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- 1. “What are the chances ofthat happening?”Theory of Probability
- 2. Weather forecast predicts 80% chance of rain.You know based on experience that goingslightly over speed increases your chance ofpassing through more green lights.You buy a ticket for the Euro Millions becauseyou figure someone has to win.A restaurant manager thinks about theprobability of how many customers will come inin order to prepare accordingly.
- 3. 17th Century GamblingChevalier de Mere gambled frequently toincrease his wealth.He bet on a roll of a die that at least one sixwould appear in four rolls.Tired of this approach he decided to change hisbet to make it more challenging.He bet that he would get a total of 12, a double6, on 24 rolls of two dice.
- 4. 17th Century GamblingChevalier de Mere gambled frequently toincrease his wealth.He bet on a roll of a die that at least one sixwould appear in four rolls.Tired of this approach he decided to change hisbet to make it more challenging.He bet that he would get a total of 12, a double6, on 24 rolls of two dice. THE OLD METHOD WAS MOST PROFITABLE!
- 5. Correspondence leads toTheoryHe asked his friend Blaise Pascal why this wasthe case.Pascal worked it out and found that theprobablity of winning was only 49.1% with thenew method compared to 51.8% using the oldapproach!Pascal wrote a letter to Pierre De Fermat, whowrote back.They exchanged their mathematical principlesand problems and are credited with the foundingof probability theory.
- 6. Probabilityis really about dealing with the unknown in asystematic way, by scoping out the most likelyscenarios, or having a backup plan in case thosemost likely scenarios don’t happen.Life is a sequence of unpredctable events, butprobability can be used to help predict thelikelihood of certain events occuring.
- 7. Careers using the“Chance Theory”Actuarial Science Industrial StatisticsAtmospheric Science MedicineBioinformatics Meteorology/Atmospheric ScienceBiostatistics Nurses/DoctorsEcological/Environmental Statistics Pharmaceutical ResearchEducational Testing and Measurement Public Healthenvironmental Health Sciences Public PolicyEpidemiology Risk AnalysisGovernment Risk Management and InsuranceFinancial Engineering/Financial Mathematics/ Social StatisticsMathematical Finance/Quantitative Finance
- 8. First you gottalearn the rulesand terms!
- 9. Problem:A spinner has fourequal sectors coloredyellow, blue, greenand red.What are the chanceson landing on blueafter spinning thespinner?What are the chancesof landing on red?
- 10. Terms Defintion ExampleAn EXPERIMENT is a situationinvolving chance or probability What color would we land on? that leads to results called outcomes.A TRIAL is the act of doing an Spinning the spinner. experiment in P! The set or list of all possible Possible outcomes are Green,outcomes in a trial is called the Blue, Red and Yellow. SAMPLE SPACE. An OUTCOME is one of the Green. possible results of a trial.An EVENT is the occurence of One event in this experiment isone or more specific outcomes landing on blue.
- 11. Getting the rules!The P! of an outcome is the percentage of timesthe outcome is expected to happen.Every P! is a number (a percentage) between 0%and 100%. [Note that statisicians often expresspercentages as proportions-no. 0 and 1]If an outcome has P! of 0% it can NEVERhappen, no matter what. If an outcome has a P!of 100% it will ALWAYS happen. Most P! areneither 0/100% but fall somewhere in between.The sum of all the Probabilities of all possibleoutcomes is 1 (or 100%).
- 12. PROBABILITY SCALE IN WORDSRANGE IN NUMBER RANGE IN PERCENTAGE 0-1 0% - 100%
- 13. Want to have a guess? 0.5 0 1 A B C D E5 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
- 14. Back to the SpinnerAFTER SPINNING THE SPINNER, WHAT IS THE PROBABILITY OF LANDING ON EACH COLOR?
- 15. Back to the SpinnerAFTER SPINNING THE SPINNER, WHAT IS THE PROBABILITY OF LANDING ON EACH COLOR? RED?
