This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Illinois Institute of Technology Department of Computer Science Second Examination CS 330 Discrete Mathematics Spring, 2007 1:50pm–3:05pm, Wednesday, April 4, 2007 238 Stuart Building Print your name and student ID, neatly in the space provided below; print your name at the upper right corner of every page. Please print legibly. Name: Student ID: This is an open book exam. You are permitted to use the textbook, any class handouts, anything posted on the web page, any of your own assignments, and anything in your own handwriting. Foreign students may use a dictionary. Nothing else is permitted : No calculators, laptops, PDAs, cell phones, etc. Do all six problems in this booklet. All problems are equally weighted, so do not spend too much time on any one question. Show your work! You will not get partial credit if the grader cannot figure out how you arrived at your answer. Question Points Score Grader 1 20 2 20 3 20 4 20 5 20 6 20 Total 120 CS 330 Second Exam—Spring, 2007 2 Name: 1. Probability of Passing CS 330 ((a):8, (b):12) (a) A student in CS 330 has to take two midterm exams and a final. To pass the course the student must pass at least one midterm and must pass the final. Let the probability of passing the first midterm be p 1 , the probability of passing the second midterm be p 2 , and the probability of passing the final be 1 / 2. Professor Reingold claims that the probability of passing the course (that is, of passing one of the midterms and the final) is 2 / 3. Is this possible? ( Hint : Express p 1 in terms of p 2 by expressing the probability of passing the course as a function of p 1 and p 2 .) (b) Professor Reingold decided to make the final easier so that probability of passing it would be 3 / 4, and he announced that the probability that a student would pass the second midterm exam was 1 / 2 if s/he failed the first midterm exam and was 3 / 4 if s/he passed the first exam. Use conditional probability and Baye’s Theorem to derive a relationship between p 1 and p 2 (you should get a quadratic relationship which you need not solve). Sol: Let M 1 , M 2 , F be the events that the student is successful in the first midterm, second midterm, and the final respectively. (a) Then, p ( M 1 ) = p 1 , p ( M 2 ) = p 2 , p ( F ) = 1 2 . p (passing the course) = p ( M 1 ∩ M 2 ∩ F ) + p ( M 1 ∩ M 2 ∩ F ) + p ( M 1 ∩ M 2 ∩ F ) = p ( M 1 ) p ( M 2 ) p ( F ) + p ( M 1 ) p ( M 2 ) p ( F ) + p ( M 1 ) p ( M 2 ) p ( F ) (since these events are indepen dent) = p ( F )( p ( M 1 ) p ( M 2 ) + p ( M 1 ) p ( M 2 ) + p ( M 1 ) p ( M 2 )) = 1 2 (( p 1 )( p 2 ) + (1 p 1 )( p 2 ) + ( p 1 )(1 p 2 )) = 1 2 ( p 1 + p 2 p 1 p 2 ) ⇒ 1 2 ( p 1 + p 2 p 1 p 2 ) = p (passing the course) ⇒ 1 2 ( p 1 + p 2 p 1 p 2 ) = 2 3 ⇒ p 1 + p 2 p 1 p 2 = 4 3 ⇒ p 1 (1 p 2 ) = 4 3 p 2 ⇒ p 1 = 4 3 p 2 1 p 2 We know that the feasible values of p 2 are in the range [0 , 1]. For any value chosen for p 2 in this interval, p 1 > 1. Hence there does not exist an assignment for both1....
View
Full Document
 Spring '08
 Reingold,EdwardM.
 Computer Science, Conditional Probability, Recurrence relation, Fibonacci number, M1 M2, K. Professor Reingold

Click to edit the document details