Mathacle. PSet Stats, Concepts In Statistics Level Number Name: Date:

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1 Mathale PSet Stats, Conepts In Statistis Change A deterministi haos V. PROBABILITY 5.1. Sets A set is a olletion of objets. Eah objet in a set is an element of that set. The apital letters are usually used to denote the sets, and the lower ase letters are used to denote the elements. In an experiment, an event is a set of outomes (elements) with ertain probability. An atomi event, or an elementary event, ontains just one outome whih an t be further deomposed. Given a set A, if u is an element of A, then u A; if u is not an element of A, then u A. 186

2 Mathale PSet Stats, Conepts In Statistis Notations Sometimes the sets an be desribed by listing the elements. For example, whole numbers are the non-negative integers: A olletion of a, b,, d is Ν = {0,1,2,3,...} T = { a, b,, d} The empty set or the null set is the set with no elements. It is denoted by φ or{}. Set A and set B are equal if they have exatly the same elements. It is denoted as A = B. Otherwise, the set A and set B are not equal, or A B. For example, if T = { a, b,, d} and R = { d, a,, b}, then T = R. If a set A is finite, then the number of elements, or the measure, of the set A is denoted as m (A). In the above example, m ( T ) = 4. Universal Set, Subsets and Power Set Universal set U is the set of all elements or atomi events related to the given problem. For example, the set of all positive integers no larger than 4 is U = {1,2,3,4 }. Set B is the subset of set A if for every element in B, it is also in set A. This is denoted as B A. For example, { 1,2} {1,2,3,4 }. For any set A, A A and φ A. The power set, denoted by S, of set A is the set that ontains all the subsets involved with A plus the empty set. 187

3 Mathale PSet Stats, Conepts In Statistis If A is a subset of U, the set of elements of U that are not in A is alled the omplement of A, denoted by A. For example, if A = {1,2 } and U = {1,2,3,4 }, then A = {3,4}. Example Let A = { a, b,, d, e}, B = { b, }, and U = { a, b,, d, e, f, g}, find A, B. Solution: A = { f, g}, B = { a, d, e, f, g}. Example If set C = { a, b,, d}, find the power set S whih ontains all the subsets of C. Solution: S = { Φ, { a}, { b},{ },{ d }, { a, b}, { a, },{ a, d},{ b, },{ b, d},{, d }, { a, b, },{ a, b, d},{ a,, d},{ b,, d }, { a, b,, d } } If the number of element is m (U ) for set U, then there are example, m ( S) = 4, and there are 4 2 = 16 subsets. m( U ) m ( S) = 2 subsets. In this Operations on the Events If A and B are two events of the universal event set U, then 1.) A does not our is the omplement or negation of A. It is denoted by A ( or A ' ) ; 2.) either A or B our is the union or disjuntion of A and B. It is denoted by A B ; 3.) both A and B our is the intersetion or onjuntion of A and B. It is denoted by A B ; 4.) A ours and B does not our is the differene of A and B. It is denoted by A B = A B. 188

4 Mathale PSet Stats, Conepts In Statistis Example In the experiment of rolling a die one, the universal set is U = {1, 2,3,4,5,6}. Let A = an odd number turns up B = the number that turns up is divisible by 3 Express the following events in terms of A and B: E = the number that turns up is even F = the number that turns up is odd or is divisible by 3 G = the number that turns up is odd and divisible by 3 H = the number that turns up is odd but not divisible by 3 Solution: A = {1,3, 5}, B = {3, 6}, and E is the omplement of A: E = A = U A = {2, 4,6} F is the union of A and B: F = A B = {1,3, 5,6} G is the intersetion of A and B: G = A B = {3} H is the differene of A and B: H = A B = A B = {1,5} Notie that A E = U and A E = {} or φ. 189

5 Mathale PSet Stats, Conepts In Statistis In general, two events A and B are exhaustive if A B = U. That is, either A or B will our. Two events A and B are disjoint or mutually exlusive if A B = φ, where φ is the impossible event. That is, if A ours, then B annot our. MECE events are mutually exlusive and olletively exhaustive. For example, A A = U and A A = φ. 190

6 Mathale PSet Stats, Conepts In Statistis Example In the experiment of tossing a oin three times, let H= Heads and T= Tails. Express the following sets in terms of H and T: The universal set U A = we throw tails exatly two times B = we throw tails at least two times C = tails did not appear before a heads appeared D = two heads E = A : F = A ( C D) : G = A D : Solution: U = { HHH, HHT, HTH, THH, HTT, TTH, THT, TTT} A = { HTT, THT, TTH} B = { HTT, THT, TTH, TTT} C = { HTT, HTH, HHT} D = { HHT, THH, HTH} E = { HHH, HHT, HTH, THH, TTT} F = { HTT, THT, TTH, HHT, HTH} G = { HTT, THT, TTH} 191

7 Mathale PSet Stats, Conepts In Statistis Example If A = it is hot and B = it is sunny, state the following sets in terms of A and B: 1.) A B 2.) A B = = 3.) A B = A B 4.) ( ) = 5.) A B = 6.) ( A B) ( A B ) Solution: = 1.) A B = it is hot and sunny 2.) A B = it is not hot but sunny 3.) A B = it is neither hot nor sunny A B 4.) ( ) = it is not true that it is neither hot nor sunny 5.) A B = it is hot or sunny A B A B = it is either not hot and sunny or hot and not sunny 6.) ( ) ( ) DeMorgan s Laws and Properties of Set Operations DeMorgan s Laws: For any two events A and B of a given universal set, ( A B) = A B and ( A B) = A B Example If Ω = {1,2,3,4,5,6,7,8,9,10 }, A = {1,3,5,7,9 } and B = {1,2,3,4,5 }, verify the DeMorgan s Laws Solution: A = {2,4,6,8,10}, = {6,7,8,9,10 } B, ( A B) = 6,8,10 } = A B ( A B) = { 2,4,6,7,8,9,10 } = A B {, 192