- 16. Back to the SpinnerAFTER SPINNING THE SPINNER, WHAT IS THE PROBABILITY OF LANDING ON EACH COLOR? RED? BLUE?
- 17. Back to the SpinnerAFTER SPINNING THE SPINNER, WHAT IS THE PROBABILITY OF LANDING ON EACH COLOR? RED? GREEN? BLUE?
- 18. Back to the SpinnerAFTER SPINNING THE SPINNER, WHAT IS THE PROBABILITY OF LANDING ON EACH COLOR? RED? GREEN? BLUE? ORANGE?
- 19. Back to the SpinnerAFTER SPINNING THE SPINNER, WHAT IS THE PROBABILITY OF LANDING ON EACH COLOR? RED? GREEN? BLUE? 1/4!!! ORANGE?
- 20. In order to measure the P!of an eventmathematicians havedeveloped a method to dothis!
- 21. Probability of an Event THE P! OF AN EVEN A OCCURINGP(A) = THE NUMBER OF WAYS EVENT A CAN OCCUR THE TOTAL NUMBER OF POSSIBLE OUTSOMES
- 22. Enter P(not A)Complement of an Event
- 23. Probability of an EventNot HappeningIf A is any event, then ‘not A’ is the event that Adoes not happen.Clearly A and ‘not A’ cannot occur at the sametime.Either A or ‘not A’ must occur.
- 24. Thus we have the following relationship betweenthe probabilities of A and ‘not A’: P(A) + P(NOT A) = 1 OR P(NOT A) = 1 - P(A)
- 25. Let’s understand!A spinner has 4 equal sectors colored yellow, blue, green and red. Whatis the probability of landing on a sector that is not red after spinning thisspinner?Sample Space: {yellow, blue, green, red}Probability: The probability of each outcome in this experiment is one fourth. Theprobability of landing on a sector that is not red is the same as theprobability of landing on all the other colors except red.P(not red) = 1/4 + 1/4 + 1/4 = 3/4
- 26. Using the rule!P(not red) = 1 - P(red)P(not red) = 1 - 1/4 = 3/4
- 27. You try please!A single card is chosen at random from astandard deck of 52 playing cards. What is theprobability of choosing a card that is not a club?
- 28. You try please!A single card is chosen at random from astandard deck of 52 playing cards. What is theprobability of choosing a card that is not a club?P(not club) = 1 = P(club)
- 29. You try please!A single card is chosen at random from astandard deck of 52 playing cards. What is theprobability of choosing a card that is not a club?P(not club) = 1 = P(club)1 - 13/52
- 30. You try please!A single card is chosen at random from astandard deck of 52 playing cards. What is theprobability of choosing a card that is not a club?P(not club) = 1 = P(club)1 - 13/52= 39/52 or 3/4
- 31. Notes:P(not A) can be also written as P(A’) or P(A).It is important NOT to count an outcome twice inan event when calculating probabilities.In Q’s on probability, objects that are identicalare treated as different objects.
- 32. The phrase ‘drawn at random’ means eachobject is equally likely to be picked.‘Unbiased’ means ‘fair’.‘Biased’ means ‘unfair’ in some way.
- 33. Guesses.
- 34. Conditional P!With this you are normally given some priorknowledge or some extra condition about theoutcome.This usually reduces the size of the samplespace.EG. In a class - 21boys&15 girls. 3boys&5girlswear glasses.
- 35. Conditional P! With this you are normally given some prior knowledge or some extra condition about the outcome. This usually reduces the size of the sample space. EG. In a class - 21boys&15 girls. 3boys&5girls wear glasses.A PUPIL PICKED AT RANDOM FROM THE CLASS IS WEARING GLASSES. WHAT IS THE P! THAT IT IS A BOY?
