7.1 Sample Space, Events, Probability
7.1 Sample space, events,probability In this chapter, we will study thetopic of probability which is used inmany different areas includinginsurance, science, marketing,government and many other areas.
Blaise Pascal-father of modernprobability http://www-gap.dcs.stand.ac.uk/ history/Mathematicians/Pascal.html Blaise PascalBorn: 19 June 1623 in Clermont (now Clermont-Ferrand),Auvergne, FranceDied: 19 Aug 1662 in Paris, FranceIn correspondence with Fermat he laid the foundation forthe theory of probability. This correspondence consisted offive letters and occurred in the summer of 1654. Theyconsidered the dice problem, already studied by Cardan,and the problem of points also considered by Cardan and,around the same time, Pacioli and Tartaglia. The diceproblem asks how many times one must throw a pair ofdice before one expects a double six while the problem ofpoints asks how to divide the stakes if a game of dice isincomplete. They solved the problem of points for a twoplayer game but did not develop powerful enoughmathematical methods to solve it for three or more players.
Pascal
Probability 1. Important in inferentialstatistics, a branch of statistics thatrelies on sample information to makedecisions about a population. 2. Used to make decisions in the faceof uncertainty.
Terminology 1.Random experiment: is a process or activitywhich produces a numberof possible outcomes. Theoutcomes which resultcannot be predicted withabsolute certainty. Example 1: Flip two coinsand observe the possibleoutcomes of heads andtails
Examples 2. Select two marbleswithout replacement from abag containing 1 white, 1 redand 2 green marbles. 3. Roll two die and observethe sum of the points on thetop faces of each die. All of the above areconsidered experiments.
Terminology Sample space: is a list of all possibleoutcomes of the experiment. Theoutcomes must be mutually exclusive andexhaustive. Mutually exclusive meansthey are distinct and non-overlapping.Exhaustive means complete. Event: is a subset of the sample space.An event can be classified as a simpleevent or compound event.
Terminology 1. Select two marbles in successionwithout replacement from a bagcontaining 1 red, 1 blue and twogreen marbles. 2. Observe the possible sums ofpoints on the top faces of two dice.
3. Select a card from an ordinary deck of playing cards (no jokers)The sample space would consist of the 52 cards, 13 of each suit. Wehave 13 clubs, 13 spades, 13 hearts and 13 diamonds. A simple event: the selected card is the two of clubs. A compoundevent is the selected card is red (there are 26 red cards and so thereare 26 simple events comprising the compound event)4.Select a driver randomly from all drivers in the age category of 18-25.(Identify the sample space, give an example of a simple event and acompound event)
More examples Roll two dice. Describe the sample space of thisevent. You can use a tree diagram todetermine the sample spaceof this experiment. There aresix outcomes on the first die{1,2,3,4,5,6} and thoseoutcomes are represented bysix branches of the treestarting from the “tree trunk”.For each of these sixoutcomes, there are sixoutcomes, represented by thebrown branches. By thefundamental countingprinciple, there are 6*6 36outcomes. They are listed onthe next slide.
Sample space of all possible outcomeswhen two dice are tossed. (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4), (5,1), (5,2), (5,3), (5,4), (6,1), (6,2), (6,3), (6,4), Quite a tedious project !!(1,5) (1,6)(2,5), (2,6)(3,5), (3,6)(4,5), (4,6)(5,5), (5,6)(6,5), (6,6)
Probability of an event Definition: sum of the probabilities of the simple eventsthat constitute the event. The theoretical probability ofan event is defined as the number of ways the event canoccur divided by the number of events of the samplespace. Using mathematical notation, we have P(E) n( E )n( S )n(E) is the number of ways the event can occur and n(S) represents the total number of events inthe sample space.