8 Mathale PSet Stats, Conepts In Statistis Properties and Laws of Sets Commutative: A B = B A Assoiative: A ( B C) = ( A B) C Distributive: A ( B C) = ( A B) ( A C), A ( B C) = ( A B) ( A C) Partition: A A ( A B) ( A B ) exlusive. Other properties: Ω = A Ω φ = A φ A = A A A = A A ( A ) = A B A = B A B ( B A) = B = Ω =, where A B and A B are mutually Proofs of some identities a.) prove A B = B ( B A) Prove: B ( B A) = B ( B A ) = B ( B A ) = B ( B A) = ( B B ) ( B A) = B A = A B 193

9 Mathale PSet Stats, Conepts In Statistis b.) Prove A ( B A) = A B Prove:.) Prove ( A B) C = A ( B C) A ( B A) = A ( B A ) = ( A B) ( A A ) = A B Prove: ( A B) C = ( A B ) C = A ( B C ) = A ( B C) = A ( B C) d.) Prove ( B ( B A)) = B Prove: ( B ( B A)) = B ( B A ) = B ( B A) = B 194

10 Mathale PSet Stats, Conepts In Statistis Quik-Chek Sets QC Assume some proper universal set. Let p = It is raining q = Mary is sik t = Bob stayed up late last night r = Paris is the apital of Frane s = John is a loud-mouth Express eah of the following statements in terms of the delarative sentenes above: Negation or Complement: 1.) It isn t raining 2.) Paris is not a apital of Frane Conjuntion: 3.) It is raining and Mary is sik 4.) Paris isn t the apital of Frane and It isn t raining 5.) It is not the ase that it is raining and Mary is sik Disjuntion: 6.) Paris is the apital of Frane and it is raining or John is a loud-mouth 7.) John is a loud-mouth or Mary is sik or it is raining Mixed statements: 8.) It is raining but Mary is not sik 9.) Neither it is raining nor Mary is sik 195

11 Mathale PSet Stats, Conepts In Statistis QC Let J = John is to blame and M = Mary is to blame. Use DeMorgan s Laws to show event A and event B are equivalent, where A = It is ertainly not true that neither John nor Mary is to blame B = John or Mary is to blame, or both Hints: find J and M first and express A and B in terms of J, M, J and M QC Prove ( A B) ( B A) = ( A B) ( A B) 196

12 Mathale PSet Stats, Conepts In Statistis Answers QC ) p 2.) r 3.) p q 4.) r p 5.) ( p q) 6.) ( r p) s 7.) ( s q) p 8.) 9.) p q p q QC J = John is not to blame, M = Mary is not to blame, A = ( J M ) = J M = B QC ( A B) ( B A) = ( A B ) ( B A ) = (( A B ) B) ( A B ) A ) = (( A B) ( B B)) (( A A ) ( B A )) = (( A B) Ω) ( Ω ( B A )) = ( A B) ( B A ) = ( A B) ( B A) = ( A B) ( B A) = ( A B) ( A B) 197

13 Mathale PSet Stats, Conepts In Statistis 5.2. Events and the Measure of Events [MATH] For any given experiment, the set of all possible outomes is alled the sample spae or probability spae. The sample spae is denoted by Ω. An event E is a subset of Ω. That is, E Ω. The ounting measure on a set E, denoted by m (E), is the number of elements in E. The olletion of all events (all subsets of sample spae), or the power set, is alled the event spae, denoted by F. [MATH] For experiment with two steps, eah step with sample spaes Ω 1 and Ω 2 respetively, the sample spae experiment is a set of paired-elements Ω = Ω1 Ω 2 = {( w 1, w2 ) : w1 Ω1, w2 Ω 2} Or simply Ω = Ω1 Ω 2 = { w 1w2 : w1 Ω1, w2 Ω 2} when it is not onfusing in the ontent. The definition an be extended to higher dimensional ases. 201

14 Mathale PSet Stats, Conepts In Statistis Example The experiment is to roll a die one. Find the sets for: Sample spae Ω Event of getting an odd numbero Event of getting an even number E Event of getting a prime number P Event of getting a non-prime number N Measures of events E, O, P, N Solution: There are six possible outomes: 1,2,3, 4,5, 6. The sample spae is Ω = {1, 2,3,4,5,6 }. The event of getting an even number is E = {2, 4,6}, the event of getting an odd number is O = {1,3,5}, the event of getting a prime number is P = {2,3,5} the event of getting a non-prime number is N = {1,4,6 }. Measures of events m ( E) = m( O) = m( P) = m( N) = 3 Notie that Ω an be partitioned by Ω = E + O or Ω = P + N. The information an also be expressed more usefully by the two-way event table: Prime\Odd-Even E O Total P 2 3, 5 2, 3, 5 N 4, 6 1 1, 4, 6 Total 2, 4, 6 1, 3, 5 1, 2, 3, 4, 5, 6 202

15 Mathale PSet Stats, Conepts In Statistis The sample spae an be expressed in a list, a table or a tree diagram: The headers of the event table are the partitions of the possible outomes of a single trial, and the ontents in the event table are the sample spae of the experiment. The two-way table an also be setup similarly for the ounting measures or probabilities. Example The experiment is to toss a oin twie. Find the sample spae and the event of getting the first toss as a Heads. Solution: The sample spae is Ω = { H, T} { H, T} = {( H, H ),( H, T ),( T, H ),( T, T )} = { HH, HT, TH, TT} and the event of getting the first toss as a Heads is E = { H} { H, T} = {( H, H ),( H, T )} = { HH, HT}. The two-way event table: 1 st Toss\2 nd Toss H T Total H [ HH ] [ HT ] HH, HT T TH TT TH, TT Total HH, TH HT, TT HH, HT, TH, TT The information an also be easily seen from the event table at row H and olumn Total. 203

16 Mathale PSet Stats, Conepts In Statistis Example Roll two die one. Sample spae Ω : E = The Sum of two die is 7 : Solution: Ω = {1, 2,3,4,5,6} {1,2,3, 4,5,6} = {(1,1),(1,2),...(6,6)}, E = {(1,6),(2,5),(3,4),(4,3),(5, 2),(6,1)} Measure of the Event Set Let A and B be the event subsets of the universal event set Ω, and A and B are the omplements of A and B, respetively. The numbers of elements in A and B, denoted by m (A) and m (B), are the ounting measures of the sets A and B. When an element in a universal set Ω is hosen, the Equally Likely property is that eah element in Ω is not disriminated. The measure of suh event set an be used to obtain useful information about the frequeny or probability of the event. 204

