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What are the chances? - Probability
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What are the chances? - Probability

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Introduction to Probability.

Introduction to Probability.

Statistics

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What are the chances? - Probability What are the chances? - Probability Presentation Transcript

  • “What are the chances of that happening?”
    • Theory of Probability
    • Weather forecast predicts 80% chance of rain.
    • You know based on experience that going slightly over speed increases your chance of passing through more green lights.
    • You buy a ticket for the Euro Millions because you figure someone has to win.
    • A restaurant manager thinks about the probability of how many customers will come in in order to prepare accordingly.
  • 17th Century Gambling
    • Chevalier de Mere gambled frequently to increase his wealth.
    • He bet on a roll of a die that at least one six would appear in four rolls.
    • Tired of this approach he decided to change his bet to make it more challenging.
    • He bet that he would get a total of 12, a double 6, on 24 rolls of two dice.
    THE OLD METHOD WAS MOST PROFITABLE!
    • He asked his friend Blaise Pascal why this was the case.
    • 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!
    • Pascal wrote a letter to Pierre De Fermat, who wrote back.
    • They exchanged their mathematical principles and problems and are credited with the founding of probability theory.
    Correspondence leads to Theory
  • Probability
    • 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.
    • Life is a sequence of unpredctable events, but probability can be used to help predict the likelihood of certain events occuring.
  • Careers using the “Chance Theory”
    • Actuarial Science
    • Atmospheric Science
    • Bioinformatics
    • Biostatistics
    • Ecological/Environmental Statistics
    • Educational Testing and Measurement
    • environmental Health Sciences
    • Epidemiology
    • Government
    • Financial Engineering/Financial Mathematics/Mathematical Finance/Quantitative Finance
    • Industrial Statistics
    • Medicine
    • Meteorology/Atmospheric Science
    • Nurses/Doctors
    • Pharmaceutical Research
    • Public Health
    • Public Policy
    • Risk Analysis
    • Risk Management and Insurance
    • Social Statistics
  • First you gotta learn the rules and terms!
  • Problem:
    • A spinner has four equal sectors colored yellow, blue, green and red.
    • What are the chances on landing on blue after spinning the spinner?
    • What are the chances of landing on red?
  • 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.
  • Getting the rules!
    • The P! of an outcome is the percentage of times the outcome is expected to happen.
    • Every P! is a number (a percentage) between 0% and 100%. [Note that statisicians often express percentages as proportions-no. 0 and 1]
    • 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.
    • The sum of all the Probabilities of all possible outcomes is 1 (or 100%).
  • Probability Scale
    • In words
    RANGE IN NUMBER RANGE IN PERCENTAGE 0 - 1 0% - 100%
  • 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
  • Back to the Spinner AFTER SPINNING THE SPINNER, WHAT IS THE PROBABILITY OF LANDING ON EACH COLOR? RED? BLUE? GREEN? ORANGE? 1/4!!!
  • In order to measure the P! of an event mathematicians have developed a method to do this!
  • 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
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  • Enter P(not A)
    • Complement of an Event
  • Probability of an Event Not Happening
    • If A is any event, then ‘not A’ is the event that A does not happen.
    • Clearly A and ‘not A’ cannot occur at the same time.
    • Either A or ‘not A’ must occur.
    • Thus we have the following relationship between the probabilities of A and ‘not A’:
    P(A) + P(NOT A) = 1 OR P(NOT A) = 1 - P(A)
    • 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?
    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
  • Using the rule!
    • P(not red) = 1 - P(red)
    • P(not red) = 1 - 1/4 = 3/4
  • You try please!
    • 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?
    • P(not club) = 1 = P(club)
    • 1 - 13/52
    • = 39/52 or 3/4
  • Notes:
    • P(not A) can be also written as P(A’) or P(A).
    • It is important NOT to count an outcome twice in an event when calculating probabilities.
    • In Q’s on probability, objects that are identical are treated as different objects.