Examples Example: Probability of a sum of 7 when two dice arerolled. First we must calculate the number of events of thesample space. From our previous example, we know thatthere are 36 possible sums that can occur when two diceare rolled. Of these 36 possibilities, how many ways can asum of seven occur? Looking back at the slide that givesthe sample space we find that we can obtain a sum ofseven by the outcomes { (1,6), (6,1), (2,5), (5,2), (4,3) (3,4)} There are six ways two obtain a sum of seven. Theoutcome (1,6) is different from (6,1) in that (1,6) means aone on the first die and a six on the second die, while a(6,1) outcome represents a six on the first die and one onthe second die. The answer is P(E) n( E ) 6 1 36 6n( S )
Meaning of probability How do we interpret this result? What does it mean to saythat the probability that a sum of seven occurs upon rollingtwo dice is 1/6? This is what we call the long-rangeprobability or theoretical probability. If you rolled twodice a great number of times, in the long run the proportionof times a sum of seven came up would be approximately one-sixth. The theoretical probability uses mathematicalprinciples to calculate this probability without doing anexperiment. The theoretical probability of an eventshould be close to the experimental probability is theexperiment is repeated a great number of times.
Some properties ofprobability 10 p( E ) 1 2.P( E1 ) P( E2 ) P( E3 ) . 1 The first property states thatthe probability of any eventwill always be a decimal orfraction that is between 0 and1 (inclusive). If P(E) is 0, wesay that event E is animpossible event. If p(E) 1, we call event E a certainevent. Some have said thatthere are two certainties inlife: death and taxes. The second property statesthat the sum of all theindividual probabilities of eachevent of the sample spacemust equal one.
ExamplesA quiz contains a multiple-choice question with fivepossible answers, only one of which is correct.A student plans to guess the answer.a) What is sample space?b) Assign probabilities to the simple eventsc) Probability student guesses the wrong answerd) Probability student guesses the correct answer.
Three approaches toassigning probabilities 1.Classical approach. This type of probabilityrelies upon mathematical laws. Assumes allsimple events are equally likely. Probability of an event E p(E) (number offavorable outcomes of E)/(number of totaloutcomes in the sample space) This approach isalso called theoretical probability. The exampleof finding the probability of a sum of seven whentwo dice are tossed is an example of the classicalapproach.
Example of classicalprobability Example: Toss two coins. Find the probability of at leastone head appearing. Solution: At least one head is interpreted as one head ortwo heads. Step 1: Find the sample space:{ HH, HT, TH, TT} There arefour possible outcomes. Step 2: How many outcomes of the event “at least onehead” Answer: 3 : { HH, HT, TH} Step 3: Use PE) n( E )n( S ) ¾ 0.75 75%
Relative Frequency Also called Empirical probability. Relies upon the long run relative frequency of anevent. For example, out of the last 1000 statisticsstudents, 15 % of the students received an A.Thus, the empirical probability that a studentreceives an A is 0.15. Example 2: Batting average of a major leagueball player can be interpreted as the probabilitythat he gets a hit on a given at bat.
Subjective Approach 1. Classical approach not reasonable 2. No history of outcomes. Subjective approach: The degree of beliefwe hold in the occurrence of an event. Examplein sports: Probability that San Antonio Spurs willwin the NBA title. Example 2: Probability of a nuclear meltdown ina certain reactor.
Example The manager of a records store has kept track ofthe number of CD’s sold of a particular type perday. On the basis of this information, themanager produced the following list of thenumber of daily sales: Number of CDsProbability 0 10.080.17 0.262 3 4 50.210.180.10
Example continued 1. define the experiment as the number of CD’ssold tomorrow. Define the sample space 2. Prob( number of CD’s sold 3) 3. Prob of selling five CD’s 4. Prob that number of CD’s sold is between 1and 5? 5. probability of selling 6 CD’s
7.2 Union, intersection,complement of an event, odds In this section, we will develop therules of probability for compoundevents (more than one event) andwill discuss probabilities involving theunion of events as well asintersection of two events.