- 36. Conditional P! With this you are normally given some prior knowledge or some extra condition about the outcome. This usually reduces the size of the sample space. EG. In a class - 21boys&15 girls. 3boys&5girls wear glasses. 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
- 37. Conditional P! With this you are normally given some prior knowledge or some extra condition about the outcome. This usually reduces the size of the sample space. EG. In a class - 21boys&15 girls. 3boys&5girls wear glasses. 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)
- 38. Conditional P! With this you are normally given some prior knowledge or some extra condition about the outcome. This usually reduces the size of the sample space. EG. In a class - 21boys&15 girls. 3boys&5girls wear glasses. 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
- 39. Combining two eventsThere are many situations where we have toconsider two outcomes. In these situations, allthe possible outcomes, the sample space canbe represented on a diagram - often called atwo-way table!
- 40. ExampleTwo fair six-sided dices, one red and one blue,are thrown. What is the P! of getting two equalscores or of the scores adding to 10?Solution:
- 41. Example 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? Solution:6 X X5 X4 X X3 X2 X1 X 1 2 3 4 5 6
- 42. Example 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? Solution: 36 POSSIBLE OUTCOMES6 X X5 X4 X X3 X2 X1 X 1 2 3 4 5 6
- 43. Example 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? Solution: 36 POSSIBLE OUTCOMES6 X X P(TWO EQUAL SCORES OR A TOTAL OF 10)5 X = 8/36 = 2/94 X X3 X2 X1 X 1 2 3 4 5 6
- 44. Example 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? Solution: 36 POSSIBLE OUTCOMES6 X X P(TWO EQUAL SCORES OR A TOTAL OF 10)5 X = 8/36 = 2/94 X X3 X2 X NOTE: (5,5) IS NOT COUNTED TWICE!1 X 1 2 3 4 5 6
- 45. Relative FrequencyExperimental Probability
- 46. Some P! cannot becalculated by just lookingat the situation!
- 47. For example you cannot work out the P! ofwinning a football match by assuming that win,draw or loose are equally likely!But we can look at previous results in similarmatches and use these results to estimate the P!of winning!
- 48. Example 1The Blues and Naoimh Martin Gaelic Teams areplaying a match tonight and you want to knowwhat is the P! that the Blues will win?They have played each other 50 times before.The Blues won 35 of those games and there wasalso 5 draws!So we can say so far the Blues have won 35/50games or 7/10!
- 49. The fraction isn’t the P! of the Blues winning butan estimate!We say that the relative frequency og the Blueswinning is 7/10.
- 50. Example 2Matthew decides to see what the P! is thatbuttered toast lands buttered side down whendropped.He drops 50 pieces of buttered toast.30 pieces land buttered side down.His relative frequency is 30/50=3/5.Therefore he would estimate that the P! oflanding buttered side down is 3/5.
- 51. Definition:Relative Frequency is a good estimate of howlikely an event is to occur, provided that thenumber of trials is sufficiently large.
- 52. Experiment - FormulaThe relative frequency of an event in anexperiment is given by:
- 53. Experiment - Formula The relative frequency of an event in an experiment is given by:P(E) = RELATIVE FREQUENCY OF AN EVENT=
- 54. Experiment - Formula The relative frequency of an event in an experiment is given by: NO. OF SUCCESSFUL TRIALSP(E) = RELATIVE FREQUENCY OF AN EVENT=
- 55. Experiment - Formula The relative frequency of an event in an experiment is given by: NO. OF SUCCESSFUL TRIALSP(E) = RELATIVE FREQUENCY OF AN EVENT=
- 56. Experiment - Formula The relative frequency of an event in an experiment is given by: NO. OF SUCCESSFUL TRIALSP(E) = RELATIVE FREQUENCY OF AN EVENT= NO. OF TRIALS
- 57. How many times you expect a particular outcome to happen in an experiment. The expected number of outcomes is calculated as follows:EXPECTED NO. OF OUTCOMES = (RELATIVE FREQUENCY) X (NO. OF TRIALS) OR EXPECTED NO. OF OUTCOMES = P(EVENT) X (NO. OF TRIALS)
- 58. Example Sarah throws her fair six sided die a total of 1,200 times. Find the expected number of times the number 3 would appear. Solution: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.
