elem. stat. week 4 ex 2

elem. stat. week 4 ex 2 - Date Course Math 201 Fall B10...

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Unformatted text preview: Date: 10/16/11 - Course: Math 201 Fall B10 Time: 6:01 PM Book: Larson: Elementary Statistics: Picturing the World, 5e Assignment: Week 4 Exercises Multiple-choice questions each have four possible answers (a, b, c, d), one of which is correct. Assume that you guess the answers to three such questions. 21. Use the multiplication rule to find P(WCC), where C denotes a correct answer and W denotes a wrong answer. By the multiplication rule, since each question is independent of the others, P(WCC) is equal to P(W) ' P(C) - P(C). If each multiple—choice question has four possible answers, what is the probability of guessing the correct answer? P(C) = 0.25‘ What is the probability of guessing a wrong answer? P(W) : .75 Now substitute the probabilities into the new expression for P(WCC) and simplify. P(WCC) P(W) -P(C) -P<C) : (0.75) ' (0.25) - (0.25) = .046875 (Type an exact answer.) b. Beginning with WCC, make a complete list of the different possible arrangements of two correct answers and one wrong answer, then find the probability for each entry in the list. The easiest way to determine the different possible arrangements is to place the one wrong answer in each possible position and then fill in the other positions. Doing this, we see that the possible arrangements are WCC, CCW, and CWC. Now find the probabilities for CCW, and CWC. To do so, follow the same process as in part a. P(CCW) = 0462375 (Type an exact answer.) P(CWC) = .046875 (Type an exact answer.) c. Based on the preceding results, what is the probability of getting exactly two correct answers when three guesses are made? To find this answer, first determine what expression this probability is equal to. Choose the correct answer below. Page 1 Assignment: Week 4 Exercises ? Date: '10/16/1 i i "‘ Course: Ma'iii B10 'Time: 6:01 PM Book: Larson: Elementary Statistics: Picturing the World, Se P(WCC) -— P(CCW) -— P(CWC) P(WCC) --P(CCW) -*P(CWC) — P(WCC) - P(CCV\0 - P(CWC) P(WCC) -P(Ccm -P(CWC) P(WCC m ccw m CWC) Now substitute in the probabilities and simplify. P(WCC) + P(CCW) +1?(_CWC) : 0.046875 + 0.046875 + 0.046875 . . @ = .140625 (T can exact answer.) Therefore, the probability of getting exactly two correct answers when three guesses are made is 0.140625. Page 2 ...
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elem. stat. week 4 ex 2 - Date Course Math 201 Fall B10...

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