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Probability

Probability. Experiments, Sample Spaces & Events. Definitions. 1. Experiment An activity with observable outcomes (results). A special case of experiments are those in which the considered outcomes involve chance. Examples:

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Probability

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  1. Probability

  2. Experiments, Sample Spaces & Events

  3. Definitions 1. Experiment An activity with observable outcomes (results). A special case of experiments are those in which the considered outcomes involve chance. Examples: 1. Tossing a coin with the considered outcomes are the coin falling head or tail. we will refer to this experiment by the usual tossing of a coin experiment. 2. Rolling a dice with the considered outcomes are the numbers of the dots (the integers from 1 to 6) showing up. we will refer to this experiment by the usual dice experiment.

  4. 2. Sample Space A simple space is the set of possible outcomes of an experiment. Examples (1): 1. In the usual experiment of tossing a coin with the considered outcomes being the coin falling head or tail, the sample space contains two elements: head and tail. We write: S = { head , tail } 2. In the usual experiment of rolling a dice with the considered outcomes are the number of dots (the integers 1 to 6) showing up (on the top face), the sample space contains 6 elements: the integers from 1 to 6. We write: S = {1,2,3,4,5,6}

  5. 2. Event An event is ny subset of the sample space of an experiment. Examples(2) : 1. In the usual experiment of tossing a coin with the considered outcomes are the coin falling head or tail, where the sample space is S = { head , tail }, each of the following subsets of S is an event: Φ, S, { tail }, { head }. 2. In the usual experiment of rolling a dice with the considered outcomes are the number of dots showing up, where the sample space is S = {1,2,3,4,5,6}, each of the following subsets of S is an event: Φ, S, {1} , {2} , {3} ,…… {6} , {1,2} , {1,3} ,….. , {5,6} , {1,2,3} , {1,2,4},…….., {4,5,6} , {1,2,3,4} , {1,2,3,5} ,……,{3,4,5,6} , {1,2,3,4,5} ,…. {1,2,3,4,6} , {2,3,4,5,6}.

  6. Notice 1. Notice that within a given experiment, the sample space is the universal set containing all of the considered outcomes for that experiments and the events are the subsets of this universal set, this includes the empty set and the sample space itself. 2. The sample space is referred to as the certain event. It is called certain, because it must occur, since there is no outcome not belonging to this set. 2. The empty set is referred to as the impossible event. It is called impossible, because it can not occur, since no outcome belongs to this set.

  7. Simple Events Let S = {s1, s2, s3,……,sn} be a sample space. A simple event in singleton subset ( a subset consisting of exactly one element) of S. Thus, {sk} is a simple event for any positive integer k ≤ n. That’s all of the following events are simple events: {s1} , {s2} , {s3} , {s4} ,….,{sn}

  8. The algebra of events 1. Since events are subsets of a universal set (the sample space, which is the universal set here), then all set operations learned in section IV-01 apply. Thus we can talk about, union and intersection of events and a complement of an event. 2. A nonintersecting events ( that’s having an empty intersection; that’s, the intersection of them is equal to Φ) are referred to as mutually exclusive, because they can not happen at the same time. For example, the event A = {1,2,3} and B = {5,6}, in the In the usual dice experiment, are mutually exclusive, since their intersection is the empty set. If the top face is one of the three numbers (of dots) belonging to the set A, it certainly, can not be one of the numbers belonging to the set B, since there is no common element of these two sets. Notice that if A is a subset of the complement of B, then no element of A belongs to B, which means that the intersection of A and B is empty, and so they are mutually exclusive. This will include the case when A is equal to BC.

