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# What are the chances? - Probability

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Introduction to Probability.

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### What are the chances? - Probability

1. 1. “What are the chances of that happening?” <ul><li>Theory of Probability </li></ul>
2. 2. <ul><li>Weather forecast predicts 80% chance of rain. </li></ul><ul><li>You know based on experience that going slightly over speed increases your chance of passing through more green lights. </li></ul><ul><li>You buy a ticket for the Euro Millions because you figure someone has to win. </li></ul><ul><li>A restaurant manager thinks about the probability of how many customers will come in in order to prepare accordingly. </li></ul>
3. 3. 17th Century Gambling <ul><li>Chevalier de Mere gambled frequently to increase his wealth. </li></ul><ul><li>He bet on a roll of a die that at least one six would appear in four rolls. </li></ul><ul><li>Tired of this approach he decided to change his bet to make it more challenging. </li></ul><ul><li>He bet that he would get a total of 12, a double 6, on 24 rolls of two dice. </li></ul>THE OLD METHOD WAS MOST PROFITABLE!
4. 4. <ul><li>He asked his friend Blaise Pascal why this was the case. </li></ul><ul><li>Pascal worked it out and found that the probablity of winning was only 49.1% with the new method compared to 51.8% using the old approach! </li></ul><ul><li>Pascal wrote a letter to Pierre De Fermat, who wrote back. </li></ul><ul><li>They exchanged their mathematical principles and problems and are credited with the founding of probability theory. </li></ul>Correspondence leads to Theory
5. 5. Probability <ul><li>is really about dealing with the unknown in a systematic way, by scoping out the most likely scenarios, or having a backup plan in case those most likely scenarios don’t happen. </li></ul><ul><li>Life is a sequence of unpredctable events, but probability can be used to help predict the likelihood of certain events occuring. </li></ul>
6. 6. Careers using the “Chance Theory” <ul><li>Actuarial Science </li></ul><ul><li>Atmospheric Science </li></ul><ul><li>Bioinformatics </li></ul><ul><li>Biostatistics </li></ul><ul><li>Ecological/Environmental Statistics </li></ul><ul><li>Educational Testing and Measurement </li></ul><ul><li>environmental Health Sciences </li></ul><ul><li>Epidemiology </li></ul><ul><li>Government </li></ul><ul><li>Financial Engineering/Financial Mathematics/Mathematical Finance/Quantitative Finance </li></ul><ul><li>Industrial Statistics </li></ul><ul><li>Medicine </li></ul><ul><li>Meteorology/Atmospheric Science </li></ul><ul><li>Nurses/Doctors </li></ul><ul><li>Pharmaceutical Research </li></ul><ul><li>Public Health </li></ul><ul><li>Public Policy </li></ul><ul><li>Risk Analysis </li></ul><ul><li>Risk Management and Insurance </li></ul><ul><li>Social Statistics </li></ul>
7. 7. First you gotta learn the rules and terms!
8. 8. Problem: <ul><li>A spinner has four equal sectors colored yellow, blue, green and red. </li></ul><ul><li>What are the chances on landing on blue after spinning the spinner? </li></ul><ul><li>What are the chances of landing on red? </li></ul>
9. 9. Terms Defintion Example An EXPERIMENT is a situation involving chance or probability that leads to results called outcomes. What color would we land on? A TRIAL is the act of doing an experiment in P! Spinning the spinner. The set or list of all possible outcomes in a trial is called the SAMPLE SPACE. Possible outcomes are Green, Blue, Red and Yellow. An OUTCOME is one of the possible results of a trial. Green. An EVENT is the occurence of one or more specific outcomes One event in this experiment is landing on blue.