The number of events in the union of A and B is equal to thenumber in A plus the number in B minus the number of eventsthat are in both A and B.Samplespace S N(A UB) n(A) n(B) – n(A B)Event AEvent B
Addition Rule If you divide both sides of the equation by n(S), thenumber in the sample space, we can convert the equationto an equation of probabilities:n( A B) n( A) n( B ) n( A B ) n( S )n( S ) n( S )n( S )P ( A B) P ( A) P( B ) P( A B)
Addition Rule A single card isdrawn from a deckof cards. Find theprobability that the card is a jack or club.Set of jacksSet ofclubs P(J or C) p(J) p(C)P(J andC)4 13 1 16 4 52 52 52 52 13Jack andclub (jackof clubs)
The events King and Queen are mutually exclusive. Theycannot occur at the same time. So the probability of a kingand queen is zero. the card is king orqueenP( K Q) p ( K ) P (Q) p( K Q)KingsQueens 4/52 4/52 – 0 8/52 2/13Kings andqueens
Mutually exclusive eventsIf A and B are mutuallyexclusive then P ( A B )The intersection of A and B is the emptysetAB p ( A) p( B)
Use a table to list outcomes of an experiment Three coins are tossed. Assume they are faircoins. Give the sample space. Tossing three coinsis the same experiment as tossing one coin threetimes. There are two outcomes on the first toss,two outcomes on the second toss and twooutcomes on toss three. Use the multiplicationprinciple to calculate the total number ofoutcomes: (2)(2)(2) 8 We can list the outcomesusing a little “trick” In the far left hand column,write four H’s followed by four T’s. In the middlecolumn, we write 2 H’s, then two T’s, two H’s ,then 2 T’s. In the right column, writeT,H,T,H,T,H,T,H . Each row of the table consists ofa simple event of the sample space. The indicatedrow, for instance, illustrates the outcome {heads,heads, tails} in that order.hhhhtttthhtthhtththththt
To find the probability of at least two tails, we mark each row(outcome) that contains two tails or three tails and divide thenumber of marked rows by 8 (number in the sample space) Sincethere are four outcomes that have at least two tails, the probability is4/8 or ½ .hhhhtttthhtthhtththththt
Two dice are tossed. What is the probability of a sum greater than 8or doubles?P(S 8 or doubles) p(S 8) p(doubles)-p(S 8 and doubles) 10/36 6/36-2/36 14/36 7/18. (1,1), (2,1), (3,1), (4,1), (5,1), ),(5,4),(6,4),Circled elementsbelong to theintersection of thetwo events.(1,5) (1,6)(2,5), (2,6)(3,5), (3,6)(4,5), (4,6)(5,5), (5,6)(6,5), (6,6)
Complement RuleMany times it is easier to compute the probability that A won’t occurthen the probability of event A. P( A) p (not A) 1 P (not A) 1 P ( A) Example: What is the probabilitythat when two dice are tossed,the number of points on each diewill not be the same?This is the same as saying thatdoubles will not occur. Since theprobability of doubles is 6/36 1/6, then the probability thatdoubles will not occur is 1 –6/36 30/36 5/6.
Odds In certain situations, such as the gaming industry, it iscustomary to speak of the odds in favor of an event E andthe odds against E.P( E ) Definition: Odds in favor of event E 'p( E )'p(E) Odds against E P( E ) Example: Find the odds in favor of rolling a seven when twodice are tossed. Solution: The probability of a sum of seven is 6/36. SoP( E ) 'p( E )636 6 130 30 536
7.3 Conditional Probability,Intersection and Independence Consider the following problem: Find the probability that a randomly chosen person in theU.S. has lung cancer. We want : p(C) . To determine the answer, we must knowhow many individuals are in the sample space, n(S). Ofthose, how many have lung cancer, n(C) and find the ratioof n( c) to n(S).n(C )P (C ) n( S )
Conditional Probability Now, we will modify the problem: Find the probability thata person has lung cancer, given that the person smokes. Do we expect the probability of cancer to be the same? Probably not, although the cigarette manufacturers maydisagree. What we now have is called conditional probability. It issymbolized byP (C S ) and means the probability of lung cancer assuming orgiven that the person smokes.
