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# Statistics lecture 5 (ch4)

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### Statistics lecture 5 (ch4)

1. 1.  What is probability?  Probability is a numerical value used to express the chance that a specific event will occur.  Probability is always in the interval 0 to 1 and can be expressed as percentages.  The greater the chance that an event will occur, the closer the probability is to 1.  The smaller the chance that an event will occur, the closer the probability is to 0.  Probabilities is needed to make generalisations about the population based 2 on a sample drawn from the population.
2. 2.  Experiment  Process that results in obtaining observations from an experimental unit.  Experimental unit is the object on which the observations are made.  Results of experiment are called outcomes.  Stochastic experiment  The results are a definite set of two or more possible outcomes.  The outcome can not be determined in advance.  Can be repeated under stable conditions. 3
3. 3.  Sample space  Collection of all possible outcomes of an experiment.  It is denoted by S.  List all possible outcomes inside braces. S={ } S 4
4. 4.  Event  Collection of some outcomes of the sample space.  It is denoted by A, B, C, etc.  List all possible outcomes inside braces. A={ }  Can have one or more outcomes. S A 5
5. 5.  Event  Can define more than one event for the same sample space.  Two events are mutually exclusive if they can not occur at the same time.  Events A and B are mutually exclusive. S B A 6
6. 6.  Event  Can define more than one event for the same sample space.  Two events are non mutually exclusive if they can occur at the same time.  Events A and C are non mutually exclusive. Probability of an event  Probability of Event A. S  P(A) A C 7
7. 7.  Properties of probability:  0 ≤ P(A) ≤ 1  P(B) = 0  B impossible event  P(C) = 1  C certain event  P(S) = 1  Compliment of Event A:  P(Ā) = 1 – P(A)  Events A and B mutually exclusive.  P(A or B) = P(A) + P(B) 8
8. 8. Three approaches to probability 1. Relative frequency approach  The probabilities of the outcomes differ.  Counting the number of times that an event occurs when performing an experiment a large number of times. number of times event A occurredP(event A) = number of times the experiment was repeated fP ( A) = n 9
9. 9. Three approaches to probability Relative frequency approach  Example  In a group of 20 tourists staying in a hotel, nine prefer to pay cash for their accommodation.  The probability that a tourist will pay cash for accommodation is: f 9 P ( A) = = = 0, 45 n 20 10
10. 10. Three approaches to probability  2. Classic approach The outcomes all have the same probabilities. Not necessary to performing an experiment a number of outcomes of experiment favourable to the eventP(eventA) = total number of outcomes of experiment fP ( A) = n 11
11. 11. Three approaches to probability Classic approach  Example  Chance to get an uneven number on a dice:  S = {1; 2; 3; 4; 5; 6}  F = {1; 3; 5} 1 1 1 3 P( F ) = + + = = 0,5 6 6 6 6 12
12. 12. Three approaches to probability 3. Subjective approach  The probabilities assigned to the outcomes of the experiment is subjective to the person who performs the experiment. 13
13. 13.  The word “or” in probability is an indication of addition  P(A or B) The word “and” in probability is an indication of multiplication  P(A and B) 14
14. 14. Addition rules for calculating probabilities Events are mutually exclusive when they have no outcomes in common. For mutually exclusive events:  P(A or B) = P(A) + P(B) = 4/21 + 3/21 = 7/21 B Events A and B A are mutually S exclusive 15
15. 15. Addition rules for calculating probabilities Events are mutually exclusive when they have no outcomes in common If events are not mutually exclusive  P(C or D) = P(C) + P(D) – P(C and D)  P(C and = 5/21 + 5/21 – 2/21 = 8/21 D)  Outcomes are included in C and D. Events C and D are not mutually exclusive S 16
16. 16. Conditional probability Need to know the probability of an event given another event has already occurred  The two events must be dependent.  Probability of A, given B has occurred: P ( A and B )  P(A|B) = , P(B) ≠ 0 P( B)  Multiplication rule: The outcome of one event affects the probability of the  P(A and B) = P (B)P(A|B)  Probability of B, given A has occurred: event. occurrence of another P ( A and B )  P(B|A) = , P(A) ≠ 0 17 P( A)
