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Part 3 Module 3 Probability

Part 3 Module 3 Probability. Suppose we create a four-letter password, such as bbca , acda, or edfb, by making four random selections, with replacement, from the set { a, b, c, d, e, f }. How many different 4 letter passwords are possible?

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Part 3 Module 3 Probability

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  1. Part 3 Module 3 Probability • Suppose we create a four-letter password, such as bbca, acda, or edfb, by making four random selections, with replacement, from the set {a, b, c, d, e, f }. • How many different 4 letter passwords are possible? • 2. How many 4 letter passwords have no repeated letters?

  2. Part 3 Module 3 Probability We can use the answers to the previous two counting problems to answer this question about probability: If we create a four letter password by making four random selections from the set {a, b, c, d, e, f }, what is the probability that the password will have no repeated letters? Answer:

  3. The previous calculation illustrates the classical or theoretical definition of the probability of an event E in a random experiment. The classical definition of probability

  4. Probability • Suppose a random experiment involves selecting two coins, without replacement, from a jar containing a penny, a nickel, a dime, a quarter, a half dollar, and a silver dollar. • What is the probability that the monetary sum of the two coins is 75¢? • To answer this question, we need to answer two prelimary questions: • How many different outcomes (monetary sums) are possible in this experiment? • 2. How many of these outcomes are favorable to our event (sum = 75¢)?

  5. Probability In much of our work in Part 3 Modules 3 through 5, we will work with experiments that involve selecting or more individuals from a specified population. The simplest situation is one in which a single individual is being selected. At the Forest Folks Gathering there are 5 elves (E) 2 hobbits (H) 8 gnomes (G) They will randomly select one person from amongst this group to serve as Grand Marshall. 1. What is the probability that the randomly selected person is a gnome (G)? 2. What is the probability that the randomly selected person is an elf (E)? Answers P(G) = 8/15 = .5333 P(E) = 5/15 = .3333

  6. Probability At the Forest Folks Gathering there are 5 elves (E) 2 hobbits (H) 8 gnomes (G) They will randomly select one person from amongst this group to serve as Grand Marshall. 1. What is the probability that the randomly selected person is not a gnome (G´)? 2. What is the probability that the randomly selected person is not an elf (E´)? Answers P(G´) = 7/15 = .4667 P(E´) = 10/15 = .6667

  7. Let E be any event. The complement of E, denoted E´ is the non-occurrence of E, or the opposite of E. In the previous examples, for instance, note that the probability of selecting a gnome was .5333 [that is, P(G) = .5333] and the probability of not selecting a gnome was .4667 [that is, P(G´) = .4667] Also note that these two probabilities have a special relationship: P(G) + P(G´) = .5333 + .4667 = 1 This illustrates an important fact called the Complements Rule. Complements

  8. The Complements Rule For any event E in any experiment, P(E) + P(E´) = 1 This rule is usually presented in a slightly different, but equivalent, form: The Complements Rule For any event E in any experiment, P(E´) = 1 – P(E) The Complements Rule

  9. Among a certain group of Vikings, 28 of them like to pillage, 18 of them like to plunder, while 10 of them like to pillage and like to plunder and 12 of them don't like to pillage and don't like to plunder. If one of these Vikings is randomly selected, find the probability that he/she likes to pillage.
 A. 0.4828 B. 0.3103 C. 0.5833 D. 0.2800 A different idea

  10. Among a certain group of Vikings, 28 of them like to pillage, 18 of them like to plunder, while 10 of them like to pillage and like to plunder and 12 of them don't like to pillage and don't like to plunder. If one of these Vikings is randomly selected, find the probability that he/she doesn’t like to pillage.
 Solution: (We will refer to the answer to the previous question, and use the complements rule) P(doesn’t like to pillage) = 1 – P(likes to pillage) = 1 – .5833 = .4167 Another exercise

  11. Two events are said to be mutually exclusive if it is impossible for the two events to occur simultaneously. If E, F are mutually exclusive events, then P(E or F) = P(E) + P(F) Mutually exclusive events