The probability of having lung cancer given that the person smokes isfound by determining the number of people who have lung cancerand smoke and dividing that number by the number of smokers.n( L S )p( L S ) n( S )People with lung cancerpeople who smoke andhave lung cancer.People who smoke
Formula for Conditional probability Derivation:n( L S )p( L S ) n( S )n( L S )n (T )p( L S ) n( S )n (T )p( L S )p( L S ) p( S ) Dividing numerator anddenominator by the totalnumber , n(T) , of thesample space allows us toexpress the conditionalprobability of L given S asthe quotient of theprobability of L and Sdivided by the probabilityof smoker.
The probability of event A given that event B has alreadyoccurred is equal to the probability of the intersection ofevents A and B divided by the probability of event B alone.p( A B )P( A B) p( B )
Example There are two majors of a particular college:Nursing and Engineering. The number of studentsenrolled in each program is given in the tableFind the following probabilities by using thefollowing table. The row total gives the totalnumber of each category and the number in thebottom-right cell gives the total number ofstudents. A single student is selected at randomfrom this college. Assuming that each student isequally likely to be chosen, find :
Joint and MarginalProbability Table 1.Prob(Nurse) 100/150 2/3UndergradsGradsNursing5347100 2. Prob(Graduate student) 60/150 2/5Engineers371350 3. Probability (Nurse andGraduate student) 47/1509060150 4. Probability ( Engineeringand Grad Student) 13/150
Joint, Marginal andConditional Probability Combinations of simple events A and B : symbolismP( A B) Probability of intersection is jointprobability. The symbolism represents theprobability of the intersection of eventsA and B.
Conditional probability Given that a under-graduatestudent is selected at random,what is the probability thatthis student is a nurse? Restricting our attention onthe column representingundergrads, we find that ofthe 90 undergrad students,53 are nursing majors.Therefore, P(N/U) 9060150
Using the table Given that an engineeringstudent is selected, find theprobability that the student isa under-graduate student.Restricting the sample spaceto the 50 engineeringstudents, 37 of the 50 areundergrads, indicated by thered cell. Therefore,P(U/E) 37/50 60150
Derivation of general formulas for P( A and B) usingbasic algebra Algebra:P ( B A)P ( B A) P ( A) SinceP( A B) p(B A) We haveP( B A) P( A) p( B A)p( A B )P( A B) p( B )P ( A B ) p( B ) P ( A B )P ( A B ) p( B A ) P ( A ) p( B A ) p( B ) p( A B )
Two cards are drawn without replacement from an ordinarydeck of cards . Find Probability (two clubs aredrawn in succession). Symbolize mathematically:P (C1 C 2 )means draw a club on the firstdraw and then a second club.P (C1 C 2 ) p(C1 ) p(C 2 C1 ) 13 12 1 41 52 51 4 17 17 Because the selection is donewithout replacement, we haveone less card in the samplespace and one less club sincewe assume that the first carddrawn is a club, there are 12remaining clubs and 51 totalremaining cards.