10. 10. Getting the rules! <ul><li>The P! of an outcome is the percentage of times the outcome is expected to happen. </li></ul><ul><li>Every P! is a number (a percentage) between 0% and 100%. [Note that statisicians often express percentages as proportions-no. 0 and 1] </li></ul><ul><li>If an outcome has P! of 0% it can NEVER happen, no matter what. If an outcome has a P! of 100% it will ALWAYS happen. Most P! are neither 0/100% but fall somewhere in between. </li></ul><ul><li>The sum of all the Probabilities of all possible outcomes is 1 (or 100%). </li></ul>
11. 11. Probability Scale <ul><li>In words </li></ul>RANGE IN NUMBER RANGE IN PERCENTAGE 0 - 1 0% - 100%
12. 12. Want to have a guess? 0 1 0.5 A B C D E 5 EVENTS (A,B,C,D AND E ) ARE SHOWN ON A P! SCALE. COPY AND COMPLETE THE FOLLOWING TABLE: Probability Event Fifty-fifty C Certain Very unlikely Impossible Very likely
13. 13. Back to the Spinner AFTER SPINNING THE SPINNER, WHAT IS THE PROBABILITY OF LANDING ON EACH COLOR? RED? BLUE? GREEN? ORANGE? 1/4!!!
14. 14. In order to measure the P! of an event mathematicians have developed a method to do this!
15. 15. Probability of an Event THE P! OF AN EVEN A OCCURING P(A) = THE NUMBER OF WAYS EVENT A CAN OCCUR THE TOTAL NUMBER OF POSSIBLE OUTSOMES
16. 19. Enter P(not A) <ul><li>Complement of an Event </li></ul>
17. 20. Probability of an Event Not Happening <ul><li>If A is any event, then ‘not A’ is the event that A does not happen. </li></ul><ul><li>Clearly A and ‘not A’ cannot occur at the same time. </li></ul><ul><li>Either A or ‘not A’ must occur. </li></ul>
18. 21. <ul><li>Thus we have the following relationship between the probabilities of A and ‘not A’: </li></ul>P(A) + P(NOT A) = 1 OR P(NOT A) = 1 - P(A)
19. 22. <ul><li>A spinner has 4 equal sectors colored yellow, blue, green and red. What is the probability of landing on a sector that is not red after spinning this spinner? </li></ul>Let’s understand! Sample Space:  {yellow, blue, green, red} Probability:   The probability of each outcome in this experiment is one fourth. The probability of landing on a sector that is not red is the same as the probability of landing on all the other colors except red. P(not red) = 1/4 + 1/4 + 1/4 = 3/4
20. 23. Using the rule! <ul><li>P(not red) = 1 - P(red) </li></ul><ul><li>P(not red) = 1 - 1/4 = 3/4 </li></ul>
21. 24. You try please! <ul><li>A single card is chosen at random from a standard deck of 52 playing cards. What is the probability of choosing a card that is not a club? </li></ul><ul><li>P(not club) = 1 = P(club) </li></ul><ul><li>1 - 13/52 </li></ul><ul><li>= 39/52 or 3/4 </li></ul>
22. 25. Notes: <ul><li>P(not A) can be also written as P(A’) or P(A). </li></ul><ul><li>It is important NOT to count an outcome twice in an event when calculating probabilities. </li></ul><ul><li>In Q’s on probability, objects that are identical are treated as different objects. </li></ul>
23. 26. <ul><li>The phrase ‘ drawn at random ’ means each object is equally likely to be picked. </li></ul><ul><li>‘ Unbiased ’ means ‘ fair ’. </li></ul><ul><li>‘ Biased ’ means ‘ unfair ’ in some way. </li></ul>
24. 27. Guesses.