  9. Example (3) 1. In the usual dice experiment, let A, B and C, be the following events: A = {1,4,6} and B = {1,2,4,5} and C = {3,5} Find the following: 1. The union and intersection of the events A and B. 2. The complement of the event A. 3. Which two of the sets A, B and C are mutually exclusive. Solution: 1. AUB = {1,4,6} U {1,2,4,5} = {1,2,4,5,6} A∩B = {1,4,6} ∩ {1,2,4,5} = {1,4} 2. AC = S - {1,4,6} = {1,2,3,4,5,6} - {1,4,6} = {2,3,5} 3. Since, A∩C = {1,4,6} ∩ {3,5} = Φ, then A and C are mutually exclusive. Notice that C is a subset of the complement of A C = {3,5} is a subset of AC = {2,3,5}

  10. Example (4) In the experiment consisting of tossing a coin three times and considering the sequence of heads and tails outcomes. Find the following: 1. The sample space 2. The event E that exactly two heads appear. 3. The event F that at least one tail appears. 4. The event G that two heads appear in a sequence. 5. The event A that three heads appear. Solution: First let’s study the tree diagram of this experiment.

  11. H (H,H,H) H T (H,H,T) H H (H,T,H) T T (H,T,T) Third Toss Second Toss H (T,H,H) First Toss H T (T,H,T) T H (T,T,H) T T (T,T,T)

  12. Solution 1. The sample space S = {(H,H,H), (H,H,T), (H,T,H) , (H,T,T), , (T,H,H), , (T,H,T), , (T,T,H), , (T,T,T } Convention: For the sake of brevity, we will use the notation HHH for (H,H,H), HHT for (H,H,T)……etc. Thus: S = { HHH, HHT, HTH, HTT,THH, THT, TTH, TTT} 2. E (the event that exactly 2 heads appear) = { HHT,HTH,THH} 3. F (the event that at least one tail appears) = {HHT, HTH, HTT,THH, THT, TTH, TTT}

  13. 4. The event G that two heads appear in sequence = { HHT,THH,HHH} 5. The event A that three heads appear = {HHH}

  14. Example (5) In the experiment consisting of tossing a pair of dice and considering the numbers of dots falling uppermost on each dice . Find the following: 1. The sample space 2. The events En that the sum of the numbers of dots falling uppermost is n, for n = 2,3,4, ….,12 First study the table on the next slide

  15. Sum ofuppermost numbers Event

  16. Solution 1. The sample space S = { (k,m) : where k and m are the numbers of dots appearing uppermost on the first and the second dice respectively} = {(1,1), (1,2), ….(1,6), (2,1), (2,2),…..,(2,6),…..(6,1), (6,2),…..(6,6) } 2. E2 = {(1,1)} E3 = {(1,2), (2,1)} E4 = {(1,3), (2,2), (3,1)}, Find , E5 and E6 E7 = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} E10 = {(4,6), (5,5), (4,6)}, Find E8 , E9 E11 = {(5,6), (6,5)} E12 = {(6,6)}

  17. Example (6) In the experiment consisting of guessing the number of tickets that will be sold for a play . If there are 40 seats in the theater. Find the following: 1. The sample space 2. The event E that fewer than 30 tickets will be sold. 3. The event F that at least half of tickets will be sold.

  18. Solution The number of the tickets that can be sold could run from 0 to 40 1. The sample space S = { m : m the number of tickets that could be sold} = {0,1,2,3,4,5……….,40 } 2. E (the event that fewer than 30 tickets will be sold) = { 0,1,2,3,…….29} 3. E (the event that the theater will be at least half full, i.e. at least half of the tickets will be sold) = { 20,21,22,23,…….40}

  19. Example (7)Infinite Sample Space In the experiment consisting of guessing the distance that an animal travels far away from its original home. 1. The sample space 2. The event E that the animal travels no more than 1000 KM away from home. 3. The event F that the animal travels more than 1000 KM away from home. 4. The event G that the animal travels between 500 KM and 1000 Km (inclusive) away from home.