17 Mathale PSet Stats, Conepts In Statistis [MATH] The Priniple of Inlusion Exlusion (PIE) states Where the relationships among some of the subsets are The Venn diagram an be expressed in a two-way or ontingeny table of two partitions of the universal set Ω = A A and Ω = B B : B / A A A Total B m( A B) m( A B) m (B) B m( A B ) m( A B ) m ( B ) Total m (A) m ( A ) m (U ) Example In 1997, Exeutive magazine surveyed the CEOs of 500 largest orporations in the United States. Of these 500 people, 310 had degrees (of any sort) in business, 238 had undergraduate degrees in business, and 184 had graduate degree in business. How many CEOs had both undergraduate and graduate degrees in business? 205

18 Mathale PSet Stats, Conepts In Statistis Solution: Let S = CEOs with undergraduate degree in business T= CEOs with a graduate degree in business Example [MC1323M] In a ertain shool, 17 perent of the students are enrolled in a psyhology ourse, 28 perent are enrolled in a foreign language ourse, and 32 perent are enrolled in either a psyhology ourse or a foreign language ourse or both. What is the perent of students who are enrolled in both a foreign language ourse and a psyhology ourse? Solution: Let F = students enrolled in foreign language ourse P = students enrolled in psyhology ourse. Note that m( F P ) = m( U ) m( F P) = = F\ P P P Total F F = Total From the table, m ( F P) = That is, 13% of students are enrolled in both ourses. For three sets A, B and C, 206

19 Mathale PSet Stats, Conepts In Statistis m( A B C) = m( A) + m( B) + m( C) m( A B) m( A C) m( B C) + m( A B C) m( A B C) = m( Ω) m( A B C ) Example Dogs in the GoodDog obediene shool win a blue ribbon for learning how to sit, a green ribbon for learning how to roll over, and a white ribbon for learning how to stay. There are 100 dogs in the shool. Suppose: 73 have blue ribbons, 39 have green ribbons, and 62 have white ribbons. 21 have a blue ribbon and a green ribbon; 28 have a green ribbon and a white ribbon; 41 have a blue ribbon and a white ribbon. 14 have all three ribbons. How many dogs have not learned any triks? Solution: 100 = N N = 2 207

20 Mathale PSet Stats, Conepts In Statistis Counting Methods [MATH] Here are some of the ommonly used methods for ounting equally likely events. Fatorial of a positive integer n is defined as the produt of n ( n 1) ( n 2) 2 1. It is denoted as n!. That is n! = n ( n 1) ( n 2) 2 1 Note that 0! = 1as defined. The four main methods for ounting are Multipliation Priniple ---- Suppose that a ounting of objets is omposed of t onseutive operations. If operation 1 an be ounted in m1 ways and for eah of those, operation 2 an be ounted in m 2 ways, and so forth. Then the omplete ounting an be performed in m1 m2 mt ways. Addition Priniple ---- If the objets to be ounted are separated into t ases with m1, m2,, mt as the numbers of ways in eah ase, respetively. The total number of ounts is the sum m1 + m2 + + mt of the various ases. Permutation ---- For the ordered arrangements of objets. The formula to arrange r distint objets less than or equal to the n given distint objets is n! n Pr = ( n r)! Note that P = n!. The formula annot be used to solve all permutation problems. n n Combination ---- For arrangements of objets with regard to no order. The formula to selet r distint objets less than or equal to the n given distint objets is n C r n! = r! ( n r)! Note that ncn = nc0 = 1. The same as n P r in permutation problems, the formula annot be used to solve all ombinational problems. 208

21 Mathale PSet Stats, Conepts In Statistis Example Eah of 5 men daned with eah of 5 women, and the eah woman daned with eah of the other women, How many danes were there? Solution: There were total of 5(5) + 5C2 = = 35 danes. Example A TV diretor is sheduling a ertain sponsor s ommerials for an upoming broadast. There are six slots available for ommerials. In how many ways may the diretor shedule the ommerials in eah of the following ases? a.) If the sponsor has six different ommerials, eah to be shown one? b.) If the sponsor has three different ommerials, eah to be shown twie? 209

22 Mathale PSet Stats, Conepts In Statistis.) If the sponsor has two different ommerials, eah to be shown three times? d.) If the sponsor has three different ommerials, the first of whih is to be shown three times, the seond two times, and the third one? Solution: a.) 6 P 6 = 6! = 720 b.) There are 6 C 2 to hoose the 1 st ommerial, 4 C 2 to hoose the 2 nd and 2 C 2 to hoose the 3 rd ommerial. So, the total number of ways is 6C2 4C2 2C2 = 90 1st 2nd 3rd.) There are 6 C 3 to hoose the 1 st ommerial, 3 C 3 to hoose the 2 nd. So, the total number of ways is C C = st 2nd 210

23 Mathale PSet Stats, Conepts In Statistis d.) There are 6 C 3 to hoose the 1 st ommerial, 3 C 2 to hoose the 2 nd and 1 C 1 to hoose the 3 rd ommerial. So, the total number of ways is 6C3 3C2 1C1 = 60 1st 2nd 3rd 211

24 Mathale PSet Stats, Conepts In Statistis Quik-Chek Events and the Measure of Events QC Toss a oin ontinuously until one gets a Heads. Assume H= Heads and T= Tails. Sample spae Ω : Event E = Get a Heads on the 1 st try or the 3 rd try : QC You toss a oin three times. Write the tree diagram for the experiment QC If 20 girls are on JPS soer team, 25 girls are on the hokey team, and 11 girls play both sports, then how many girls play soer or hokey? 212

25 Mathale PSet Stats, Conepts In Statistis QC A dotor is studying the relationship between blood pressure and heartbeat abnormalities in her patients. She tests a random sample of patients and notes their blood pressures (high, low, normal) and their heartbeats (regular or irregular). She finds that: 1.) 14% have high blood pressure. 2.) 22% have low blood pressure. 3.)15% have an irregular heartbeat. 4.) Of those with an irregular heartbeat, one-third have high blood pressure. 5.) Of those with normal blood pressure, one-eight have an irregular heartbeat. Calulate the portion of the patients seleted who have a regular heartbeat and low blood pressure. QC CHS High Shool has 85 senior boys, eah of whom plays on at least one of the shool s three boys varsity sports teams: football, baseball, and larosse. It so happens that 74 are on the football team; 26 are on the baseball team; 17 are on both the football and larosse teams; 18 are on both the baseball and football teams; and 13 are on both the baseball and larosse teams. Compute the numbers of senior boys playing all three sports, given that twie this number are members of the larosse team. QC How many ways an 12 people be partitioned into three groups of four people? 213