25. 29. Conditional P! <ul><li>With this you are normally given some prior knowledge or some extra condition about the outcome. </li></ul><ul><li>This usually reduces the size of the sample space. </li></ul><ul><li>EG. In a class - 21boys&15 girls. 3boys&5girls wear glasses. </li></ul>A PUPIL PICKED AT RANDOM FROM THE CLASS IS WEARING GLASSES. WHAT IS THE P! THAT IT IS A BOY? WE ARE CERTAIN THAT THE PUPIL PICKED WEARS CLASSES. THERE ARE 8 PUPILS THAT WEAR GLASSES AND 3 OF THOSE ARE BOYS P(WHEN A PUPIL WHO WEARS GLASSES IS PICKED, THE PUPIL IS A BOY) = 3/8
26. 30. Combining two events <ul><li>There are many situations where we have to consider two outcomes. In these situations, all the possible outcomes, the sample space can be represented on a diagram - often called a two-way table! </li></ul>
27. 31. Example <ul><li>Two fair six-sided dices, one red and one blue, are thrown. What is the P! of getting two equal scores or of the scores adding to 10? </li></ul><ul><li>Solution: </li></ul>36 POSSIBLE OUTCOMES P(TWO EQUAL SCORES OR A TOTAL OF 10) = 8/36 = 2/9 NOTE: (5,5) IS NOT COUNTED TWICE! 6 X X 5 X 4 X X 3 X 2 X 1 X 1 2 3 4 5 6
28. 32. Relative Frequency <ul><li>Experimental Probability </li></ul>
29. 33. Some P! cannot be calculated by just looking at the situation!
30. 34. <ul><li>For example you cannot work out the P! of winning a football match by assuming that win, draw or loose are equally likely! </li></ul><ul><li>But we can look at previous results in similar matches and use these results to estimate the P! of winning! </li></ul>
31. 35. <ul><li>The Blues and Naoimh Martin Gaelic Teams are playing a match tonight and you want to know what is the P! that the Blues will win? </li></ul><ul><li>They have played each other 50 times before. The Blues won 35 of those games and there was also 5 draws! </li></ul><ul><li>So we can say so far the Blues have won 35/50 games or 7/10! </li></ul>Example 1
32. 36. <ul><li>The fraction isn’t the P! of the Blues winning but an estimate! </li></ul><ul><li>We say that the relative frequency og the Blues winning is 7/10. </li></ul>
33. 37. <ul><li>Matthew decides to see what the P! is that buttered toast lands buttered side down when dropped. </li></ul><ul><li>He drops 50 pieces of buttered toast. </li></ul><ul><li>30 pieces land buttered side down. </li></ul><ul><li>His relative frequency is 30/50=3/5. </li></ul><ul><li>Therefore he would estimate that the P! of landing buttered side down is 3/5. </li></ul>Example 2
34. 38. Definition: <ul><li>Relative Frequency is a good estimate of how likely an event is to occur, provided that the number of trials is sufficiently large. </li></ul>
35. 39. Experiment - Formula <ul><li>The relative frequency of an event in an experiment is given by: </li></ul>P(E) = RELATIVE FREQUENCY OF AN EVENT= NO. OF SUCCESSFUL TRIALS NO. OF TRIALS
36. 40. How many times you expect a particular outcome to happen in an experiment. <ul><li>The expected number of outcomes is calculated as follows: </li></ul>EXPECTED NO. OF OUTCOMES = (RELATIVE FREQUENCY) X (NO. OF TRIALS) OR EXPECTED NO. OF OUTCOMES = P(EVENT) X (NO. OF TRIALS)
37. 41. Example <ul><li>Sarah throws her fair six sided die a total of 1,200 times. Find the expected number of times the number 3 would appear. </li></ul><ul><li>Solution: </li></ul>IF THE DIE IS FAIR, THEN THE P! OF A SCORE OF 3 WOULD BE 1/6! THUS THE EXPECTED NO OF 3’S = P(EVENT) X (NO. OF TRIALS) = 1/6 X 1200 =200.