  20. Solution 1. The distance covered by the animal away from home is a nonnegative number The sample space S = {x: x = the distance the animal travels away from home} = { x: x ≥ 0 } = [0,∞) This is an infinite set 2. E (The event E that the animal travels no more than 1000 KM away from home) = {x: 0 ≤ x ≤ 1000 } = [0 , 1000] 3. F (the event that the animal travels more than 1000 KM away from home) = {x: x > 1000 } = (1000,∞) 4. G (the event that the animal travels between 500 KM and 1000 Km (inclusive) away from home) = {x: 500 ≤ x ≤ 1000 } = [500 , 1000]

  21. Probability

  22. Relative Frequency & Empirical Probabilityof an Event & Probability Distribution 1. In an experiment, repeatable under independent and similar condition, if in n trails, an event E occurs m times, then the ratio m/n is called the relative frequency of the event E. 2. If the relative frequency of an event E approaches some value as n becomes larger and larger, then this value is called the empirical probability of the event E, and denoted by p(E). An event’s probability is a measure of the proportion of the time that the event will occur. The meaning of the phrase “the relative frequency of E approaches some value”, can be understood by studying the table on the next slide.

  23. The relative frequency of a coin’s toss results in showing tail – approaching 0.5 = ½

  24. Probability Function Let S = {sk : k = 1, 2, 3, ….,n} = {s1 , s2 , s3 , s4 ,………., sn } be a sample space 1. The function p assigning probability to an event E is called a probability function. The probability p({sk}) of the simple event {sk} is written in short as p(sk) , for any k = 1, 2, 3, ….,n. 2. The property function has the following rules: a. 0 ≤ p(sk) ≤ 1 ; k = 1, 2, 3, ….,n b. p(s1) + p(s2) +……. + p(sn) = 1 C. p( {si} U {sj} ) = p(si) + p(sj) ; i not equal j

  25. Probability distribution tableThe table assigning probabilities to each of the simple events is called a Probability distribution for the experiment.

  26. General Rules of Probability 1. The property function has the following rules: a. p(Φ) = 0 ≤ p(E) ≤ 1 = p(S) ; for any event E in the sample space S 2. p(EUF) = p(E) + p(F) – p(E∩F) Special case: If E & F are mutually exclusive, then: p(EUF) = p(E) + p(F) (Why?) 3. p(EC) = 1 – p(E)

  27. Probability in a Uniform (Equiprobable) Sample Space 1. A sample space is said to be uniform (equiprobable) if simple events are equally likely to occur. 2. If S is uniform (equiprobable) and n(S) = k, then: p({s}) = 1 / k; s in S.

  28. Example (8) For the experiment of the usual rolling of the dice: 1. Exhibit the probability distribution table 2. Compute the probability that the dice shows an odd number of dots. 3. Compute the probability that the dice shows less than 5 dots. Solution: 1. The sample space S = {1,2,3,4,5,6} consists of six outcomes. Thus each simple event is to be assigned a probability of 1/6. See the probability distribution table on the next slide

  29. 2. The event E that the dice shows an odd number of dots = {1 , 3 , 5 } → p(E) = p({1}) + p({3}) + p({5}) =1/6 +1/6 + 1/6 = 3(1/6) = ½ 3. The event F that the dice shows less than 5 dots = { 1, 2 , 3 , 4 } → p(E) = p({1}) + p({2}) + p({3}) + p({4}) =1/6 +1/6 + 1/6 + 1/6 = 4(1/6) = 2/3

  30. Example (9) A pair of dice is rolled. Calculate the following probabilities: 1. The probability that the dice show the same number. 2. The probability that the sum of the numbers shown by the two dice is 10. 3. The probability that the sum of the numbers shown by the two dice is 12. 4. The probability that the number shown on one dice is exactly five times that shown on the other.

  31. Solution: See Example (5) for the sample space S of this experiment. The sample spaces S contains (consists of) 36 outcomes, hence each simple event is to be assigned the probability 1/36. 1. The event E that the two dice show the same number is: E = { ( 1, 1 ) , ( 2, 2 ) , ( 3, 3 ) , ( 4, 4 ) , ( 5, 5 ) , ( 6, 6 ) } This event contains 6 element outcomes). Since (E is the union of 6 simple events), then p(E) = p({( 1, 1 ) }) + p({( 2, 2 ) }) + p({( 1, 1 ) }) + …….p({( 6, 6 ) }) = 1/36 + 1/36+ ……+1/36 ( 6 term) = 6(1/36) = 1/6