    • The phrase ‘ drawn at random ’ means each object is equally likely to be picked.
    • ‘ Unbiased ’ means ‘ fair ’.
    • ‘ Biased ’ means ‘ unfair ’ in some way.
  • Guesses.
  •  
  • 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
  • Combining two events
    • 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!
  • 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 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
  • Relative Frequency
    • Experimental Probability
  • Some P! cannot be calculated by just looking at the situation!
    • For example you cannot work out the P! of winning a football match by assuming that win, draw or loose are equally likely!
    • But we can look at previous results in similar matches and use these results to estimate the P! of winning!
    • 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?
    • They have played each other 50 times before. The Blues won 35 of those games and there was also 5 draws!
    • So we can say so far the Blues have won 35/50 games or 7/10!
    Example 1
    • The fraction isn’t the P! of the Blues winning but an estimate!
    • We say that the relative frequency og the Blues winning is 7/10.
    • Matthew decides to see what the P! is that buttered toast lands buttered side down when dropped.
    • 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! of landing buttered side down is 3/5.
    Example 2
  • Definition:
    • Relative Frequency is a good estimate of how likely an event is to occur, provided that the number of trials is sufficiently large.
  • Experiment - Formula
    • The relative frequency of an event in an experiment is given by:
    P(E) = RELATIVE FREQUENCY OF AN EVENT= NO. OF SUCCESSFUL TRIALS NO. OF TRIALS
  • 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)
  • 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.
  • Example 2
    • 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.
    • (I) Write down the value of B.
    • (ii) If the spinner is spun 200 times, how many fives would expect?
    Number 1 2 3 4 5 Probability 0.25 0.2 0.25 0.15 B
  • Solution:
    • (i) Since one of the no. 1-5 must appear, the sum of all the P! must add to 1!
    • Expected no. of 5’s.
    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
  • Combined Events
    • 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:
    • It is often called the ‘ or ’ rule!
    • 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.
    P(A OR B) = P(A) + P(B) - P(A AND B) REMOVES DOUBLE COUNTING
  • Mutually Exclusive Events
    • Events A and B are said to be mutually exclusive events if they cannot occur at the same time.
    • 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.
    • If A and B are mutually exclusive events, then P(A ∩ B) = 0. There is no overlap of A and B.
    • For mutually exclusive events, P(A ∪ B) = P(A) + P(B).
    • A and B are two events such that P(A ∪ B) = 9/10, P(A) = 7/10 and P(A ∩ B) = 3/20
    • Find: (i) P(B) (ii) P(B’) (iii) P[(A ∪ B)’]
    • Solution:
    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)
  • Example 2
    • An unbiased 20-sided die, numbered 1 to 20, is thrown.
    • (i) What is the P! of obtaining a no. divisible by 4 or 5?
    • (ii) Are these events mutually exclusive?
  • 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! ∕
  • Example 3
    • 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?
  • Solution:
    • 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)
    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! ∕
  • Q. 8 Pg: 73 Active Math
    • Given the Venn diagram, write down:
    • P(E) =
    • P(F) =
    • P(E ∩ F) =
    • P(E ∪ F) =
    • Verify that P(E ∪ F) = P(E)+P(F)-P(E ∩ F)
    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
  • 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.
    • 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:
    33/50 20/50 = 2/5 5/50 = 1/10 27/50 Boys Girls Aged 17 18 12 Aged 18 15 5
  • Conditional Probability
    • In a class there are 15 male students, 5 of whom wear glasses and 10 female students, 3 of whom wear glasses.
    • We will let M = {male students}, F = {female students} and G = {students who wear glasses}.
    • A student is picked at random from the class. What is the P! that the student is female, given that the student wears glasses?
    • 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
    • In general the P(A|B) =
    # (A ∩ B) P(A ∩ B) #B P(B) =
    • 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.
    Q 4. Pg: 79 {GGG, GGB, BGG, GBG, BBG, GBB, BGB, BBB} 2/8 = 0.25
  • 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 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