38. 42. Example 2 <ul><li>A spinner numbered 1-5 is biased. The P! that the spinner will land on each of the numbers 1 to 5 is given in the P! distribution table below. </li></ul><ul><li>(I) Write down the value of B. </li></ul><ul><li>(ii) If the spinner is spun 200 times, how many fives would expect? </li></ul>Number 1 2 3 4 5 Probability 0.25 0.2 0.25 0.15 B
39. 43. Solution: <ul><li>(i) Since one of the no. 1-5 must appear, the sum of all the P! must add to 1! </li></ul><ul><li>Expected no. of 5’s. </li></ul>Therefore 0.25 + 0.2 + 0.25 + 0.15 + B = 1 0.85 + B =1 thus B = 0.15 = P(5) X (no. of trials) = 0.15 X 200 = 30
40. 44. Combined Events <ul><li>If A and B are two different events of the same experiment, then the P! that the two events, A or B, can happen is given by: </li></ul><ul><li>It is often called the ‘ or ’ rule! </li></ul><ul><li>It is important to remember that P(A or B) means A occurs, or B occurs, or both occur. By subtracting p(A and B), the possibility of double counting is removed. </li></ul>P(A OR B) = P(A) + P(B) - P(A AND B) REMOVES DOUBLE COUNTING
41. 45. Mutually Exclusive Events <ul><li>Events A and B are said to be mutually exclusive events if they cannot occur at the same time. </li></ul><ul><li>Consider the following event of drawing a single card from a deck of 52 cards. Let A be the even a king is drawn and B the event a Queen is drawn. The single card drawn cannot be a King and a Queen. The events A and B are said to be mutually exclusive events. </li></ul>
42. 46. <ul><li>If A and B are mutually exclusive events, then P(A ∩ B) = 0. There is no overlap of A and B. </li></ul><ul><li>For mutually exclusive events, P(A ∪ B) = P(A) + P(B). </li></ul>
43. 47. <ul><li>A and B are two events such that P(A ∪ B) = 9/10, P(A) = 7/10 and P(A ∩ B) = 3/20 </li></ul><ul><li>Find: (i) P(B) (ii) P(B’) (iii) P[(A ∪ B)’] </li></ul><ul><li>Solution: </li></ul>Example 1 (I) P(A ∪ B) = P(A) + P(B) - P(A ∩ B) 9/10 = 7/10 + P(B) - 3/20 P(B) = 9/10 - 7/10 + 3/20 = 7/20 (II) P(B’) = 1 - P(B) = 1 - 7/20 = 13/20 (III) P[(A ∪ B)’] = 1 - P(A ∪ B) = 1 - 9/10 = 1/10 A B 3/20 1/10 U 11/20 2/5 (7/10 - 3/20 = 11/20) (7/20 - 3/20 = 2/5)
44. 48. Example 2 <ul><li>An unbiased 20-sided die, numbered 1 to 20, is thrown. </li></ul><ul><li>(i) What is the P! of obtaining a no. divisible by 4 or 5? </li></ul><ul><li>(ii) Are these events mutually exclusive? </li></ul>
45. 49. Solution: <ul><li>There are 20 possible outcomes: </li></ul><ul><li>(i) No ÷ by 4 are 4, 8, 12, 16 or 20 ∴ P(no÷4) = 5/20 No ÷ by 5 are 5, 10, 15 or 20 ∴ P(no÷5) = 4/20 No ÷ by 4 and 5 is 20 ∴ P( no divisible by 4 and 5 ) = 1/20 </li></ul><ul><li>P(no÷4/5) = P(no÷4) + P(no÷5) - P(no÷4 and 5) </li></ul>= 5/20 + 4/20 - 1/20 = 8/20 = 4/5 THE NO 20 IS COMMON TO BOTH EVENTS, AND IF THE PROBABILITIES WERE SIMPLY ADDED, THEN THE NO 20 WOULD HAVE BEEN COUNTED TWICE. (II) P(NUMBER DIVISIBLE BY 4 AND 5) = 0 ∴ THE EVENTS ARE Not MUTUALLY EXCLUSIVE! ∕
