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Probability and Statistics: College Assignment on Random Experiments, Events, and Distributions

Assignment on probability theory, event distributions, and statistical applications.

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    Probability and Statistics: College Assignment on Random Experiments, Events, and Distributions Week 3 7. You flip a coin three times. The possible outcomes of this random experiment are, S = {(HHH), (HHT), (HTH), (THH), (HTT), (TTH), (THT), (TTT)} n (S) = 8 (a) What is the probability of getting heads on only one of your flips? The cases favorable to the event E 1 of getting heads on only one of your flips are, E 1 = {(HTT), (TTH), (THT) } n (E 1 ) = 3 P (E 1 ) = n (E 1 ) / n (S) = 3/8 = 0.375 (b) What is the probability of getting heads on at least one flip? The cases favorable to the event E 2 of getting heads on only one of your flips are, E 2 = {(HHH), (HHT), (HTH), (THH), (HTT), (TTH), (THT)} n (E 2 ) = 7 P (E 2 ) = n (E 2 ) / n (S) = 7/8 = 0.875 25. You are to participate in an exam for which you had no chance to study, and for that reason cannot do anything but guess for each question (all questions being of the multiple choice type, so the chance of guessing the correct answer for each question is 1/d, d being the number of options (distractors) per question; so in case of a 4 - choice question, your guess chance is 0.25). Your instructor offers you the opportunity to choose amongst the following exam formats: I. 6 questions of the 4 - choice type; y ou pass when 5 or more answers are correct; II. 5 questions of the 5 - choice type; you pass when 4 or more answers are correct; III. 4 questions of the 10 - choice type; you pass when 3 or more answers are correct. Rank the three exam formats according to

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    their attractiveness. It should be clear that the format with the highest probability to pass is the most attractive format. Which would you choose and why? Rank Exam Format Probability 1 II       5 6 6 0.25 0.75 0.25   = 0.0067 2 I       4 5 4 0.20 0.80 0.20   = 0.0046 3 III       3 4 4 0.10 0.90 0.10   = 0.0037 27 . A refrigerator contains 6 apples, 5 oranges, 10 bananas, 3 pears, 7 peaches, 11 plums, and 2 mangos. a. Imagine you stick your hand in this refrigerator and pull out a piece of fruit at random. What is the probability that you will pull out a pear? Total number of fruits in the refrigerator, n = 6 + 5 +10 + 3 + 7 + 11 + 2 = 44 Number of pears, m = 3 The required probability, p = m / n = 3/44 = 0.0682 b. Imagine now that you put your hand in the refrigerator and pull out a piece of fruit. You decide you do not want to eat that fruit so you put it back into the refrigerator and pull out another piece of fruit. What is the probability that the first piece of fruit you pull out is a banana and the second piece you pull out is an apple? Total number of fruits in the refrigerator, n = 6 + 5 +10 + 3 + 7 + 11 + 2 = 44 Number of banana s, m 1 = 10 The required probability of selecting banana first , p 1 = m 1 / n = 10 /44 = 0. 2273 Number of apple s, m 2 = 6 The required probability of selecting apple second , p 2 = m 2 / n = 6 /44 = 0. 1364

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