chapter 5- probability review
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Chapter 5- Probability Review. Section 5.1. An event is the set of possible outcomes Probability is between 0 and 1 The event A has a complement, the event not A. Together these two probabilities sum 1. ex. At least one and none are complements Probability of an event = - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 5- Probability Review
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Section 5.1
An event is the set of possible outcomes
Probability is between 0 and 1 The event A has a complement, the
event not A. Together these two probabilities sum 1.ex. At least one and none are complements
Probability of an event = number of outcomes in event / number of equally
likely outcomes
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Section 5.1
Probability Distributions give all values resulting from a random process.
Sample space is the complete list of disjoint outcomes. All outcomes in a sample space must have total probability equal to 1.
Ex. The sample space for rolling a die is {1,2,3,4,5,6} The sample space for rolling two die is the table of 36 outcomes we’ve seen.
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Disjoint
Disjoint and mutually exclusive mean the same thing.
Disjoint means two different outcomes can’t occur on the same opportunity
Ex. Can’t roll an extra credit and no collect on the same roll. Can’t get a heads and tail on the same flip.
These items on a Venn diagram would have no intersection
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Disjoint Continued
Flipping a coin and getting a head and tail is disjoint
Flipping a coin twice and getting a head and tail is disjoint.
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Law of Large Numbers
In random sampling, the larger the sample, the closer the proportion of successes in the sample tends to be to the population proportion.
The difference between a sample proportion and the population proportion must get smaller as the sample size gets larger.
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Fundamental Principle of Counting Processes can be split into stages (Flip
coin once, then again) If there are k stages with different
possible outcomes for each stage, the number of total possible outcomes is
n1*n2*n3*n4… nk
Ex. How many total outcome for rolling a die and flipping a coin?
6*2 = 12
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Probability Simulations
1) Assumptions
What probability are you assuming? Are events independent?
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Probability Simulations
2) Model*How specifically will you use your table of random digits? *Make sure to say what to do with repeats, unallocated numbers*Fully describe what constitutes a run and what statistics you’re collecting.*Make a table to show how digits /groups are assigned
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Probability Simulations
3) Repetition
Run the simulations and record the results in a frequency table.
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Probability Simulations
4) Conclusion
Write the conclusion in context of the situation. Be sure to say the probability is ESTIMATED.
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Addition Rule of Probability Remember first that “or” means one
or the other or both
P(A or B) = P(A) + P(B) – P(A and B)
If A and B are disjoint, there is no intersection. Therefore, P(A and B)= 0.
If A and B are disjoint: P(A or B) = P(A) + P(B)
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Conditional Probability and the Multiplication Rule The probability of an event A and B
both happening isP(A and B) = P(A) * P(B|A)
P(A and B) = P(B) * P(A|B)
The probability of an event changes based on what happened before.
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Conditional Probability
Rearranging means P(A|B) = P(A and B) / P(B)
Probability of both divided by the probability of the first event.
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Independent Events
The occurrence of one event doesn’t change the probability of the second event occurring
Test for Independence: IS P(A|B) = P(A) or P(B|A) =
P(B) ?If yes, you have independent events.
Sometimes this isn’t obvious that one has an effect without checking
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Multiplication Rule for Independent Events Remember if events are
independent,P(A|B) = P(A) or P(B|A) = P(B)
Therefore, since P(A and B) = P(A) * P(B|A)
P(A and B) = P(A) * P(B) for independent events
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Review Questions
If events A and B are independent and P(A) = 0.3 and P(B) = 0.5, then which of these is true?
A. P(A and B) = 0.8B. P(A or B) = 0.15C. P(A or B) = 0.8D. P(A | B) = 0.3E. P(A | B) = 0.5
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Answer
If events A and B are independent and P(A) = 0.3 and P(B) = 0.5, then which of these is true?
