The next few questions refer to the following scenario. They are designed to ill
ID: 3209266 • Letter: T
Question
The next few questions refer to the following scenario. They are designed to illustrate how combinatorial reasoning is often important in characterizing the sample space for a probability distribution, which is critical to your research/experimental design. Your campus organization is made up of 10 students: {a, b, c, d, e, f, g, h, i, j}. You will soon each take a test and receive your scores. The test is very detailed, so there won’t be any ties; i.e. different students always receive different scores. In all cases, you will be asked to specify the sample space S, or some features of S, given particular basic outcomes (aka atoms, elementary events, etc.) of interest. [Alternatively, this “club” might be a set of 10 portfolios being evaluated in terms of their forecasted 2-year yields, or a set of 10 loan applicants being evaluated for their creditworthiness, etc.] 106. What is S if the basic outcomes are the top-scoring student? How many elements are in this set? 107. What is S if the basic outcomes are the number of students scoring above a 75? How many elements are in S? 108. How many elements are in S if the basic outcomes are the students ordered by their scores (highest to lowest)? 109. How many elements are in S if the basic outcomes are the top 3 scorers (in order)? 110. Suppose S is the set of possible top 3 scorers (in order), and the associated probability distribution assigns the same probability to each element of S. What is the probability of any one of these outcomes? 111. How many elements are in S if the basic outcomes are the top 3 scorers (unordered)? 112. Suppose S is the set of possible top 3 scorers (unordered), and the associated probability distribution assigns the same probability to each element of S. What is the probability of any one of these outcomes?
Explanation / Answer
Answers
106
S = {a, b, c, d, e, f, g, h, i, j}; n(S) = 10 [because any one of the 10 students could be top scorer.]
107
S = {n: n W, n < 11}; n(S) = 11 [because no student may get above 70. It may be just one student scoring above 70, ……, or all10 may get over 70]
108
S will have only 1 element, namely an ordered set of {a, b, c, d, e, f, g, h, i, j}
109
S will have 10C3 elements because any 3 out of 10 could be the top 3 scorers.
[note: answer is not 10P3 since top scores are in order and hence permuting is not allowed.]
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