26 Mathale PSet Stats, Conepts In Statistis QC [AMC ] A box ontains exatly five hips, three red and two white. Chips are randomly removed one at time without replaement until all red hips are drawn or all white hips are drawn. What is the probability that the last hip drawn is white? 214

27 Mathale PSet Stats, Conepts In Statistis Answers QC Ω = { H, TH, TTH,...}, E = { H, TTH} QC QC team. Then 34. Let S be the member of soer team and H be the member of hokey n( S H ) = n( S) + n( H ) n( S H ) = = 34 QC Heartbeats\Blood Pres High Low Normal Total Regular? = = = 0.85 Irregular i = = i = 0.15 Total = boxed numbers are given. 20% of the patients have regular heartbeats and low blood pressure. QC T= = T T T = C4 8 C4 QC = ! QC Let R = read and W = white. There are total of 5 C2 = 10 ways to arrange the five RRRWW hips. The favorable outomes the following 6 ases: RRRWW, RWRRW, RRWRW, WWRRR, WRWRR and WRRWR. So, the probability is 6/10 or 3/5. 215

28 Mathale PSet Stats, Conepts In Statistis 5.3. Probability of Equally Likely Events [MATH] The Equally Likely property of a universal set Ω is that eah element in Ω is not disriminated. An equally likely event E is a subset of the total possible outomes T, where E T Ω. The probability of an equally likely event E is the ratio defined as m ( E) P ( E) =. Note that all three things -- Probability P, Event E, Event Spae F are m ( T ) involved when the probability is alulated, and E Ω F. That is, the probability of an equally likely event an be determined by the measure of favorable outomes and the measure of the total possible outomes. Example The experiment is to toss a oin one. Find the sets for Sample spae Ω Event of getting a head E Event spae F Measure of event spae m (F) Possible outomes T Probability P (E) 220

29 Mathale PSet Stats, Conepts In Statistis Solution: There are two possible outomes: Heads (H) or Tails (T). Sample spae is Ω = { H, T}. The event of getting a Heads is E = {H}. The event spae F = { ϕ,{ H},{ T},{ H, T}}. Note that F Ω. Measure of event spae m ( F) = 4. That is, the number of elements in F is 4. Possible outomes T = { H, T}, note that T F. Probability m( E) P ( E) = = m( T ) 1 2 Example The experiment is to toss a die first and then to toss a oin. a.). Find the sample spae by using the list of events, two-way table and tree diagram. Ω = 221

30 Mathale PSet Stats, Conepts In Statistis b.) Let A= the outome is an even number and H= the outome is a Heads, find the following probabilities. 1.) P ( A H ) = 2.) P ( A H ) = 3.) P (( A H ) ) = 4.) P ( A H ) = Solution: a.) The sample spae is Ω = {( 1, H ),(2, H ),(3, H ),(4, H ),(5, H ),(6, H ),(1, T ),(2, T ),(3, T ),(4, T ),(5, T ),(6, T )} b.) 1.) 2.) 3 P ( A H ) = = P ( A T ) = = = 12 12, the probability of getting an even number and a Heads. 3, the probability of an even number or a Tails

31 Mathale PSet Stats, Conepts In Statistis 3.) 4.) 3 P (( A H ) ) = = 12 3 P ( A H ) == = , the probability of NOT an even or a Heads., the probability of an odd number AND a Tails. Here are more examples: Experiment Sample Spae Ω Event E Possible outomes T Probability P Rolling a die one. The probability of odd number faes up. Ω = {1,2,3,4,5,6} E = {1,3,5 } m ( E) 3 1 P( E) = = = T = {1,2,3,4,5,6} m ( T ) 6 2 Tossing a oin twie. The probability of two heads -up. Ω = { HH, HT, TH, TT} The order is from left to right. Two heads E = {HH} T = { HH, HT, TH, TT} m ( E) P ( E) = = m ( T ) 1 4 Tossing a oin twie. Probability of the 2 nd is heads-up when the 1 st is a tails-up. Ω = { HH, HT, E = {TH } m ( E) 1 P ( E) = = TH, TT} T = { TH, TT} Ω m ( T ) 2 Tossing a oin Ω = { HHH, HHT, E = { HHT, HTH} m ( E) 2 1 three times. P ( E) = = = HTH, HTT, T = { HHH, HHT, m ( T ) 4 2 The probability of THH, THT, HTH, HTT} Ω two heads -up if the 1 st is heads-up. TTH, TTT} 223

32 Mathale PSet Stats, Conepts In Statistis Example What is the probability that a teaher divides 10 students into a group of 4 and a group of 6, where Sam and Daniel, two of the students, are in separate groups? Solution: Let F = Sam and Daniel are not in the same group. Then, there are m ( Ω ) = C possible 10 4 ways of grouping students, and m( F ) = C + C 112 ways of not grouping them = together. Therefore, the probability that they are in separate group is Example You roll a pair of die. m( F) 112 P ( F) = = = m 8 ( Ω) a.) What is probability of rolling a 7 if they are a pair of fair die? b.) What is probability of rolling a 7 if they are a pair of peuliar die whose probabilities of rolling 1, 2, 3, 4, 5 and 6 on eah die are in the ratio 1:2:3:4:5:6? Solution: 6 1 a.) This is an equally likely event. P ( sum = 7) = = D1\D b.) This is not an equally likely event ( P(1) P(6) + P(2) P(5) + P(3) (4)) = = P( sum = 7) = 2 P

33 Mathale PSet Stats, Conepts In Statistis Quik-Chek Probability of Equally Likely Events QC [MC1338] Eah of the faes of a fair six-sided number of ube is numbered with one the numbers 1 through 6, with a different number appearing on eah fae. Two suh number ubes will be tossed, and the sum of the numbers appearing on the faes that land up will be reorded. What is the probability that the sum will be 4, given that the sum is less than or equal to 6? QC When five ordinary six-sided die are tossed simultaneously, what is the probability that the 1 faes up on exatly two of the die? QC A bag has 3 red and k white marbles, where k is an (unknown) positive integer. Two of the marbles are hosen at random from the bag. Given that the probability that the two marbles are the same olor is 0.5, find k. 225

