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Introduction to Probability & Statistics Code 104 ريض

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## Introduction to Probability & Statistics Code 104 ريض

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**Introduction to Probability & Statistics**Code 104 ريض CREDIT HOURS: 4 UNIT Lecture 3.0 hours/week Tutorial 1.0 hour/on every week This course introduces the students to the concept of probability and random variables Also deals with the conditional probability The course also looks into probability distribution of some discrete and continuous variables ..**Learning Objectives:**• Identify the role of probability in our life by learning the principal and rules of probability, conditional probability, independent events and Bayes Theorem. • Demonstrate understanding discrete and continuous probability distributions • References: • H.Hsu, 2nd Edition “Probability, Random Variables, Random Processes”, Schaum’s Outline Series, McGraw Hill, 2011 • Statistics and probability Dr Anies Ismaaeil Kinjo 2nd Edition King Saud University**Starting From Week 2**Sunday [10:00 – 12:00]-[1-2] (9-04-1435 h) – LEC. 3[ H] Monday 1:00 – 2:00 (10-04-1435 h) – Tutorial 1 [H]**9/4/1435 h Sunday**Lecture 1 Jan 2009**Q:Define the following**• Probability Probability refers to the study of randomness and uncertainty. • Random Variable numerical measurements or observations that have uncertain variability each time they are repeated are called random variables**.**• Distribution the term “distribution” refers to how probability is spread out. • Random Process any process whose possible results are known but actual results cannot be predicted with certainty in advance. • Experiment: process by which an observation or measurement is obtained (yield outcomes) Jan 2009**.**• Outcome : each possible result for a random process • Sample Space: Is the set of all possible outcomes of an experiment. denoted by S, • Event : Is any collection (subset) of outcomes contained in the sample space S. Jan 2009**Types of eventsQ: Mention and define types of events?**• simple An event is if it consists of exactly one outcome • Compound An event that consists of more than one outcome. • Null event: An event with no outcomes (= impossible event, empty set). Jan 2009**16/4/1435 h Sunday**Lecture 2 Sample Space and Events Jan 2009**Number of Elements in a Sample Space**Number of elements in = number of elements inn a sample space a sample space in one experiment Where: n is a number of time the experiment is repeated Examples: • Toss a coin • Once = 21 =2 elements • Twice = 22 =4 elements Jan 2009**.**• Three times = 23 = 8 elements • Roll a fair die: • Once: = 61= 6 elements • Twice: = 62= 36 elements Jan 2009**Sample Space and Events For Simple Events**1.Roll a die Once A six-sided die has this sample space Number of elements in the sample space = = 61= 6 elements • Sample space: S = {1, 2, 3, 4, 5, 6} • Simple events (or outcomes): E1= {1} , E2 = {2} , E3 = {3} E4 = {4} E5 = {5} E6 = {6}**Toss a coin once**Q: Show the sample space • Sample space:Number of elements in the sample space = = 21 = 2 Where: H = Head T = Tail • Simple events (or outcomes): E1( Head) = {H} E2(Tail} = {T} Jan 2009**0**• Sample Space and Events For Compound Events • Roll a die Q: Observe these events A: odd numbers A = {1, 3, 5} B: observe an even number B= {2, 4, 6} C: observe a number greater than or equal to 4 C= {4, 5, 6} D: observe a number less than or equal to 4 D = {4, 3, 2,1} Jan 2009**.**• The roll of a Two dice Q: Show the sample space when two unbiased dice were rolled or (if one die is rolled twice) Number of elements in the sample space = = 62 = 36 elements**.