MA MATHEMATICAL ASSOCIATION OF AMERICA Log Kwebwork / 12017stat213101 / discrete
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MA MATHEMATICAL ASSOCIATION OF AMERICA Log Kwebwork / 12017stat213101 / discrete_distributions lab_exercises /3 Discrete Distributions Lab Exercises: Problem 3 Previous ProblemProblem List (1 point) Suppose I give you a list of 20 problems to study, from which I will randomly pick 14 questions for your first midterm exam. For whatever reason, you prepare for the midterm exam by completing and understanding how to solve 11 questions of the 20, so there are 9 Next Problem questions you do not know how to solve. Part (a) Can you identify the underlining probability model? What is the probability that you completely solve 8 of the 14 questions appearing on the midterm exam? Ell If rounding use at least four digits after the decimal in your answer) Part (b) To pass the midterm, you must correctly solve at least half of the 14 questions. What is the probability of you passing the midterm exam? EE (If rounding use at least four digits after the decimal in your answer) Preview My Answers Submit Answers You have attempted this problem O times You have unlimited attempts remaining ated at 10/24/2 7 at 12:05pm MDTExplanation / Answer
(a) The underlying probability model is the hypergeometric distribution. The hypergeometric distribution is a discrete probability distributionthat describes the probability of k successes (random draws for which the object drawn has a specified feature) in n draws, without replacement, from a finite population of size N that contains exactly K objects with that feature, wherein each draw is either a success or a failure.
To solve 8 questions out of the 14 asked in the examination, those 8 questions must be from the 11 questions for which preparation has been done. This can happen in C(11,8) ways. The other 6 questions are from the 9 unprepared questions and this can happen in C(9,6) ways. But 14 questions can be chosen from the 20 possible questions in C(20,14) ways.
So the probability of solving exactly 8 questions out of the 14 asked = C(11,8) * C(9,6)]/C(20,14) = 165*84/38760 = 0.357585.
(b) Probability of passing the mid term examination = probability of solving 7 questions + probability of solving 8 questions + probability of solving 9 questions + probability of solving 10 questions + probability of solving 11 questions + probability of solving 12 questions + probability of solving 13 questions + probability of solving 14 questions = C(11,7) * C(9,7)]/C(20,14) + C(11,8) * C(9,6)]/C(20,14) + C(11,9) * C(9,5)]/C(20,14) + C(11,10) * C(9,4)]/C(20,14) + C(11,11) * C(9,3)]/C(20,14) + 0 + 0 + 0 = 330*36/38760 + 165*84/38760 + 55*126/38760 + 11*126/38760 + 1*84/38760 = 0.306502 + 0.357585 + 0.178793 + 0.035759 + 0.002167 = 0.8808
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