  32. 2. The probability that the sum of the numbers shown by the two dice is 10. The event F that the sum of the numbers shown by the two dice is 10 is: F = { ( 4, 6 ) , ( 5, 5 ) , ( 6, 4 ) } This event contains 3 element. Since (F is the union of 3 simple events), then p(F) = p({( 4, 6 ) }) + p({( 5, 5 ) }) + p({( 6, 4 ) }) = 1/36 + 1/36 + 1/36 = 3(1/36) = 1/12

  33. 3. The probability that the sum of the numbers shown by the two dice is 12. The event G that the sum of the numbers shown by the two dice is 12 is the singleton set: G = { ( 6, 6 ) } This event contains only one element. p(G) = p({( 6, 6 ) }) = 1/36

  34. 4. The probability that the number shown on one dice is exactly five times that shown on the other. The event H that the number shown on one dice is exactly 5 times that shown on the other is the set: H = { ( 1, 5) , ( 5, 1 )} This event contains 2 elements (outcomes). p(H) = p({ ( 1, 5 )}) + p({ ( 5, 1 )}) = 1/36 + 1/36 = 2(1/36) = 1/18

  35. Example (10) The next slide shows the data gathered from an experiment of 200 test runs of the distance a prototype electric car is able to cover, using a fully charged battery of a certain brand. 1. Find the sample space S of this experiment. 2. Find the empirical property distribution of this experiment. 3. Compute the probability that the such a car will be able to cover more than 150 KM. 4. Compute the probability that the such a car will be able to cover no more than 150 KM

  36. Solution 1. The sample space S of this experiment = { s1 , s2 , s3 , s4 , s5 , s6 } S1 denotes the outcome that the car was able to cover a distance no more than 50 KM S2 denotes the outcome that the car was able to cover a distance greater than 50 KM, but less or equal 100 KM. S3 denotes the outcome that the car was able to cover a distance greater than 100 KM, but less or equal 150 KM S4 denotes the outcome that the car was able to cover a distance greater than 150 KM, but less or equal 200 KM S5 denotes the outcome that the car was able to cover a distance greater than 200 KM, but less or equal 250 KM S6 denotes the outcome that the car was able to cover a distance more than 250 KM

  37. 2. Find the empirical probability distribution of this experiment. p(s1) = relative frequency of s1 = number of trails in which s1 occurs / total number of trails = 4 / 200 = 2 / 100 = 0.02 p(s2) = 10 / 200 = 5 / 100 = 0.05 p(s3) = 30 / 200 = 15 / 100 = 0.15 p(s4) = 100 / 200 = 50 / 100 = 0.50 p(s5) = 40 / 200 = 20 / 100 = 0.20 p(s6) = 16 / 200 = 8 / 100 = 0.08 This results in the probability distribution table of the next slide

  38. 3. The event E that such a car will be able to cover more than 150 KM is: E = {s4 , s5 , s6 } → p(E) = p(s4) + p(s5) + p(s6) = 0.50 + o.20 + 0.08 = 0.78 4. The event F that such a car will be able to cover no more than 150 KM is: F = {s1 , s2 , s3 } → p(E) = p(s1) + p(s2) + p(s3) = 0.02 + o.05 + 0.15 = 0.22 Notice that F = EC and that p(F) = 1 – p(E) = 1 0.78 = 0.22

  39. Example (11) The next slide shows the probability distribution with a final exam scores of a Math course. If we select at random a student who has done the exam, what is the probability that her score will be: 1. More than 40. 2. Less than or equal 50 3. greater than 40 but less or equal 70 4. greater than 40 but less or equal 60

  40. Solution 1. The sample space S of this experiment = { s1 , s2 , s3 , s4 , s5 , s6 } S1 denotes the outcome that the score is greater than 70 S2 denotes the outcome that the score is greater than 60, but less or equal 70 S3 denotes the outcome that the score is greater than 50, but less or equal 60 S4 denotes the outcome that the score is greater than 40, but less or equal 50 S5 denotes the outcome that the score is greater than 30, but less or equal 40 S6 denotes the outcome that the score is less or equal 30