17. 17. Conditional probability Need to know the probability of an event given another event has already occurred.  If sampling without replacement takes place.  The events are considered dependent.  Example – Three students need to pick a biscuit from a plate with 10 biscuits.  1st student can pick any of the 10 biscuits.  2nd student has only 9 biscuits to pick from.  3rd student has only 8 biscuits to pick from.  The probability to pick the 1st biscuit is 1/10, the 2nd is 1/9 and the 3rd is 1/8. 18
18. 18. Independent Events Events are statistically independent if the outcome of one event does not affect the probability of occurrence of another event. P(A|B) = P(A) P(B|A) = P(B) Multiplication rule for dependent events:  P(A and B) = P(B)P(A|B) Multiplication rule for independent events:  P(A and B) = P(A)P(B) 19
19. 19. Counting rules Multi-step experiments  The number of outcomes for ‘k’ trails each with the same ‘n’ possible outcomes.  The number of outcomes in S = nk.  Example  How many ways can 10 multiple choice question with 4 possible answers be answered:  410 = 1 048 576 ways 20
20. 20. Counting rules Multi-step experiments  The number of outcomes for ‘j’ trails each with a different number of ‘n’ outcomes.  The number of outcomes in S = n1 x n2 … X nj  Example  Need to order a meal where you can pick ‘1’ burger from ‘8’, ’1’ cool drink from ’10’, ‘1’ ice cream from ‘5’.  Number of possible orders: 8×10×5 = 400 21
21. 21. Counting rules The factorial  The number of ways in which ‘r’ objects can be arranged in a row, without replacement.  r! = r×(r – 1)×(r – 2)× …×3×2×1  Note r! = 0! = 1  Example  Six athletes compete in a race. The number of order arrangements for completing the race.  6! = 720 different ways 22
22. 22. Counting rules Combination  Select r objects without replacement from a larger set of n objects, order of selection not important.  n! n Cr = r !(n − r )!  Example  Six lotto numbers should be selected form a possible 49 – order of selection not important.  C = C = 49! n r 49 6 = 13 983 816 6!(49 − 6)! 23
23. 23. Counting rules Permutation  Select r objects without replacement from a larger set of n objects, order of selection is important.  n! nP = r (n − r )!  Example  Six athletes in a race, how many ways to compete for the gold, silver and bronze medals.  6! n Pr = 6 P3 = = 120 (6 − 3)! 24
25. 25. Experiment: Draw a card from a play pack of cards Experimental unit: Play pack of cards Outcome of experiment: Any card from the packSample space:S={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A } 52 Cards in the pack 26
26. 26. AS={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A }Event - some outcomes of the sample spaceEvent A – get a ‘4’ if you pick one card from thepack of cardsA = {♠4 ♣4 ♥4 ♦4}P(A) = 4/52 27
27. 27. BS={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A }Event - some outcomes of the sample spaceEvent B – get a picture card if you pick one cardfrom the pack of cardsB = {♠JQKA ♣ JQKA ♥ JQKA ♦ JQKA}P(B) = 16/52 28
28. 28. BS={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A }Complement of an eventEvent B – get a picture card if you pick one cardfrom the pack of cardsWhat is the probability not to get a picture card?P( B) = 1 – P(B) = 1 – 16/52 = 36/52 29
29. 29. B AS={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A }Additional ruleAre events A and B mutually exclusive?P(A or B) = P(A) + P(B) = 4/52 + 16/52 = 20/52 30
30. 30. S={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A C ♦ 2 3 4 5 6 7 8 9 10 J Q K A }Event C – get a red card if you pick one cardfrom the pack of cardsC = { ♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A }P(C) = 26/52 31
31. 31. BS={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A C ♦ 2 3 4 5 6 7 8 9 10 J Q K A }Additional ruleAre events B and C mutually exclusive? 32
32. 32. BS={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A C ♦ 2 3 4 5 6 7 8 9 10 J Q K A }Additional ruleAre events B and C mutually exclusive?P(B or C) = P(B) + P(C) – P(B and C)= 16/52 + 26/52 – 8/52 = 34/52 33
