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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**

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 is P(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) = 0.8 B. P(A or B) = 0.15 C. P(A or B) = 0.8 D. P(A | B) = 0.3 E. 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) = 0.8 B. P(A or B) = 0.15 C. P(A or B) = 0.8 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 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. 0.28 B. 0.49 C. 0.58 D. 0.69 E. The answer cannot be determined from the information given.

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**Answer 0.49 10,741/22,086 = 0.49 Crash Type Single Vehicle**

Multiple Vehicles Total Alcohol Related 10,741 4,887 15,628 Not Alcohol Related 11,345 11,336 22,681 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? 0.49 10,741/22,086 = 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 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. 0.28 B. 0.49 C. 0.58 D. 0.69 E. The answer cannot be determined from the information given

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**Answer 0.28 Because 10,741/38,309 = 0.28 Crash Type Single Vehicle**

Multiple Vehicles Total Alcohol Related 10,741 4,887 15,628 Not Alcohol Related 11,345 11,336 22,681 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? 0.28 Because 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 P(B) · P(A | B) This is the multiplication rule.**

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) 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. 7/20 B. 5/12 C. 7/12 D. 5/6 E. The answer cannot be determined from the information given.

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Answer D. 5/6 Easy Not Easy Total Good 5 7 12 Not Good 1 8 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

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Answer First, 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 Total Male 25 15 40 Female 20 N 20+N 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? Events A and B are independent and mutually exclusive. Events A and B are independent but not mutually exclusive. Events A and B are mutually exclusive but not independent. Events A and B are neither independent nor mutually exclusive. 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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**Answer Events A and B are independent but not mutually exclusive.**

If P(A) = 0.4, P(B) = 0.2, and P(A and B) = 0.08, which of these is true? Events A and B are independent and mutually exclusive. Events A and B are independent but not mutually exclusive. Events A and B are mutually exclusive but not independent. Events A and B are neither independent nor mutually exclusive. 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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Applying the ideas: Probability

Applying the ideas: Probability

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