A. P(A and B) = 0.8B. P(A or B) = 0.15C. P(A or B) = 0.8D. P(A | B) = 0.3 Probability of A is not changed
based on the occurrence of event B
E. P(A | B) = 0.5
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Two Way Table Questions
Crash Type
Single Vehicle Multiple Vehicles
Total
Alcohol Related 10,741 4,887 15,628
Not Alcohol Related
11,345 11,336 22,681
Total 22,086 16,223 38,309
If a fatal auto crash is chosen at random, what is the approximate probability that the crash was alcohol related, given that it involved a single vehicle?A. 0.28B. 0.49C. 0.58D. 0.69E. The answer cannot be determined from the information given.
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AnswerCrash Type
Single Vehicle Multiple Vehicles
Total
Alcohol Related 10,741 4,887 15,628
Not Alcohol Related
11,345 11,336 22,681
Total 22,086 16,223 38,309
If a fatal auto crash is chosen at random, what is the approximate probability that the crash was alcohol related, given that it involved a single vehicle?
B. 0.4910,741/22,086 = 0.49
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Two Way Table Questions
Crash Type
Single Vehicle Multiple Vehicles
Total
Alcohol Related 10,741 4,887 15,628
Not Alcohol Related
11,345 11,336 22,681
Total 22,086 16,223 38,309
What is the approximate probability that a randomly chosen fatal auto crash involves a single vehicle and is alcohol related?A. 0.28B. 0.49C. 0.58D. 0.69E. The answer cannot be determined from the information given
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AnswerCrash Type
Single Vehicle Multiple Vehicles
Total
Alcohol Related 10,741 4,887 15,628
Not Alcohol Related
11,345 11,336 22,681
Total 22,086 16,223 38,309
What is the approximate probability that a randomly chosen fatal auto crash involves a single vehicle and is alcohol related?A. 0.28Because 10,741/38,309 = 0.28
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Review Question
For all events A and B, P(A and B) =A. P(A) · P(B)B. P(B | A)C. P(A | B)D. P(A) + P(B)E. P(B) · P(A | B)
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Answer
For all events A and B, P(A and B) =A. P(A) · P(B)B. P(B | A)C. P(A | B)D. P(A) + P(B)E. P(B) · P(A | B)This is the multiplication rule.If they are independent, P(A|B) = P(A)
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Review Question
The mathematics department at a school has twenty instructors. Six are easy graders. Twelve are considered to be good teachers. Seven are neither. If a student is assigned randomly to one of the easy graders, what is the probability that the instructor will also be good?
A. 7/20B. 5/12C. 7/12D. 5/6E. The answer cannot be determined from the
information given.
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Answer
D. 5/6
Easy Not Easy
Total
Good 5 7 12Not Good
1 7 8
Total 6 14 20
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Review Question The management of Young & Sons Sporting
Supply, Inc., is responding to a claim of discrimination. If the company has employed the 60 people in this table, how many females over the age of 40 must the company hire so that the age and sex of its employees are independent? 40 Years > 40 Years Total
Male 25 15
Female 20
Total
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AnswerFirst, let N be the number of females
older than 40 to be hired.
Set up a proportion so P(male |<40) = P(Female|< 40)N=12
40 Years > 40 Years TotalMale 25 15 40Female 20 N 20+NTotal 45 15+N 60+N
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Review Question
If P(A) = 0.4, P(B) = 0.2, and P(A and B) = 0.08, which of these is true?
A. Events A and B are independent and mutually exclusive.
B. Events A and B are independent but not mutually exclusive.
C. Events A and B are mutually exclusive but not independent.
D. Events A and B are neither independent nor mutually exclusive.
E. Events A and B are independent, but whether A and B are mutually exclusive cannot be determined from the given information.
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AnswerIf P(A) = 0.4, P(B) = 0.2, and P(A and B) = 0.08, which of
these is true?A. Events A and B are independent and mutually
exclusive.B. Events A and B are independent but not
mutually exclusive.C. Events A and B are mutually exclusive but not
independent.D. Events A and B are neither independent nor mutually
exclusive.E. Events A and B are independent, but whether A and
B are mutually exclusive cannot be determined from the given information.