155S4.3_3 Addition Rule
September 15, 2010
MAT 155
Dr. Claude Moore
Cape Fear Community College
Key Concept
Chapter 4 Probability
4­1 Review and Preview
4­2 Basic Concepts of Probability
To find more information about 4­3 Addition Rule
4­4 Multiplication Rule: Basics statistics and probability, visit Rice Virtual Lab in Statistics at 4­7 Counting
http://davidmlane.com/hyperstat/probability.html
The following lesson is in Course Documents of CourseCompass.
S2.D1.MAT 155 Chapter 4 ­ Probability
155Chapter 4 ( Package file )
This contains some notes for Chapter 4 Probability.
It contains the following topics relative to probability: Fundamentals; Addition Rule; Multiplication Rule: Basics; Multiplication Rule: Complements and Conditional Probability; Probabilities through Simulations; Counting
Compound Event ­ any event combining 2 or more simple events.
Notation P(A or B) = P (in a single trial, event A occurs or event B occurs or they both occur)
General Rule for a Compound Event
When finding the probability that event A occurs or event B occurs, find the total number of ways A can occur and the number of ways B can occur, but find that total in such a way that no outcome is counted more than once.
This section presents the addition rule as a device for finding probabilities that can be expressed as P(A or B), the probability that either event A occurs or event B occurs (or they both occur) as the single outcome of the procedure.
The key word in this section is “or.” It is the inclusive or, which means either one or the other or both.
Compound Event ­ Formal Addition Rule
P(A or B) = P(A) + P(B) – P(A and B)
where P(A and B) denotes the probability that A and B both occur at the same time as an outcome in a trial of a procedure.
Intuitive Addition Rule
To find P(A or B), find the sum of the number of ways event A can occur and the number of ways event B can occur, adding in such a way that every outcome is counted only once. P(A or B) is equal to that sum, divided by the total number of outcomes in the sample space.
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155S4.3_3 Addition Rule
September 15, 2010
Disjoint or Mutually Exclusive
Events A and B are disjoint (or mutually exclusive) if they cannot occur at the same time. (That is, disjoint events do not overlap.)
Complementary Events
P(A) and P(A)
are disjoint
It is impossible for an event and its complement to occur at the same time.
Rule of Complementary Events
Venn Diagram for Events That Are Not Disjoint
Venn Diagram for Disjoint Events
Recap
Venn Diagram for the Complement of Event A
In this section we have discussed:
• Compound events.
• Formal addition rule.
• Intuitive addition rule.
• Disjoint events.
• Complementary events.
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155S4.3_3 Addition Rule
September 15, 2010
Determining Whether Events Are Disjoint. For Exercises 5–12, determine whether the two events are disjoint for a single trial. Hint: (Consider “disjoint” to be equivalent to “separate” or “not overlapping.” )
Determining Whether Events Are Disjoint. For Exercises 5–12, determine whether the two events are disjoint for a single trial. Hint: (Consider “disjoint” to be equivalent to “separate” or “not overlapping.” )
163/6. Conducting a Pew Research Center poll and randomly selecting a subject who is a Republican Conducting a Pew Research Center poll and randomly selecting a subject who is a Democrat
163/8. Randomly selecting a fruit fly with red eyes Randomly selecting a fruit fly with sepian (dark brown) eyes
Disjoint, because a person would not be both a Republican and a Democrat.
163/10. Randomly selecting someone treated with the cholesterol­ reducing drug Lipitor Randomly selecting someone in a control group given no medication
Disjoint, because a person would not be a person who used the drug and did not use the drug.
163/14. Colorblindness Women have a 0.25% rate of color blindness. If a woman is randomly selected, what is the probability that she does not have color blindness? (Hint: The decimal equivalent of 0.25% is 0.0025, not 0.25.)
P(not color blind) = 1 ­ P(color blind)
= 1 ­ 0.0025 = 0.9975
or 0.998 according to the round off rule.
Disjoint, because a fruit fly would not have both red eyes and dark brown eyes.
163/12. Randomly selecting a college graduate Randomly selecting someone who is homeless
Not disjoint, because a person could be both a college graduate and a homeless person.