**Jan 2009**.**Examples that shows compound events of rolling two dice: Q: show the event of Sum of 6 A: = { (1,5),(5,1),(2,4),(4,2),(3,3)} Q: show the event that shows similar faces B= {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6,)} Jan 2009**Toss a coin twice Q:Show the sample space?The number of**elements in the sample space = 22 = 4 elements Show these events:A: Observe the number of heads A = { 0, 1, 2, }B : Having exactly two head B={(H,H)}C: Having at least one headC = {(H,H),(H,T),(T,H)} Jan 2009**.**D: Observe the number of tailsD = { 0, 1, 2, } E: Having exactly two tail E={(T,T)} F: Having at least one tail F = {(T,T),(H,T),(T,H)} • Toss a coin three timesQ: Show the sample space The number of elements in the sample space = 23 = 8 elements S={(H,H,H),(H,H,T),(H,T,H),(T,H,H),(H,T,T),(T,H,T),(T,T,H),(T,T,T)} Q: Observe the number of heads S = { 0, 1, 2,3 } Jan 2009**.**Q: Observe these events A: number of Tails A = { 0, 1, 2,3 } B: Having at least one head B={(T,H,T), (H,T,T),(T,T,H), (H,T,H),(T,H,H), (H,H,T),(H,H,H),} C: Show the sample space that represents the working state of a machine C = { working, fail } D: Show the sample space that represents the number of calls arriving at a telephone exchange during a specific time interval D = { 0, 1, …} Jan 2009**Exercise 1**• Q:Define the following Probability Event Distribution • Mention and define types of events? • Toss a coin twice Q:Show the sample space? Show these events:A: Observe the number of headsB : Having exactly two head Jan 2009**1/5/1435 h Sunday**Lecture 3 Jan 2009**More Definitions**• union Union of events A and B is the event consisting of all outcomes that are either in A or in B or in both events. denoted by A U B and read “A or B” • intersection intersection of A and B, is the event consisting of all outcomes that are in both A and B. denoted by A ∩ B and read “A and B”,**.**• complement Complements of event A is the event of all outcomes in the sample space S that are not contained in event A. Denoted by A’ • mutually exclusiveordisjoint events If two events A and B have no outcomes in common they are said to be mutuallyexclusive or disjoint events. This means if one of the event occurs the other cannot. Jan 2009**Probability of Events**Q: What is the objective of probability? • The objective of probability is to assign to each event, say E, a number P(E), called the probability of E which will give a exact measure of the chance that E will occur.**Define a Probability**A probability P is a rule which assigns a number between 0 and 1 to each event and satisfies these probability axioms: • 0 ≤ P(E) ≤ 1 for any event E • P(Ø ) = 0 , P(S) = 1, • If A1 , A2 , … is an infinite collection of mutually exclusive events, then**.**• The probability of the complement of any event A is given as • Example: If P(rain tomorrow) = 0.6 then P(no rain tomorrow) = 0.4 • Other notations for complement: Ac or Ā**General Probability Laws**• Let A and B be two events defined in a sample spaceS. • If two events A and B are mutually exclusive, then Thus This can be expanded to consider more than two mutually exclusive events.**Q: How can we calculate the probability of an event**The probability of any event E, where n: number of times we observe the event (frequency) N: a very large number of trials P(E) = n/N Jan 2009**.**Q1: In the experiment tossing a die repeatedly, in the long run, what would we expect P(E1) (probability of number 1 occurs) equal to? P ( A) 1 =(1/6) Exercise: Q2: In a year calendar, what would we expect, P(Em), date 30th occurs in a year if it a leap year? Jan 2009**8/5/1435 Sunday**Leture4 Jan 2009**Examples**Q:In an experiments of tossing a piece of fair coin once: • Show the sample space S= {H,T} • Probability of having a head P(H) = ½ • Probability of having a tail P(T) = ½ Jan 2009**.