- 59. Example 2A spinner numbered 1-5 is biased. The P! thatthe spinner will land on each of the numbers 1 to5 is given in the P! distribution table below. Number 1 2 3 4 5 Probability 0.25 0.2 0.25 0.15 B(I) Write down the value of B.(ii) If the spinner is spun 200 times, how manyfives would expect?
- 60. Solution:(i) Since one of the no. 1-5 must appear, the sumof all the P! must add to 1!Therefore 0.25 + 0.2 + 0.25 + 0.15 + B = 10.85 + B =1 thus B = 0.15Expected no. of 5’s.= P(5) X (no. of trials)= 0.15 X 200 = 30
- 61. Combined EventsIf A and B are two different events of the sameexperiment, then the P! that the two events, A orB, can happen is given by: P(A OR B) = P(A) + P(B) - P(A AND B) REMOVES DOUBLE COUNTINGIt is often called the ‘or’ rule!It is important to remember that P(A or B) meansA occurs, or B occurs, or both occur. Bysubtracting p(A and B), the possibility of doublecounting is removed.
- 62. Mutually ExclusiveEventsEvents A and B are said to be mutually exclusiveevents if they cannot occur at the same time.Consider the following event of drawing a singlecard from a deck of 52 cards. Let A be the evena king is drawn and B the event a Queen isdrawn. The single card drawn cannot be a Kingand a Queen. The events A and B are said to bemutually exclusive events.
- 63. If A and B are mutually exclusive events, thenP(A ∩ B) = 0. There is no overlap of A and B.For mutually exclusive events,P(A ∪ B) = P(A) + P(B).
- 64. Example 1A and B are two events such that P(A ∪ B) = 9/10,P(A) = 7/10 and P(A ∩ B) = 3/20Find: (i) P(B) (ii) P(B’) (iii) P[(A ∪ B)’]Solution: U(I) P(A ∪ B) = P(A) + P(B) - P(A ∩ B) A B 9/10 = 7/10 + P(B) - 3/20 P(B) = 9/10 - 7/10 + 3/20 11/20 3/20 2/5 = 7/20 1/10(II) P(B’) = 1 - P(B) = 1 - 7/20 = 13/20 (7/10 - 3/20 = 11/20) (7/20 - 3/20 = 2/5) (III) P[(A ∪ B)’] = 1 - P(A ∪ B) = 1 - 9/10 = 1/10
- 65. Example 2An unbiased 20-sided die, numbered 1 to 20, isthrown.(i) What is the P! of obtaining a no. divisible by 4or 5?(ii) Are these events mutually exclusive?
- 66. Solution: There are 20 possible outcomes: (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 P(no÷4/5) = P(no÷4) + P(no÷5) - P(no÷4 and 5) = 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!
- 67. Example 3A bag contains five red, three blue and yellowdiscs. The red discs are numbered 1, 2, 3, 4 and5; the blue are numbered 6, 7, and 8; and theyellow are 9 and 10. A single disc is drawn atrandom from the bag. What is the P! that thedisc is blue and even? Are these mutuallyexclusive?
- 68. Solution: 8
- 69. Solution:1 8
- 70. Solution:1 2 8
- 71. Solution:1 2 3 8
- 72. Solution:1 2 3 4 8
- 73. Solution:1 2 3 4 5 8
- 74. Solution:1 2 3 4 5 6 8
- 75. Solution:1 2 3 4 5 6 7 8
- 76. Solution:1 2 3 4 5 6 7 8 9
- 77. Solution:1 2 3 4 5 6 7 8 9 10
- 78. Solution:1 2 3 4 5 6 7 8 9 10 Let B represent that a blue disc is chosen and E represent that a disc with an even no is chose. P(B or E) = P(B) + P(E) - P(B and E)
- 79. Solution:1 2 3 4 5 6 7 8 9 10 Let B represent that a blue disc is chosen and E represent that a disc with an even no is chose. P(B or E) = P(B) + P(E) - P(B and E) = 3/10 + 5/10 - 2/10
- 80. Solution:1 2 3 4 5 6 7 8 9 10 Let B represent that a blue disc is chosen and E represent that a disc with an even no is chose. P(B or E) = P(B) + P(E) - P(B and E) = 3/10 + 5/10 - 2/10 = 6/10 = 3/5
- 81. Solution:1 2 3 4 5 6 7 8 9 10 Let B represent that a blue disc is chosen and E represent that a disc with an even no is chose. P(B or E) = P(B) + P(E) - P(B and E) = 3/10 + 5/10 - 2/10 = 6/10 = 3/5 P(B AND E) = 2/10 = 0 ∴ THE EVENTS ARE NOT MUTUALLY EXCLUSIVE!