46. 50. Example 3 <ul><li>A bag contains five red, three blue and yellow discs. The red discs are numbered 1, 2, 3, 4 and 5; the blue are numbered 6, 7, and 8; and the yellow are 9 and 10. A single disc is drawn at random from the bag. What is the P! that the disc is blue and even? Are these mutually exclusive? </li></ul>
47. 51. Solution: <ul><li>Let B represent that a blue disc is chosen and E represent that a disc with an even no is chose. </li></ul><ul><li>P(B or E) = P(B) + P(E) - P(B and E) </li></ul>1 2 3 4 5 6 7 8 9 10 = 3/10 + 5/10 - 2/10 = 6/10 = 3/5 P(B AND E) = 2/10 = 0 ∴ THE EVENTS ARE NOT MUTUALLY EXCLUSIVE! ∕
48. 52. Q. 8 Pg: 73 Active Math <ul><li>Given the Venn diagram, write down: </li></ul><ul><li>P(E) = </li></ul><ul><li>P(F) = </li></ul><ul><li>P(E ∩ F) = </li></ul><ul><li>P(E ∪ F) = </li></ul><ul><li>Verify that P(E ∪ F) = P(E)+P(F)-P(E ∩ F) </li></ul>s S E F 0.1 0.3 0.5 0.1 0.4 0.6 0.1 0.9 0.9 = 0.4 + 0.6 - 0.1
49. 53. Q. 18 Pg: 74 <ul><li>The data shows the number of boys and girls aged either 17 or 18 on a school trip, in which only these 50 students took part. </li></ul><ul><li>A students is chosen at random from the party. Find the P! that the student: </li></ul><ul><li>(i) Is a boy: </li></ul><ul><li>(ii) is aged 18: </li></ul><ul><li>(iii) is a girl of 18 years: </li></ul><ul><li>(iv) is a girl aged 17 or a boy aged 18: </li></ul>33/50 20/50 = 2/5 5/50 = 1/10 27/50 Boys Girls Aged 17 18 12 Aged 18 15 5
50. 54. Conditional Probability <ul><li>In a class there are 15 male students, 5 of whom wear glasses and 10 female students, 3 of whom wear glasses. </li></ul><ul><li>We will let M = {male students}, F = {female students} and G = {students who wear glasses}. </li></ul><ul><li>A student is picked at random from the class. What is the P! that the student is female, given that the student wears glasses? </li></ul>
51. 55. <ul><li>So we write this as: P(F|G) [The P! of F, given G] </li></ul><ul><li>8 students wear glasses. </li></ul><ul><li>3 of these are girls. </li></ul><ul><li>Hence P(F|G) = 3/8 </li></ul><ul><li>In general the P(A|B) = </li></ul># (A ∩ B) P(A ∩ B) #B P(B) =
52. 56. <ul><li>A family has three children. Complete the outcome space: {GGG, GGB, ...}, where GGB means the first two are girls and the third is a boy. </li></ul><ul><li>Find the P! that all 3 children are girls, given that the family has at least two girls. </li></ul>Q 4. Pg: 79 {GGG, GGB, BGG, GBG, BBG, GBB, BGB, BBB} 2/8 = 0.25
53. 57. Q. 9 PG: 80 <ul><li>E and F are two events such that P(E) = 2/5, P(F) = 1/2, and P(E|F) = 1/9. </li></ul><ul><li>Find: (i) P(E ∩ F) (ii) P(F|E) (iii) P(E ∪ F) </li></ul>P(E|F) = P(E ∩ F) P(F) 1/9 = P(E ∩ F) 1/2 ∴ P(E ∩ F) = 1/18 P(F|E) = P(F ∩ E) P(E) P(F P(F|E) = 1/18 2/5 P(F|E) = 5/36 P(E ∪ F) = P(E) + P(F) - P(E ∩ F) 2/5 + 1/2 - 1/18 38/45