**• Probability of head and tail = The event head and tail = ( H and T) = {Ѳ} P {Ѳ} = 0 mutually exclusive, events • Probability of head or tail The event head or tail = HUT = S P(HUT) = P(H)+ P(T) = ½+1/2 = 1 Jan 2009**Q:In an experiment of rolling a fair die once show the**probability of the following events 1st the sample space is: S = {1,2,3,4,5,6} • Probability of a number from 1-6 on the upper face: P(1)=p(2)=p(3)=p(4)=p(5)=p(6)=1/6 • The probability of having the number 3 A1={3} (simple event) P{A1} = P{3} = 1/6 Jan 2009**Event of having even numbers**A3 = {2,4,6} Compound event P(A3) = 3/6 = ½ • Event of having odd numbers A4 = {1,3,5} Compound event P(A4) = 3/6 = ½ • Event of having a numbers greater than 4 A5 = {5,6} P(A5 )=P(5,6) = 2/6 = 1/3 2009**Event of having numbers greater than 6**A6 = {Ф} P(Ф) = 0 • Event of having numbers divisible by 3 A7 = {3,6} P(A7) = P(3,6) = 2/6 = 1/3 • Event of having numbers not divisible by 3 A8 = (1,2,4,5) P(A8) = 4/6 = 2/3 Jan 2009**OR طريقة اخرى للحل**= Ac7 = 1- P(A7) = 1-1/3 = 2/3 • The event of having odd number and divisible by 3 A9= (A4П A7) A4= {1,3,5} A7 = {3,6} {A4ПA7} = {3} P(A9) = P(3) = 1/6 Jan 2009**The event of having an odd number or divisible by 3**A10 ={ A4UA7} P(A10) = P(A4UA7) =P(A4)+P(A7) – P(A4A7) Because A4 and A7 are not mutually exclusive =3/6 + 2/6 – 1/6 = 4/6 = 2/3 • The event of having even number and odd number A11 = {A3ПA4) = {Ф} P(Ф) = 0 Jan 2009**The event of having odd number or even number**A12 ={A3UA4} = {1,2,3,4,5,6} =P(A3) + P(A4)- P(A3ПA4) = ½ + ½ - 0 = 1 i.e P(A12) = P(S) = 1 Jan 2009**9/5/1435 h Monday**Lecture 5 Jan 2009**Q:In rolling a fair die twice find the following**probabilities • A1: The sum of the two numbers is 10 First the sample space is Number of elements in the sample spa = 62 = 36 elements Jan 2009**أمثلة**فراغ العينة يتألف من 36 نقطة. الوجه الثاني الوجه الأول**A1 = {(4,6),(6,4),(5,5)}**P(A1) = 3/36 = 1/12 • A2: having sum of two numbers that is greater than 10 A2 = {(5,6),(6,5),(6,6)} P(A2) = 3/36 = 1/12 • A3: Having two numbers of sum = 7 A3 = {(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)} P(A3) = 6/36 = 1/6 • A4:Having 1 in the first roll Jan 2009**A4 = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)}P(A4) = 6/36 = 1/6**• A5: Having a sum less than 2 A5 = {Ф} P(A5) = P{Ф} = 0 • A6: The difference between the two numbers is equal to the absolute value A6 = {(1,2),(2,1),(2,3),(3,2),(3,4),(4,3),(4,5),(5,4),(5,6),(6,5)} P(A6) = 10/36 • A7: Having equal numbers A7 = {(1,1),(2,2),(3,3(,(4,4),(5,5),(6,6)} P(A7) = 6/36 = 1/6 Jan 2009**A8 : Having multiplication that equal at most 6**A8 ={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6) (2,1),(2,2),(2,3),(3,1),(3,2)(4,1),(5,1),(6,1)} P(A8) = 14/36= 7/18 Jan 2009**Q:In an experiment of tossing a piece of fair coin twice:**Find the following probabilities: 1st we have to define the sample space No. of elements = 22 = 4 elements S = {(H,H),(H,T),(T,H),(T,T)} The probability of each event = ¼ • A1: Having similar faces AI = {(H,H),(T,T)} P(A1) = 2/4 = ½ Jan 2009**A2: Having H once**A2 = {(H,T),(T,H)} P(A2) = 2/4 = ½ • A3: Having H at least once A3 = {(H,T),(T,H),(H,H)} P(A3) = ¾ • A4: Having H at most once A4 = {(H,T),(T,H),(T,T)} P(A4) = ¾ • A5: Having H twice Jan 2009**A5= {(H,H)}**P(A5) = ¼ • A6: Having no H A6 = {(T,T)} P(A6) = ¼ • A7: Having H three times A7: {(H,H,H)} A7 = {Ф} P(A7) = 0 Jan 2009