34 Mathale PSet Stats, Conepts In Statistis QC [FRQ1402] Nine sales representatives, 6 men and 3 women, at a small ompany wanted to attend a national onvention. There were only enough travel funds to send 3 people. The manager seleted 3 people to attend and stated that the people were seleted at random. The 3 people seleted were women. There were onerns that no men were seleted to attend the onvention. (a) Calulate the probability that randomly seleting 3 people from a group of 6 men and 3 women will result in seleting 3 women. (b) Based on your answer to part (a), is there reason to doubt the manager s laim that the 3 people were seleted at random? Explain. QC On a standard die one of the dots is removed at random with eah dot equally likely to be hosen. The die is rolled. What is the probability that the top fae has an odd number of dots? 226

35 Mathale PSet Stats, Conepts In Statistis Answers QC QC QC P( same) = 3 P ( Sum = 4 Sum <= 6) = 15 D1\D P ( two " one") = C = k ( k 6)( k 1) = 0 5 C2 C 2 + k 3+ k C2 C k = 6, 2 = 1 2 k = 1 3(2) k( k + 1) + ( k + 3)( k + 2) ( k + 3)( k + 2) = 1 2 QC C3 1 a.). P( three women) = = C3 84 b.) Yes, the hane of seleting three women is only about 1.2%. 11 QC Total = 21 dots. The probability of removing a dot from an 21 odd number is = probability of removing a dot from an even number is and the probability of an odd number on the top is 6 2 ; the =, and the probability P ( odd) = + = of an odd number on the top is 6 4. So,

36 Mathale PSet Stats, Conepts In Statistis 5.4. Probability of Events [MATH] A set funtion P defined over the sample spae Ω is a probability funtion if 1.) For any event A, P ( A) 0 2.) P ( Ω) = 1, and 3.) ( A A A ) = P( A ) + P( A ) + P( ) + P, if A A,, are disjoint or A3 mutually exlusive events. That is A A = φ for all i j. i j 1, 2 A3 [MATH] Some properties of probability: 1.) ( Ω) = P( A A ) = P( A) + P( A ) = 1 P, so that P( A) 1for any A Ω. 2.). if A B, then B = A ( B A), P( B) = P( A) + P( B A), or P( B A) = P( B) P( A) 3.) In general, A B = B ( A B ), A = ( A B) ( A B ), and hene P( A B) = P( B) + P( A B ), P( A) = P( A B) + P( A B ). Then, 4.) When there are three events: P( A B) = P( A) + P( B) P( A B) P( A B C) = P( A) + P( B) + P( C) P( A B) P( A C) P( B C) + P( A B C) 231

37 Mathale PSet Stats, Conepts In Statistis If P( A B) = P( A) P( B), then A and B are independent. If A B = φ, then If P ( A B) = P( φ) = 0 P ( A B) = P( A) + P( B) P( A B) = P( A) + P( B), then A and B are mutually exlusive or disjoint. Mutually exlusive and independene are very different onepts. The partitions of sample spae an be expressed in a two-way or a ontingeny table. Let C C = Ω and R R = Ω, the two different partitions of the sample spae, then the two-way table is Row\Column C C Total R P( R C) P( R C ) P (R) R ( R C) P P( R C ) P ( R ) Total P (C) P ( C ) 1 Example Event A ours with probability 0.4. If events A and B are mutually exlusive, then (A) P( B) 0.4 (B) P( B) 0.4 (C) P( B) 0.6 (D) P( B) 0.6 (E) P( B ) = 0.6 Solution: The answer is D. The sample spae may inlude other events. For example, Ω = A + B + C 1 = P( A) + P( B) + P( C) P( B)

38 Mathale PSet Stats, Conepts In Statistis Example [MC0223] Whih of the following statements is true for two events, eah with probability greater than 0? (A) If the events are mutually exlusive, they must be independent. (B) If the events are independent, they must be mutually exlusive. (C) If the events are not mutually exlusive, they must be independent. (D) If the events are not independent, they must be mutually exlusive. (E) If the events are mutually exlusive, they annot be independent. Solution: The answer is E. Let the events A and B, then answer E implies P ( A B) = 0, and independent means P ( A B) = P( A) P( B) = 0. Sine neither of P(A) nor P (B) is zero, so the two events annot be independent. Example [MC9713] Joe and Matthew plan to visit to this bookstore, the probability distribution of the number of books they will buy are given below. Assume that Joe and Matthew make their deisions independently, what is the probability that they will purhase no books on this visit to the bookstore? Solution: P ( no _ book) = 0.5(0.25) =

39 Mathale PSet Stats, Conepts In Statistis Example Roll a pair of fair die. a.) What is the probability that the sum of the numbers is 7 or 11? b.) What is the probability that both die either turn up the same number or that the sum of the numbers is less than 5? Solution: a.) Let A = sum is 7, B = sum is 11. Then, A = {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1 )}, B = {(5,6),(6,5)}, and A B = φ. Beause of the outomes are equally likely, then m( A) m( B) P( A B) = P ( A) + P( B) = m( Ω) + m( Ω) = = 9 b.). Let A = both die turn up the same number, B = sum is less than 5. Then, A = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6 )}, B = {(1,1),(1,2),(1,3),(2,1),(2,2),(3,1 )}, and A and B are not mutually exlusive. That is, A B = {(1,1),(2,2 )}. The probability P( A B) is then m( A) m( B) m( A B) P( A B) = P ( A) + P( B) P( A B) = + = + = m( Ω) m( Ω) m( Ω) Example A fair oin is tossed four times. What is the probability of getting at least a Tails? Solution: Let A = at least a Tails. Then A = all Heads. Ω = { HHHH, HHHT,, TTTT} and A = {HHHH}, and 234