- 82. Solution:1 2 3 4 5 6 7 8 9 10 Let B represent that a blue disc is chosen and E represent that a disc with an even no is chose. P(B or E) = P(B) + P(E) - P(B and E) = 3/10 + 5/10 - 2/10 = 6/10 = 3/5 ∕ P(B AND E) = 2/10 = 0 ∴ THE EVENTS ARE NOT MUTUALLY EXCLUSIVE!
- 83. Q. 8 Pg: 73 Active MathGiven the Venn Sdiagram, write down: E FP(E) = 0.3 0.1s 0.5P(F) =P(E ∩ F) = 0.1P(E ∪ F) =Verify thatP(E ∪ F) = P(E)+P(F)-P(E ∩ F)
- 84. Q. 8 Pg: 73 Active MathGiven the Venn Sdiagram, write down: E FP(E) = 0.4 0.3 0.1s 0.5P(F) =P(E ∩ F) = 0.1P(E ∪ F) =Verify thatP(E ∪ F) = P(E)+P(F)-P(E ∩ F)
- 85. Q. 8 Pg: 73 Active MathGiven the Venn Sdiagram, write down: E FP(E) = 0.4 0.3 0.1s 0.5P(F) = 0.6P(E ∩ F) = 0.1P(E ∪ F) =Verify thatP(E ∪ F) = P(E)+P(F)-P(E ∩ F)
- 86. Q. 8 Pg: 73 Active MathGiven the Venn Sdiagram, write down: E FP(E) = 0.4 0.3 0.1s 0.5P(F) = 0.6P(E ∩ F) = 0.1 0.1P(E ∪ F) =Verify thatP(E ∪ F) = P(E)+P(F)-P(E ∩ F)
- 87. Q. 8 Pg: 73 Active MathGiven the Venn Sdiagram, write down: E FP(E) = 0.4 0.3 0.1s 0.5P(F) = 0.6P(E ∩ F) = 0.1 0.1P(E ∪ F) = 0.9Verify thatP(E ∪ F) = P(E)+P(F)-P(E ∩ F)
- 88. Q. 8 Pg: 73 Active MathGiven the Venn Sdiagram, write down: E FP(E) = 0.4 0.3 0.1s 0.5P(F) = 0.6P(E ∩ F) = 0.1 0.1P(E ∪ F) = 0.9Verify thatP(E ∪ F) = P(E)+P(F)-P(E ∩ F) 0.9 = 0.4 + 0.6 - 0.1
- 89. Q. 18 Pg: 74 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. Boys Girls Aged 17 18 12 Aged 18 15 5 A students is chosen at random from the party. Find the P! that the student: (i) Is a boy: (ii) is aged 18: (iii) is a girl of 18 years: (iv) is a girl aged 17 or a boy aged 18:
- 90. Q. 18 Pg: 74 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. Boys Girls Aged 17 18 12 Aged 18 15 5 A students is chosen at random from the party. Find the P! that the student: (i) Is a boy: 33/50 (ii) is aged 18: (iii) is a girl of 18 years: (iv) is a girl aged 17 or a boy aged 18:
- 91. Q. 18 Pg: 74 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. Boys Girls Aged 17 18 12 Aged 18 15 5 A students is chosen at random from the party. Find the P! that the student: (i) Is a boy: 33/50 (ii) is aged 18: 20/50 = 2/5 (iii) is a girl of 18 years: (iv) is a girl aged 17 or a boy aged 18:
- 92. Q. 18 Pg: 74 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. Boys Girls Aged 17 18 12 Aged 18 15 5 A students is chosen at random from the party. Find the P! that the student: (i) Is a boy: 33/50 (ii) is aged 18: 20/50 = 2/5 (iii) is a girl of 18 years:5/50 = 1/10 (iv) is a girl aged 17 or a boy aged 18:
