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Probability

Probability. The Study of Chance!. Part 3. In this powerpoint we will continue our study of probability by looking at: Conditional Probability Independence Reversing the Condition. Conditional Probabilities.

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Probability

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  1. Probability The Study of Chance! Part 3

  2. In this powerpoint we will continue our study of probability by looking at: • Conditional Probability • Independence • Reversing the Condition

  3. Conditional Probabilities • What about when we know something up front about the individual that we have chosen? • How does this change our calculations? • Let’s look a the employment/credit card data that we used in Part 2.

  4. Credit Card J O B Yes No Total Yes No Total Conditional Probabilities • What is the probability that a respondent has a credit card given that they are employed? • We write this type of question like this: • P(C E) *the line between is read “given” and indicates that we already know the outcome of the event that comes after the line. • Since we already know that the respondent has a job, this means that our population of interest is only the 47 respondents that have jobs. • 18 of those have credit cards. • So P(C E) = 18/47 18 29 47 28 34 62 46 63 109

  5. Credit Card J O B Yes No Total Yes No Total Conditional Probability • Find the probability that a respondent does not have a job given that they have a credit card. • P(Ec C) = • Since we already know they have a credit card, our population of interest has become they 46 individuals who have a credit card. • Our “successes” are the 28 of those people who do not have a job. • P(Ec C) = 28/46 18 29 47 28 34 62 46 63 109

  6. Disjoint vs Independent • Two very important concepts in probability are disjoint events and independent events. The ability to tell the difference between these are very important when calculating probabilities.

  7. Disjoint Events(Mutually Exclusive) • We’ll start with disjoint events. • When events are disjoint this means they have no outcomes in common. • Mutually Exclusive is another term used to describe disjoint events. • Examples: • The relationships between disjoint and joint events may be best shown using Venn diagrams. • Let the event A: Democratic voter • Let the event B: Republican voter A B Republican Voter Democratic Voter Since a voter can not be registered as both a Democratic Voter and a Republican Voter, there is no overlap of the events. This can either be shown with non-overlapping Venn diagrams or with the overlap being empty.

  8. Independence • Events are independent when knowing the outcome of one event does NOT change the probability of the other. • For Example: • Consider the two events: flip a coin, then roll a die • We wish to know the probability of the event • Head and a number less than 2 • The probability of getting heads on a fair coin is ½ • The probability of getting a number less than 2 on a fair die is 2/6 or 1/3 • Knowing the outcome of the coin toss does NOT change the probability of getting a number less than 2 on the die. • These two events are independent, since knowing the outcome of the coin toss didn’t change the probability of the dice roll.

  9. Independence • The formal mathematical definition tells us that two events are independent if: • P(A B) = P(A) • In other words, this “insider” information (knowing that “B” has happened doesn’t gives us NO new information about “A” happening! • Let’s look at an example.

  10. Tests for Drunk Driving • Police report that 78% of drivers are given a breath test, 36% a blood test, and 22% both tests • We wish to know if giving the two test are independent? • Let’s construct a contingency table for this situation

  11. Independent???? • If the two tests are independent then the following should be true: • P(blood test breath test ) = P(blood test) • Let’s find out if they are.

  12. P(blood test breath test ) = P(blood test) • Remember from our study of conditional probabilities that the P(blood test breath test) = .22/.78= .28 • P(blood test) = .36 • Since .28 ≠ .36, these events are NOT independent. • Knowing the outcome of one gives changes the probability of the other

  13. Disjoint vs Independent • Can disjoint events ever be independent? • Remember that for events to be independent, knowing the outcome of one does NOT change the outcome of the other. • Also remember that disjoint events can NOT happen at the same time, which means knowing that one happened makes the probability of the other happening zero (regardless of its original probability) • In short: Disjoint events can NEVER be independent!!!!!

  14. Reversing the condition • Sometimes the information we know and the information we are looking for doesn’t come in such a way as to be a straight forward question. • Let’s find out how we can use tree diagrams to answer these kinds of questions.

  15. Binge drinking and Accidents • Our question of interest is this: • What is the probability that if a college student has been involved in an alcohol-related accident, s/he is a binge drinker? • Here is what we know: • A study on binge drinking on Campus found that 44% of college students engage in binge drinking (5+ for men, and 4+ for women), 37% drink moderately, and 19% abstain entirely. • Another study found that among binge drinkers aged 21-34, 17% have been involved in an alcohol-related accident, while among non-bingers of the same age, only 9% have been involved in such accidents. • Let’s draw a tree diagram to help answer our question

  16. Binge and Acc .44(.17)=.0748 Accident .17 • Remember that our question can be written: • P(binge accident) • What we know is the second event, (accident). This essentially changes our sample space (our denominater) • We are only concerned with the branches where accidents occurred. • Our numerator is the branch among these in which the driver is a binge drinker. Binge Drinkers .44 None .83 Binge and None .44(.83)=.3652 Accident .09 Mod and Acc .37(.09)=.0333 Moderate .37 None .91 Mod and None .37(.91)=.3367 Abstain .19 Abstain and Acc .19(.00) = 0 Accident .00 None 1.0 Abstain and None .19(1.0) = .19 P(Binge Acc) = .0748 / (.0748 + .0333 + 0) = .0748/.1081 = .6920

  17. What you learned • Disjoint Events • Independent Events • Disjoint vs Independent • Reversing the Condition

  18. Additional Resources • The Practice of Statistics—YMM • Pg 341 -363 • The Practice of Statistics—YMS • Pg 359-387 • Against All Odds—Video #15 • http://www.learner.org/resources/series65.html

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