40 Mathale PSet Stats, Conepts In Statistis 1 P ( A) = 1 P( A ) = 1 = 16 Example Seven blue marbles and six red marbles are held in a single ontainer. Marbles are randomly seleted one at a time and not returned to the ontainer. If the first two marbles seleted are both blue, what is the probability that at least two red marbles will be hosen in the next three seletions? Solution: Let T 2 = two marbles hosen, T 3 = Three marbles hosen. Then 6C2 5C1 + 6C3 P( T2 T3 ) = P( T2 ) + P( T3 ) P( T2 T3 ) = P( T2 ) + P( T3 ) = = C Example Jon and Jim eah sits in a row of 7 hairs. They hoose their seats at random. What is the probability that they don t sit next to eah other? Solution: The problem beomes hoose two seats from seven seats, exept six ases that they sit 7C2 6C1 5 next to eah other. So, p = =. C Example You are given P( A B) = 0.7 and P( A B ) = 0.9, what is P( A ) =? 235

41 Mathale PSet Stats, Conepts In Statistis Solution: P( A) = 1 P( A ) = 1 [ P( A B) + P( A B )] = 1 [1 P(( A B) ) + 1 P(( A B ) )] = P( A B ) 1 + P( A B) = = 0.6 A A B = 0.1 B = 0.3 Total = = Total Example An urn ontains 10 balls: 4 red and 6 blue. A seond urn ontains 16 red balls and an unknown number of blue balls. A single ball is drawn from eah urn. The probability that both balls are the same olor is Calulate the number of blue balls in the seond urn. Solution: Let B 1 = selet blue ball from the 1 st urn, B = selet blue ball from the 2 nd urn, 2 R = selet red ball from the 1 st urn, 1 R 2 = selet red ball from the 2 nd urn, x = number of blue balls in 2 nd urn. 236

42 Mathale PSet Stats, Conepts In Statistis Then, Therefore, x = 4. That is, there are 4 blue balls in the 2rd urn. Example Four students are seleted from a group of two 6 th graders, three 7 th graders and four 8 th graders. What is the probability that at least one student is seleted from eah grade? Solution: There are three ases: 1.) two 6 th graders, one 7 th grader and one 8 th grader: 2 C 2 3 C 1 4 C 1, 2.) one 6 th grader, two 7 th graders and one 8 th grader: 2 C 1 3 C 2 4 C 1, 3.) one 6 th grader, one 7 th grader and two 8 th graders: 2 C 1 3 C 1 4 C 2, p C C C + C C C + C C C C = =

43 Mathale PSet Stats, Conepts In Statistis Quik-Chek Probability of Events QC What is the probability that a number seleted at random from the first 50 positive integers is exatly divisible by 3 or 4? QC A bag has four red and six blue marbles. A marble is seleted and not replaed, and then a seond marble is seleted. What is the probability that both are the same olor? QC [MC1216] A omplex eletroni devie ontains three omponents, A, B and C. The probabilities of failure for eah omponent in any one year are 0.01, 0.03, and If any one omponent fails, the devie will fail. If the omponents fail independently of one another, what is the probability that the devie will not fail in one year? QC Given two events T and A, if P( T ) = 0.24, P( A ) = 0.50 and P( T A) = 0.56, what is P( T A)? Use both formula and the probability table to answer the question. 238

44 Mathale PSet Stats, Conepts In Statistis QC [MC1518] Suppose study partiipants are asked two questions. Based on a large number of partiipants, the probability distribution for the responses to Questions 1 and 2 is given below. Q2\Q1 Corret Inorret Corret Inorret For one partiipant, let A=Question 1 is answered orretly and B= Question 2 is answered orretly. Whih of the following must be true? (A) A and B are independent beause eah question is orret with probability 0.50, and the probability that both questions are answered orretly is (0.5)(0.5)=0.25. (B) A and B are not independent beause the probability in eah ell is not (C) A and B are not independent beause ( 0.33)(0.45) (D) A and B are independent beause one partiipant s answers do not affet another partiipant s answers. (E) There is not enough information provided to determine whether A and B are independent. 239

45 Mathale PSet Stats, Conepts In Statistis Answers QC Let A = the number that is divisible by 3, B = the number that is divisible by 4, and 50 E = A B. The sample spae is Ω = { 1, 2,3,,50}. There are = 16 3 integers that are divisible by 3, = 12 4 integers that are divisible by 4, and = 4 12 integers that are divisible by P ( E) = P( A) + P( B) P( A B) = + = QC C2 6C p = P( two red) + P( two blue) = + = = = C C QC Let A= A fails, B= B fails and C= C fails. P( A ) = 0.01, P( B ) = 0.03, P( C ) = 0.04, P( A B C) = P( A) P( B) P( C). The question is P( A B C ) =? P( A B C ) = P( A ) P( B ) P( C ) = (1 0.01)(1 0.03)(1 0.04) = QC = P( T A) P( T A) = = From the given, the probability table an be onstruted (the boxed numbers are given): A \ T T T Total A = = A = = = 0.5 Total = From the table P( T A) = QC C. P( A) = 0.33, P( B) = 0.45, P( AB) = 0.25 P( A) P( B) P( AB). 240

46 Mathale PSet Stats, Conepts In Statistis 5.5. Conditional Probability Let events A and B be in sample spae Ω, and suppose that P( A ) > 0. The onditional probability of B when A is given is: P( A B) P( B A) = P( A) or written as the multipliation rule: P( A B) = P( B A) P( A) P( A ), the unonditional probability, is alled marginal probability. Similar formula an be obtained for the probability of A when B is given: P ( A B) = P( A B) P( B) When event B is independent of event A, P( A B) = P( A) P( B) or P( B) = P( B A) or P ( A) = P( A B). Note that P ( A B) P( B) = ( B A) P( A). The information of onditional probability an be expressed in a tree diagram: 243

47 Mathale PSet Stats, Conepts In Statistis The onditional probabilities an also be alulated from the information provided from the two-way ontingeny table. Example A sample of 150 plasti panels was seleted and tested for shok resistane and srath resistane. The results are summarized in the table below Srath \ Shok H H Total C C Total A panel is seleted at random. If H denotes the event that it has high shok resistane, and C denotes the event that it has high srath resistane, explain and alulate the following probabilities. Write the results in the fration form without simplifiation. a.) P( H ) =, omment: b.) P( H ) =, omment:.) P( C ) =, omment: d.) P( C ) =, omment: e.) P( H C) =, omment: f.) P( H C ) =, omment: g.) P( H C) =, omment: h.) P( H C ) =, omment: 244