- 93. Q. 18 Pg: 74 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. Boys Girls Aged 17 18 12 Aged 18 15 5 A students is chosen at random from the party. Find the P! that the student: (i) Is a boy: 33/50 (ii) is aged 18: 20/50 = 2/5 (iii) is a girl of 18 years:5/50 = 1/10 (iv) is a girl aged 17 or a boy aged 18: 27/50
- 94. Conditional ProbabilityIn a class there are 15 male students, 5 of whomwear glasses and 10 female students, 3 of whomwear glasses.We will let M = {male students}, F = {femalestudents} and G = {students who wear glasses}.A student is picked at random from the class.What is the P! that the student is female, giventhat the student wears glasses?
- 95. =
- 96. So we write this as: P(F|G) [The P! of F, given G] =
- 97. So we write this as: P(F|G) [The P! of F, given G]8 students wear glasses. =
- 98. So we write this as: P(F|G) [The P! of F, given G]8 students wear glasses.3 of these are girls. =
- 99. So we write this as: P(F|G) [The P! of F, given G]8 students wear glasses.3 of these are girls.Hence P(F|G) = 3/8 =
- 100. So we write this as: P(F|G) [The P! of F, given G]8 students wear glasses.3 of these are girls.Hence P(F|G) = 3/8In general the P(A|B) = =
- 101. So we write this as: P(F|G) [The P! of F, given G]8 students wear glasses.3 of these are girls.Hence P(F|G) = 3/8 # (A ∩ B) P(A ∩ B)In general the P(A|B) = #B = P(B)
- 102. Q 4. Pg: 79 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. Find the P! that all 3 children are girls, given that the family has at least two girls. {GGG, GGB, BGG, GBG, BBG, GBB, BGB, BBB} 2/8 = 0.25
- 103. Q. 9 PG: 80 E and F are two events such that P(E) = 2/5, P(F) = 1/2, and P(E|F) = 1/9. Find: (i) P(E ∩ F) (ii) P(F|E) (iii) P(E ∪ F)
- 104. Q. 9 PG: 80 E and F are two events such that P(E) = 2/5, P(F) = 1/2, and P(E|F) = 1/9. Find: (i) P(E ∩ F) (ii) P(F|E) (iii) P(E ∪ F)
- 105. Q. 9 PG: 80 E and F are two events such that P(E) = 2/5, P(F) = 1/2, and P(E|F) = 1/9. Find: (i) P(E ∩ F) (ii) P(F|E) (iii) P(E ∪ F) P(E|F) = P(E ∩ F) P(F)
- 106. Q. 9 PG: 80 E and F are two events such that P(E) = 2/5, P(F) = 1/2, and P(E|F) = 1/9. Find: (i) P(E ∩ F) (ii) P(F|E) (iii) P(E ∪ F) P(E|F) = P(E ∩ F) P(F) 1/9 = P(E ∩ F) 1/2