  41. 1. The event E that the score is greater than 40 E = { s1 , s2 , s3 , s4 } → p(E) =p(s1) + p(s2) + p(s3) + p(s4) = 0.01 + 0.07 + 0.19 +0.23 = 0.5 = ½ 2. The event F that the score is less than or equal 50 F = { s4 , s5 , s6 } → p(F) =p(s4) + p(s5) + p(s6) = 0.23 + 0.31 + 0.19 = 0.73 3. The event G that the score is greater than 40 but less than or equal 70 F = { s4 , s3 , s2 } → p(F) = p(s4) + p(s5) + p(s6) = 0.23 + 0.19 + 0.07 = 0.49 4. The event H that the score is greater than 40 but less or equal 60 H = {s3 , s4 , } → p(H) = p(s3) + p(s4) = 0.19 +0.23 = 0.42

  42. Example (12) A card is drawn from a deck of 52 playing cards. What’s the probability that it is a king or a diamond. Solution: The sample space consists of 52 outcomes. Thus, there are 52 simple events, each with a property equal to 1/52. The event E that the outcome is a king consists of 4 outcomes → p(E) = 1/52 + 1/52 + 1/52 + 1/52 = 4 (1/52) = 4/52 = 1/13 The event F that the outcome is a diamond consists of 13 outcomes → p(F) = 1/52 + 1/52 + …… + 1/52 = 13 (1/52) = 13/52 = ¼ Notes: 1. The event H that the outcome is a king or a diamond = EUF We have: p(EUF) = p(E) + p(F) – p(E∩F) = 4/52 + 13/52 – 1/52 = 16/52 = 4/13 2. The event G that the outcome is a king and a diamond consists of one outcome → p(G) = 1/52. Notice that G = E∩F.

  43. Example (13) The data shows that 3% of a brand of TV sold experience video problems, 1% audio problems and 0.1% both type of problems before the expiration of the warranty. What’s the probability that a purchased such TV 1. will experience video or audio problems before the expiry of the warranty. 1. will not experience video or audio problems before the expiry of the warranty Solution: 1. Let E denote the event that the TV will experience video problems and F the event that the TV will experience audio problems. Then E∩F is the event that the TV will experience both video and audio problems, while EUF is the event that the TV will experience video or audio problems. → p(E) = 0.03 , p(F) = 0.01 and p(E∩F) = 0.001 We have; p(EUF) = p(E) = p(F) – p(E∩F) = 0.03 + 0.01 – 0.001 = 0.040 -0.001 = 0.039 2. Let G denote the event that the TV will not experience video or audio problems, then G = (EUF)C , and so p(G) = 1 - p(EUF) = 1 – 0.039= 0.961

  44. Example (14) Let E & F be mutually exclusive events and that p(E) = 0.1 and p(F) = 0.6. Compute: 1. p(E∩F) 2. p(EUF) 3. p(EC) 4. p(EC∩FC) 5. p(ECUFC) Solution: 1. p(E∩F) = 0, since E∩F = Φ 2. p(EUF) = p(E) + p(F) = 0.1 + 0.6 = 0.7 3. p(EC) = 1 – P(E) = 1 – 0.1 = 0.9 4. EC∩FC = (EUF)C → p(EC∩FC)= p(EUF)C = 1 - p(EUF) = 1 – 0.7 = 0.3 5. ECUFC = (E ∩ F)C =(Φ) C = S → p(EC∩FC)= p(S) = 1 Another way: ECUFC = (E ∩ F)C =(Φ) C → p(EC∩FC) = p (Φ) C= 1 -p (Φ) = 1 – 0 =1

  45. Example (16) Let p(E) = 0.2 , p(F) = 0.1 and p(E∩F) = 0.05. Compute: 1. p(EUF) 2. p(EC∩FC) Solution: 1. p(EUF) = p(E) + p(F) - p(E∩F) = 0.2 + 0.1 - 0.05 = 0.25 2. EC∩FC = (EUF)C → p(EC∩FC)= p(EUF)C = 1 - p(EUF) = 1 – 0.25 = 0.75

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