd:Tung một con xúc xắc là một phép thử, còn việc lật lên mặt nào đó là biến

cố -Không gian mẫu là tập hợp tất cả các kết quả có thể có của một biến vd: tung một con

xúc xắc có sáu mặt thì, không gian mẫu là tập {1, 2, 3, 4, 5, 6}

- Biến cố là tập con của không gian mẫu, tập hợp tất cả các kết cục tạo ra một kết quả xác định.vd:Tung một con xúc xắc là một phép thử, còn việc lật lên mặt nào đó là biến cố

-Không gian mẫu là tập hợp tất cả các kết quả có thể có của một biến vd: tung một con xúc xắc có sáu mặt thì, không gian mẫu là tập {1, 2, 3, 4, 5, 6}