163/16. Sobriety Checkpoint When the author observed a sobriety checkpoint conducted by the Dutchess County Sheriff Department, he saw that 676 drivers were screened and 6 were arrested for driving while intoxicated. Based on those results, we can estimate that P( I ) = 0.00888, where I denotes the event of screening a driver and getting someone who is intoxicated. What does P( I ) denote and what is its value?
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155S4.3_3 Addition Rule
September 15, 2010
In Exercises 17–20, use the polygraph test data given in Table 4­1, which is included with the Chapter Problem.
In Exercises 17–20, use the polygraph test data given in Table 4­1, which is included with the Chapter Problem.
163/18. Polygraph Test If one of the test subjects is randomly selected, find the probability that the subject did not lie.
163/20. Polygraph Test If one of the subjects is randomly selected, find the probability that the subject had a negative test result or lied.
In Exercises 21– 6, use the data in the accompanying table, which summarizes challenges by tennis players (based on data reported in USA Today). The results are from the first U. S. Open that used the Hawk­ Eye electronic system for displaying an instant replay used to determine whether the ball is in bounds or out of bounds. In each case, assume that one of the challenges is randomly selected.
In Exercises 21– 6, use the data in the accompanying table, which summarizes challenges by tennis players (based on data reported in USA Today). The results are from the first U. S. Open that used the Hawk­ Eye electronic system for displaying an instant replay used to determine whether the ball is in bounds or out of bounds. In each case, assume that one of the challenges is randomly selected.
164/22. Tennis Instant Replay If M denotes the event of selecting a challenge made by a man, find P(not M).
164/24. Tennis Instant Replay Find the probability that the selected challenge was made by a woman or was successful.
Total
489
P(not M) = P(W) = (126 + 224) / 839
= 350 / 839 = 0.417
350
Total 327 512 839
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155S4.3_3 Addition Rule
September 15, 2010
In Exercises 27–32, refer to the following table summarizing results from a study of people who refused to answer survey questions (based on data from “I Hear You Knocking but You Can’t Come In,” by Fitzgerald and Fuller, Sociological Methods and Research, Vol. 11, No. 1). In each case, assume that one of the subjects is randomly selected.
164/28. Survey Refusals A pharmaceutical company is interested in opinions of the elderly, be­cause they are either receiving Medicare or will receive it soon. What is the probability that the selected subject is someone 60 and over who responded?
In Exercises 27–32, refer to the following table summarizing results from a study of people who refused to answer survey questions (based on data from “I Hear You Knocking but You Can’t Come In,” by Fitzgerald and Fuller, Sociological Methods and Research, Vol. 11, No. 1). In each case, assume that one of the subjects is randomly selected.
164/32. Survey Refusals A market researcher is not interested in refusals or subjects below 22 years of age or over 59. Find the probability that the selected person refused to answer or is below 22 or is older than 59.
Total
1049
56
Total 74 275 278 152 165 251 1205
P(60 and over; who responded) = 202 / 1205 = 0.168 (rounded off)
Look at the 251 who are 60 and over; count those of the 251 who responded. Thus,we have 202 as the numerator.
In Exercises 33–38, use these results from the “ 1­Panel­THC” test for marijuana use, which is provided by the company Drug Test Success: Among 143 subjects with positive test results, there are 24 false positive results; among 157 negative results, there are 3 false negative results. (Hint: Construct a table similar to Table 4­1, which is included with the Chapter Problem.)
In Exercises 33–38, use these results from the “ 1­Panel­THC” test for marijuana use, which is provided by the company Drug Test Success: Among 143 subjects with positive test results, there are 24 false positive results; among 157 negative results, there are 3 false negative results. (Hint: Construct a table similar to Table 4­1, which is included with the Chapter Problem.)
164/34. Screening for Marijuana Use If one of the test subjects is randomly selected, find the probability that the subject tested positive or used marijuana.
The underlined numbers were calculated after the other numbers were inserted into the table.
165/36. Screening for Marijuana Use If one of the test subjects is randomly selected, find the probability that the subject actually used marijuana. Do you think that the result reflects the marijuana use rate in the general population?
P(+ or Y) = (143 + 3) / 300 = 146 / 300 = 0.487
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