- 107. Q. 9 PG: 80 E and F are two events such that P(E) = 2/5, P(F) = 1/2, and P(E|F) = 1/9. Find: (i) P(E ∩ F) (ii) P(F|E) (iii) P(E ∪ F) P(E|F) = P(E ∩ F) P(F) 1/9 = P(E ∩ F) 1/2 ∴ P(E ∩ F) = 1/18
- 108. Q. 9 PG: 80 E and F are two events such that P(E) = 2/5, P(F) = 1/2, and P(E|F) = 1/9. Find: (i) P(E ∩ F) (ii) P(F|E) (iii) P(E ∪ F) P(E|F) = P(E ∩ F) P(F) 1/9 = P(E ∩ F) 1/2 ∴ P(E ∩ F) = 1/18
- 109. Q. 9 PG: 80 E and F are two events such that P(E) = 2/5, P(F) = 1/2, and P(E|F) = 1/9. Find: (i) P(E ∩ F) (ii) P(F|E) (iii) P(E ∪ F) P(E|F) = P(E ∩ F) P(F) P(F|E) = P(F ∩ E) P(E) 1/9 = P(E ∩ F) 1/2 ∴ P(E ∩ F) = 1/18
- 110. Q. 9 PG: 80 E and F are two events such that P(E) = 2/5, P(F) = 1/2, and P(E|F) = 1/9. Find: (i) P(E ∩ F) (ii) P(F|E) (iii) P(E ∪ F) P(E|F) = P(E ∩ F) P(F) P(F|E) = P(F ∩ E) P(E) P(F 1/9 = P(E ∩ F) 1/2 P(F|E) = 1/18 2/5 ∴ P(E ∩ F) = 1/18
- 111. Q. 9 PG: 80 E and F are two events such that P(E) = 2/5, P(F) = 1/2, and P(E|F) = 1/9. Find: (i) P(E ∩ F) (ii) P(F|E) (iii) P(E ∪ F) P(E|F) = P(E ∩ F) P(F) P(F|E) = P(F ∩ E) P(E) P(F 1/9 = P(E ∩ F) 1/2 P(F|E) = 1/18 2/5 ∴ P(E ∩ F) = 1/18
- 112. Q. 9 PG: 80 E and F are two events such that P(E) = 2/5, P(F) = 1/2, and P(E|F) = 1/9. Find: (i) P(E ∩ F) (ii) P(F|E) (iii) P(E ∪ F) P(E|F) = P(E ∩ F) P(F) P(F|E) = P(F ∩ E) P(E) P(F 1/9 = P(E ∩ F) 1/2 P(F|E) = 1/18 2/5 ∴ P(E ∩ F) = 1/18 P(F|E) = 5/36
- 113. Q. 9 PG: 80 E and F are two events such that P(E) = 2/5, P(F) = 1/2, and P(E|F) = 1/9. Find: (i) P(E ∩ F) (ii) P(F|E) (iii) P(E ∪ F) P(E|F) = P(E ∩ F) P(F) P(F|E) = P(F ∩ E) P(E) P(F 1/9 = P(E ∩ F) 1/2 P(F|E) = 1/18 2/5 ∴ P(E ∩ F) = 1/18 P(F|E) = 5/36 P(E ∪ F) = P(E) + P(F) - P(E ∩ F)
- 114. Q. 9 PG: 80 E and F are two events such that P(E) = 2/5, P(F) = 1/2, and P(E|F) = 1/9. Find: (i) P(E ∩ F) (ii) P(F|E) (iii) P(E ∪ F) P(E|F) = P(E ∩ F) P(F) P(F|E) = P(F ∩ E) P(E) P(F 1/9 = P(E ∩ F) 1/2 P(F|E) = 1/18 2/5 ∴ P(E ∩ F) = 1/18 P(F|E) = 5/36 P(E ∪ F) = P(E) + P(F) - P(E ∩ F) 2/5 + 1/2 - 1/18
- 115. Q. 9 PG: 80 E and F are two events such that P(E) = 2/5, P(F) = 1/2, and P(E|F) = 1/9. Find: (i) P(E ∩ F) (ii) P(F|E) (iii) P(E ∪ F) P(E|F) = P(E ∩ F) P(F) P(F|E) = P(F ∩ E) P(E) P(F 1/9 = P(E ∩ F) 1/2 P(F|E) = 1/18 2/5 ∴ P(E ∩ F) = 1/18 P(F|E) = 5/36 P(E ∪ F) = P(E) + P(F) - P(E ∩ F) 2/5 + 1/2 - 1/18 38/45

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