33. 33. BS={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A C ♦ 2 3 4 5 6 7 8 9 10 J Q K A }Conditional ruleWhat is the probability to get a red card if thecard must be a picture card? 8 P (C and B ) 52 = 8P(C | B ) = = P( B) 16 16 34 52
34. 34. AS={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A C ♦ 2 3 4 5 6 7 8 9 10 J Q K A }Conditional ruleWhat is the probability to get a ‘4’ if the cardmust be a red card? 2 P( A and C ) 52 = 2P( A | C ) = = P(C ) 26 26 52 35
35. 35. A S={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A C ♦ 2 3 4 5 6 7 8 9 10 J Q K A }Conditional ruleSampling with replacement takes placeIf we pick two cards, what is the probability that thefirst one is red and the second one is ‘4’We know: P(C) = 26/52 and P(A|C) = 2/26 2P( A and C ) = P (C ) × P ( A | C ) = 26 ×2 = 36 52 26 52
36. 36. A B S={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A }Conditional ruleSampling without replacement takes placeIf we pick two cards, what is the probability that thefirst one is a picture card and the second one is ‘4’P(B) = 16/52 and P(A|B) = 4/51 64P( B and A) = P ( B ) × P ( A | B ) = 16 ×4 = 37 52 51 2652
37. 37. Independent eventsA soccer team is playing two matches on a specificday.•The chance of winning the first match is: •P(1st) = ½•The chance of winning the second match is: •P(2nd) = ½•Will the outcome of the first match influence theoutcome of the second match?•The chance of winning both matches is: •P(1st and 2nd) = P(1st) x P(2nd |1st) •P(1st and 2nd) = P(1st) x P(2nd) = ½ X ½ = ¼ 38
38. 38. In chapter 2 we looked at an example to organisequalitative data into a frequency distribution table. 39
39. 39. Organising and graphing qualitative data in afrequency distribution table.Example:The data below shows the gender of 50 employees and thedepartment in which they work at ABC Ltd. HR – Human resources Emp. no. Gender Dept. Emp. no. Gender Mark. – Marketing Dept ….. 1 M HR 6 M Fin. – Finance Fin. …..M – Male2 F Mark. 7 M Mark. …..F – Female 3 M Fin. 8 M Fin. ….. 4 F HR 9 F HR ….. 5 F Fin. 10 F Fin. ….. 40
40. 40. Organising and graphing qualitative data in a frequencydistribution table. HR Marketing Finance Total M 4 10 5 19 F 10 16 5 31 Total 14 26 10 50 41
41. 41. If one employee is chosen from the 50 employees, what isthe probability that the employee will be male? HR Marketing Finance Total M 4 10 5 19 F 10 16 5 31 Total 14 26 10 50 P(M) = 19/50 42
42. 42. If one employee is chosen from the 50 employees, what isthe probability that the employee will be from theMarketing department? HR Marketing Finance Total M 4 10 5 19 F 10 16 5 31 Total 14 26 10 50 P(Mark) = 26/50 43
43. 43. If one employee is chosen from the 50 employees, what isthe probability that the employee will be from theMarketing or Finance departments? HR Marketing Finance Total M 4 10 5 19 F 10 16 5 31 Total 14 26 10 50 Are the marketing P(Mark or Fin) = 26/50 + 10/50 = 36/50 and finance departments 44 mutually exclusive?
44. 44. If one employee is chosen from the 50 employees, what isthe probability that the employee will be Female andfrom the Finance department? HR Marketing Finance Total M 4 10 5 19 F 10 16 5 31 Total 14 26 10 50 Are female and the P(F and Fin) = 5/50 finance department mutually exclusive? 45
45. 45. If one employee is chosen from the 50 employees, what isthe probability that the employee will be Female or fromthe Finance department? HR Marketing Finance Total M 4 10 5 19 F 10 16 5 31 Total 14 26 10 50 Are female and the P(F or Fin) = 31/50 + 10/50 – 5/50 = 36/50 finance department mutually exclusive? 46
46. 46. If one employee is chosen from the 50 employees, what isthe probability that the employee will be from theMarketing and Finance departments? HR Marketing Finance Total M 4 10 5 19 F 10 16 5 31 Total 14 26 10 50 Are the marketing P(F and Fin) = 0 and finance departments 47 mutually exclusive?
47. 47. If one employee is chosen from the 50 employees, what isthe probability that the employee will not be from the HRdepartment? HR Marketing Finance Total M 4 10 5 19 F 10 16 5 31 Total 14 26 10 50 P(HR ) = 1 - P(HR) = 1 - 14/50 = 36/50 48
48. 48. If one employee is chosen from the 50 employees, what isthe probability that the employee will female if she isfrom the HR department? HR Marketing Finance Total M 4 10 5 19 F 10 16 5 31 Total 14 26 10 50 P(F|HR) = P(F and HR)/P(HR) = 10/50 / 14/50 = 10/14 49