48 Mathale PSet Stats, Conepts In Statistis i.) P( H C ) =, omment: j.) P( H C ) =, omment: k.) P( C H ) =, omment: l.) P( C H ) =, omment: m.) P( H C ) =, omment: n.) P( H C ) =, omment: o.) P( C H ) =, omment: p.) P( C H ) =, omment: q.) P( C H ) =, omment: r.) P( C H ) =, omment: s.) P( C H ) =, omment: t.) P( C H ) =, omment: 245

49 Mathale PSet Stats, Conepts In Statistis u.) Fill in the above probabilities/onditional probabilities in the tree diagram: Solution: a.) b.).) d.) e.) f.) g.) 132 P( H ) =, probability of high shok P( H ) =, probability of low shok P( C ) =, probability of high srath P( C ) =, probability of low srath P( H C) =, probability of high shok and high srath P( H C ) =, probability of high shok and low srath P( H C) =, probability of low shok and high srath

50 Mathale PSet Stats, Conepts In Statistis h.) i.) j.) k.) l.) 6 P( H C ) =, probability of low shok and low srath P( H C ) =, probability of high shok when it is high srath P( H C ) =, probability of high shok when it is low srath P( C H ) =, probability of high srath when it is high shok P( C H ) =, probability of high srath when it is low shok. 18 m.) n.) o.) p.) q.) 12 P( H C ) =, probability of low shok when it is high srath P( H C ) =, probability of low shok when it is low srath P( C H ) =, probability of low srath when it is high shok P( C H ) =, probability of low srath when it is low shok P( C H ) = =, probability of high shok or it is high srath r.) s.) t.) P( C H ) = =, probability of high shok or it is high srath P( C H ) = =, probability of low srath or it is high shok P( C H ) = =, probability of low srath or it is low shok

51 Mathale PSet Stats, Conepts In Statistis u.) Example Three tall and two short men went on a pini with four tall and four short women. Let M = man T = tall person then, if a person is randomly seleted, write the formula and alulate the following probabilities. Write the results in frations without simplifiation. a.) the person is tall, P( T ) = b.) the person is short,.) the person is a man, d.) the person is a woman, e.) the person is a tall man, 248

52 Mathale PSet Stats, Conepts In Statistis f.) the person is a short man, g.) the person is a tall woman, h.) the person is a short woman, i.) the person is a man, given the person is tall j.) the person is a woman, given the person is tall k.) the person is a man, given the person is short l.) the person is a woman, given the person is short m.) the person is tall, given the person is a man n.) the person is short, given the person is a man o.) the person is tall, given the person is a woman p.) the person is short, given the person is a woman q.) the person is either tall or a man r.) the person is either short or a man s.) the person is either tall or a woman t.) the person is either short or a woman 249

53 Mathale PSet Stats, Conepts In Statistis Solution: Height\People M M Total T T Total a.) the person is tall, 7 P( T ) = 13 b.) the person is short,.) the person is man, 6 P( T ) = 13 5 P( M ) = 13 d.) the person is woman, 8 P( M ) = 13 e.) the person is a tall man, 3 P( T M ) = 13 f.) the person is a short man, 2 P( T M ) = 13 g.) the person is a tall woman, 4 P( T M ) = 13 h.) the person is a short woman, 4 P( T M ) = 13 3 i.) the person is a man, given the person is tall P( M T ) = 7 4 j.) the person is a woman, given the person is tall P( M T ) = 7 250

54 Mathale PSet Stats, Conepts In Statistis k.) the person is a man, given the person is short 2 P( M T ) = 6 l.) the person is a woman,, given the person is short 4 P( M T ) = 6 m.) the person is tall, given the person is a man 3 P( T M ) = 5 n.) the person is short,, given the person is a man o.) the person is tall, given the person is a woman 2 P( T M ) = 5 4 P( T M ) = 8 p.) the person is short, given the person is a woman 4 P( T M ) = 8 q.) the person is either tall or a man P( T M ) = = r.) the person is either short or a man P( T M ) = = s.) the person is either tall or a woman P( T M ) = = t.) the person is either short or a woman P( T M ) = =

55 Mathale PSet Stats, Conepts In Statistis Example Suppose that a migraine sufferer has a history that indiates the following probabilities. The probability that she will have a migraine on a random day is 25%. If she has a migraine one day, the hane that she will have a migraine the next day is 40%. If she does not have a migraine, the hane that she will have a migraine the next day is 20%. What is the probability that she has a migraine on exatly one of the two days? Solution: Let M 1 = have a migraine on day one and M 2 = have a migraine on day two. P( M 1) = 0.25, P( M 2 M 1) = 0.40, P( M 2 M 1 ) = The question is ( ) ( ) P ( M M ) ( M M ) = P M M + P( M M ) =? From the tree diagram, P ( M1 M 2 M 2 M1 ) ( ) ( ) = =

56 Mathale PSet Stats, Conepts In Statistis Example Following is a breakdown of the favorite sport and the gender of the respondents to a survey. Let F be the event that a randomly seleted respondent is female, and let S be the event that a randomly seleted respondent s favorite sport is soer. Football Basketball Hokey Baseball Soer Other Total Male Female Total a.) Given that a respondent is female, what is the probability that her favorite sport is baseball? b.) Given that the favorite sport is football, what is the probability that the respondent is male?.) Are events F and S independent? Solution: a.) b.).) 205 P( Baseball Female ) = P( Male Football ) = P( Female Soer) = P( Female) = P( Female Soer) P( Female) No, they are not independent events. 253

57 Mathale PSet Stats, Conepts In Statistis Example In a ertain town, 48% of the registered voters onsider themselves to be republian, and the rest onsiders themselves to be demorat. Overall, it is known that 64% of demorats favor a ertain proposal, while only 47% of republians favor the same proposal. Let R be the republian and let F be the person who favor the proposal, then a.) Write the givens in the two-way table and tree diagram. R b.) Given that a randomly seleted individual is against the proposal, what is the probability that the individual onsiders himself/herself to be demorat? Solution: the boxed numbers are given, Favor\Part y R R Total F = = = F = = = Total =

58 Mathale PSet Stats, Conepts In Statistis b.) P( R F ) = = Example If A and B are two independent events with P( A) = 0. 2 and P ( B) = 0. 4, then P( A B ) is? Solution: ( P B ) = 0.6 and P ( A B ) = P( A) P( B ) = =