Two machines are in operation. Machine A produces 60% of theitems whereas machine B produces the remaining 40%. MachineA produces 4% defective items whereas machine B produces 5%defective items. An item is chosen at random.a)P (item is defective) P(D and machine A) or P(D and Machine B)p(D) P ( A D ) p( B D ) 0.60(0.04) 0.40(0.05) 0.044A0.04Def60%40%0.96good0.05 defB0.95goodp( A ) p( D A ) p( B ) p( D B )
Independence If two events are independent, thenp ( A B) p ( A)P ( B A) p ( B)
A coin is tossed and a die is rolled. Find the probability thatthe coin comes up heads and the die comes up three. The number of outcomes for the coin is 2: { H , T}. Thenumber of outcomes for the die is 6: {1,2,3,4,5,6} . Usingthe fundamental principle of counting, we find that thereare 2(6) 12 total outcomes of the sample space. H p(H and 3) 1/12123456T123456
Now, let’s look at the same problem in a slightly different way To find the probability of Heads and then a three on a dice,we havep( H 3) p( H ) p(3 H ) using the rule for conditional probability. However, theprobability of getting a three on the die does not dependupon the outcome of the coin toss. We say that thesetwo events are independent, since the outcome ofeither one of them does not affect the outcome of theremaining event.p( H 3) p( H ) p(3 H ) p( H ) p(3)
Joint Probability Rule If events A and B are independent:P ( A B ) P ( A) P ( B ) If events A and B are dependent:p ( A B) p ( A)i p ( B A)
Examples of Independence 1. Two cards are drawn in succession withreplacement from a standard deck of cards. What is theprobability that two kings are drawn?P ( K 1 K 2 ) p( K 1 ) p( K 2 )4 41 52 52 169 2.Two marbles are drawn with replacement from a bagcontaining 7 blue and 3 red marbles. What is theprobability of getting a blue on the first draw and a red onthe second draw?p( B R ) p( B ) p( R ) 7 321 0.2110 10 100
Dependent Events Two events are dependent when the outcome of one eventaffects the outcome of the second event. Example: Draw two cards in succession withoutreplacement from a standard deck. Find the probability of aking on the first draw and a king on the second draw. Answer:P ( K 1 K 2 ) p( K 1 ) p( K 2 K 1 )4 31 52 51 221
Are smoking and lungdisease related? 1. Find the probability oflung disease. P(L) 0.15 (row total) 2. Find p(L/S) , probabilityof lung disease givensmoker. P(L/S) 0.12/0.19 0.63.The probability of having lungdisease given that you are asmoker is considerablyhigher than the probabilityof lung disease in thegeneral population, so wecannot say that smokingand lung disease areindependent events.SmokerNonsmokerHasLungDisease0.120.03No lungDisease0.190.66
7.4 Baye’s FormulaIn this section, we will find the probability ofan earlier event conditioned on theoccurrence of a later event.
Probability of an Earlier Event Given a Later EventThe procedures used in this example can be employed forother problems using Bayesian probability. A survey of middle-aged menreveals that 28% of them arebalding at the crown of theirhead. Moreover, it is knownthat such men have an 18%probability of suffering a heartattack in the next 10 years.Men who are not balding inthis way have an 11%probability of a heart attack.If a middle-aged man israndomly chosen, what is theprobability that he is baldinggiven that he suffered a heartattack? See tree diagramnext slide.
We want P(B/H) (probability that he is balding given that hesuffered a heart attack. Tree diagram:Heartattack 0.18Balding(0.28)Notbalding0.720.82 no heartattackHeartattack0.110.89 noheartattack
Using the rule for conditional probability, we derive thegeneral formula to solve the problem:p( B H )P(B H ) p( H )p( H ) p( B H ) p( NB H ) P(H) can be found byfinding the union of P(B and H) and p(NB and H)p( H ) p( B ) p( H B ) p( NB ) p( H NB )p( B H )P(B H ) p( B ) p( H B ) p( NB ) p( H NB ) Rule for conditionalprobability Substitution for p(H)
Using the tree diagram, substitute the probabilities in the formula .The answer reveals that the probability that a middle-aged man isbalding given he suffered a heart attack is 0.389.1.2.3.4.P(B H ) p( B H ) p( B H )p( B ) p( H B ) p( NB ) p( H NB )p( B ) p( H B )p( B ) p( H B ) p( NB ) p( H NB )p( B H ) 0.3890.28 (0.18)0.28(0.18) 0.72 (0.11)
probability or theoretical probability. If you rolled two dice a great number of times, in the long run the proportion of times a sum of seven came up would be approximately one-sixth. The theoretical probability uses mathematical principles to calculate this probability without doing an experiment. The theoretical probability of an event
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Careful! The sample space is a set (of outcomes) An outcome is an element of a sample space An event is a set (a subset of the sample space) – It can be empty (the null event { } or , which never happens) – It can contain a single outcome (simple/elementary event) – It can be the entire sample space (certain to ha
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