  12. At the Forest Folks Gathering there are 5 elves (E) 2 hobbits (H) 8 gnomes (G) They will randomly select one person from amongst this group to serve as Grand Marshall. What is the probability that the randomly selected person is a gnome or an elf? Note that these events (G, E) are mutually exclusive: it is possible that the selected person might be a gnome, and it is possible that the selected person might be an elf, but it is impossible for the selected person to be both a gnome and an elf. P(G or E) = P(G) + P(E) = .5333 + .3333 = .8666 Mutually exclusive events

  13. At the Forest Folks Gathering there are 5 elves (E) 2 hobbits (H) 8 gnomes (G) They will randomly select three people from amongst this group to serve as the Lollipop Guild. What is the probability that all three selectees are gnomes? We must answer two questions: 1. In how many ways is it possible to select three people from this group of 15? This is the total number of equally likely outcomes to the experiment. 2. In how many ways is it possible to selec three people who are all gnomes? This is the number of outcomes favorable to our event. A more complicated problem

  14. Statistical Probability Certain statistics are also probabilities. In this course, we will work frequently with population statistics. Statistical probability is inferred from population statistics.

  15. Statistical Probability A few years ago, for example, the Natural Resources Defense Council conducted a study of bottled water (this example is non-fiction). They found that 40% of bottled water samples were merely tap water. They also found that 30% of bottled water samples were contaminated by substances such as arsenic and fecal bacteria.

  16. Statistical Probability 40% of bottled water samples are merely tap water. 30% of bottled water samples are contaminated by substances such as arsenic and fecal bacteria. These unappetizing statistics are probabilities. Let T be the event that a randomly selected sample of bottled water is tap water. Let C be the event that a randomly selected sample of bottled water is contaminated. Then P(T) = 40% = .4 P(C) = 30% = .3

  17. 40% of bottled water samples are merely tap water. 30% of bottled water samples are contaminated by substances such as arsenic and fecal bacteria. Let T be the event that a randomly selected sample of bottled water is tap water. Let C be the event that a randomly selected sample of bottled water is contaminated. 1. What is the probability that a randomly selected sample of bottled water is not merely tap water? 2. What is the probability that a randomly selected sample of bottled water is not contaminated? Answers: 1. P(T´) = 1 – P(T) = 1 – .4 = .6 2. P(C´) = 1 – P(C) = 1 – .3 = .7 The Complements Rule, again

  18. The table below shows the distribution of scores on Test 1 in Partial Differential Equations for Liberal Arts. Score percent 0 - 49 2% 50 - 59 6% 60 - 64 6% 65 - 69 14% 70 - 79 37% 80 - 89 17% 90 -100 18% If one student is randomly selected, what is the probability that his/her test score is in the 50-59 range or the 70-79 range? A. .43 B. .97 C. .63 D. None of these Mutually exclusive events

  19. Odds, like probability, use ratios or fractions to indicate likelihood. Odds are not the same as probability, however. According to the classical definition, that the probability of an event is the ratio of the number of favorable outcomes to the total number of possible outcomes. The odds in favor of an event is the ratio of favorable outcomes to unfavorable outcomes. The odds against an event is the ratio of unfavorable outcomes to favorable outcomes. Side note: “Odds”

  20. The odds in favor of an event is the ratio of favorable outcomes to unfavorable outcomes. Refer to the experiment of rolling one die. Let E be the event that the result of the die roll is “2.” We know that P(E) = 1/6. The odds in favor of E are not 1/6, however. In this experiment, one of the six outcomes is favorable to E, and the other five outcomes are unfavorable. The odds in favor of E are 1/5, or “1 to 5.” This is also expressed as 1:5. The odds against E are “5 to 1”, or 5:1. Odds vs. probability

  21. The table below shows the distribution of scores on Test 1 in Partial Differential Equations for Liberal Arts. Score percent 0 - 49 2% 50 - 59 6% 60 - 64 6% 65 - 69 14% 70 - 79 37% 80 - 89 17% 90 -100 18% If one student is randomly selected, find the ODDS AGAINST the event that he or she has a test score in the 70 - 79 range?
 A. 63:37 B. 37:100 C. 37:63 D. 63:10 Exercise

  22. One last note: When the term “odds” is used in the context of sports betting, it refers to the odds against an event. For instance, if we are told that the odds for a certain horse winning a race are 7 to 1, that means that the odds against the horse winning are 7 to 1; the odds in favor are 1 to 7. Odds in Sports Betting

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