59 Mathale PSet Stats, Conepts In Statistis Quik-Chek Conditional Probability QC [MC1234M] The probability that a new mirowave oven will stop working in less than 2 years is The probability that a new mirowave oven is damaged during delivery and stops working in less than 2 years is The probability that a new mirowave oven is damaged during delivery is Translate eah of the following statements into events using S = stop working in less than two years D = damaged during delivery and alulate the probability of eah question. S a.) given that a new mirowave oven is damaged during delivery, what is the probability that it stops working in less than 2 years? (Original question) b.) the probability that the mirowave stopped working in less than two years and it was not damaged during delivery..) the probability that the mirowave did not stop working in less than two years and it was damaged during delivery. 256

60 Mathale PSet Stats, Conepts In Statistis d.) the probability that the mirowave did not stop working in less than two years and it was not damaged during delivery. e.) given that a new mirowave oven is not damaged during delivery, what is the probability that it stops working in less than 2 years? f.) given that a new mirowave oven stops working in less than 2 years, what is the probability that it is damaged during delivery? QC An insurane ompany examines its pool of auto insurane ustomers and gathers the following information: 1.) All ustomers insure at least one ar. 2.) 70% of ustomers insure more than one ar. 3.) 20% of ustomers insure a sports ar. 4.) Of those ustomers who insure more than one ar, 15% insure a sports ar. Calulate the probability that a randomly seleted ustomer insures exatly one ar and that ar is not a sports ar. 257

61 Mathale PSet Stats, Conepts In Statistis QC A publi health researher examines the medial reords of a group of 937 men who died in 1999 and disovers that 210 of the men died from auses related to heart disease. Moreover, 312 of 937 men had at least one parent who suffered from heart disease, and, of these 312 men, 102 died from auses related to heart disease. Calulate the probability that a man randomly seleted from this group died of auses related to heart disease, given that neither of his parents suffered from heart disease. 258

62 Mathale PSet Stats, Conepts In Statistis Answers QC The givens are P( S ) = 0.05, P( D S) = 0.04, P( D ) = Those numbers are boxed in the table a.) Damaged\Stopped S S Total D = D = = = 0.9 Total = P( D S) 0.04 P( S D) = = = 0.4, or the result an be obtained from the table P( D) 0.1 b.) P( SD ) = 0.01.) P( DS ) = 0.06 d.) ( P D S ) = e.) P( S D ) = = f.) P( D S ) = = QC The boxed numbers are given. 20.5% of ustomers insure exatly one ar and that ar is not sports ar. Type\Number One ar More than one Total Sports ar = i 0.15 = Regular ar? = = Total = QC Boxed numbers are given. 108 P( Heart _ Disease Parents _ Not _ Suffered ) = = Parents\Heart or not Heart Disease Not Heart Disease Total Parents Suffered i 0.15 = Parents Not Suffered = = 625 Total

63 Mathale PSet Stats, Conepts In Statistis 5.6. Total Probability and Bayes Theorem [MATH] The Law of Total Probability If C, C, 1 2, Cm are non-empty mutually exlusive events that partition the sample spae C1 C2 Cm = Ω, The probability of any event A an be expressed as: Bayes Theorem P( A) = P( A C ) P( C1) + P( A C2 ) P( C2 ) + + P( A C ) P( C 1 m m ) For any event Ci where i = 1, 2,, m : P( C i A) = P( A C ) P( C ) + P( A C P( A C ) P( C ) i ) P( C ) + + P( A C ) P( C m m i ) Example In Orange County, 51% of adults are males, 9.5% of male smoke igars, where 1.7 % of female smoke igars. If an adult is seleted and it is found out the subjet is a smoker, what is the probability that the subjet is a male? Solution: Let A = men, B = smokers. Then the probability that the subjet is a male if the subjet is a smoker is P ( A B). From the given, P ( A) = 0. 51, P ( B A) =. 095, P ( B A ) =.017. Use the tree diagram: 262

64 Mathale PSet Stats, Conepts In Statistis Use the Bayes Theorem: P( A B) = P( B A) P( A) = P( B A) P( A) + P( B A = (1 0.51) Use the two-way ontingeny table: P( B A) P( A) P( B) ) P( A ) Subjet\Smoke B (Smoke) B (Non-Smoke) Total A (Male) 9.5% i 51% = 4.845% 0.51 A (Female) 1.7% i 0.49 = 0.833% = 0.49 Total 4.845% % = 5.678% 1 P( A B) 4.845% P( A B) = = = P( B) 5.678% 263

65 Mathale PSet Stats, Conepts In Statistis Quik-Chek Total Probability and Bayes Theorem Use Bayes theorem, the Tree Diagram and the Two-Way Contingeny Table to solve eah of the following problems. QC [CB] All bags entering a researh faility are sreened. Ninety-seven perent of the bags that ontain forbidden material trigger an alarm. Fifteen perent of the bags that do not ontain forbidden material also trigger the alarm. If 1 out of every 1,000 bags entering the building ontains forbidden material, what is the probability that a bag that triggers the alarm will atually ontain forbidden material? QC Internet addition has been defined as a disorder, eah partiipant in a study was assessed. The survey results showed 1.). 51.8% of study partiipants were female 2.) 13.1% of female suffered from internet addition. 3.) 24.8% of the males suffered internet addition. If a partiipant in the survey was addited, what is probability that the partiipant was a female? 264

66 Mathale PSet Stats, Conepts In Statistis Answers QC P( T F ) = 0.97 P( T F ) = 0.15 P( F ) = P( T ) = P( T F) P( F) + P( T F ) P( F ) = 0.97i i = P( T F) P( F) = P( F T ) P( T ) P( T F) P( F) P( T F) P( F) P( F T ) = = = = P( T ) P( T F) P( F) + P( T F ) P( F ) Trigger\Forbidden F (Forbidden) F (Allowed) Total T (Triggered) 15% i0.999 = T (Not Triggered) Total = 1 QC Let I = The partiipant was suffered from internet addition F = A female was seleted Then, the question beomes P( F I ) =? From the given: P( F ) = 0.518, P( F ) = = 0.482, P( I F ) = 0.131, P( I F ) =