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midterm2sol

# midterm2sol - CS 70 Fall 2013 Discrete Mathematics and...

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CS 70 Discrete Mathematics and Probability Theory Fall 2013 Midterm #2 1. (30 points) Short Answer. No justifcation necessary. Please write only the answer in the space provided aFter each question. Please do any calculations on the back oF the page. (a) (2 points) How many distinct polynomials of degree (at most) 3 are there modulo 11, such that the value at the points x = 1 , 2 , 3 , 4 lie in the set { 1 , 2 , 3 , 4 , 5 } ? Solution: 5 4 . 4 points uniquely determine a degree 3 polynomial and each of the 4 points have 5 possibilities. (b) (2 points) A bridge hand is a set of 13 cards from a standard 52-card deck. A bridge hand is called balanced if it has 4 cards from one suit and 3 each from the remaining three suits. How many distinct balanced bridge hands are there? Solution: ° 4 1 ±° 13 4 ±° 13 3 ± 3 . Select one suit out of 4 to have 4 cards from that suit, and all other suits have 3 cards. (c) (2 points) Recall that an anagram of a word is a string made up from the letters of that word, in any order. (For instance, there are exactly three anagrams of BEE: namely EEB, EBE, and BEE. Note that by our de±nition anagrams need not form legal words.) How many anagrams are there of the word ANAGRAM? Solution: 7! 3! . There are 7! ways to permute the letters, and 3! accounts for over counting for the A’s. (d) (2 points) In an n dimensional hypercube, how many vertices are there at a distance of exactly 6 from a particular vertex? Solution: ° n 6 ± . The vertex can be represented by an n -bit string and all vertices at a distance of exactly 6 differ in 6 bits - we just need to choose which 6 bits. (e) (2 points) If A , B , C are event such that P [ A ]= . 5, P [ B . 4 and P [ C . 3, and such that P [ A B . 2, P [ A C . 1, P [ B C . 1 and P [ A B C . 9. What is P [ A B C ] ? Solution: . 1 . By the inclusion/exclusion principle, Pr [ A B C Pr [ A ]+ Pr [ B Pr [ C ] Pr [ A B ] Pr [ A C ] Pr [ B C Pr [ A B C ] . (f) (5 points) Pick two non-zero numbers x and y modulo 7 at random. Let z = xy mod 7. Let A be the event that x = 3, B that y = 3 and C that z = 3. Note: An alternate interpretation would be picking a random non-zero number and reducing it modulo 7, so you are picking from the set { 0 ,..., 6 } . This was also given full credit, and the solutions for this interpretation (in italics) are included below. Of course you used the alternative interpretation the solution to part v. was different. i. What is Pr [ C ] ? Solution: 1 6 . z can be any value other than zero. CS 70, Fall 2013, Midterm #2 1

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Name: SID: Alternate interpretation: 6 49 . Let x take on any non-zero value (6 possibilities) and y equal 3 x 1 . ii. What is Pr [ A B ] ? Solution: 1 36 . Both x and y are equal to 3 with probability ( 1 6 ) 2 . Alternate interpretation: 1 49 . Both x and y are equal to 3 with probability ( 1 7 ) 2 . iii. What is Pr [ A C ] ?
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midterm2sol - CS 70 Fall 2013 Discrete Mathematics and...

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