HW4Solutions (dragged) 2 - P n,m = n-m n m,n ≥ m...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
We will assume all sequences are equally likely. To calculate P 2 , 1 , the only sequence where A leads B for all counts are { AAB } . The total number of sequences are ( n + m )! n ! m ! = (2+1)! 2!1! = 3, thus, P 2 , 1 = 1 3 . To calculate P 3 , 1 , the only sequences where A leads B for all counts are { AAAB, AABA } . The total number of sequences are ( n + m )! n ! m ! = (3+1)! 3!1! = 4, thus, P 3 , 1 = 2 4 = 1 2 . To calculate P 4 , 1 , the only sequences where A leads B for all counts are { AAAAB, AAABA, AABAA } . The total number of sequences are ( n + m )! n ! m ! = (4+1)! 4!1! = 5, thus, P 3 , 1 = 3 5 . To calculate P 3 , 2 , the only sequences where A leads B for all counts are { AAABB, AABAB } . The total number of sequences are ( n + m )! n ! m ! = (3+2)! 3!2! = 10, thus, P 3 , 1 = 1 5 . Similarly, P 4 , 2 = 1 3 and P 4 , 3 = 1 7 . Theoretical Exercise 3.21 b. P n, 1 = P (A receives first 2 votes) = n n + 1 n - 1 n = n - 1 n + 1 P n, 2 = P (A receives first 2 votes and at least 1 of the next 2 = n n + 2 n - 1 n + 1 1 - 2 n 1 n - 1 = n - 2 n + 2 Theoretical Exercise 3.21 c.
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: P n,m = n-m n + m ,n ≥ m Theoretical Exercise 3.21 d. P n,m = P (A always ahead) = P (A always ahead | A receives last vote) n n + m + P (A always ahead | B receives last vote) m n + m = n n + m P n-1 ,m + m n + m P n,m-1 Theoretical Exercise 3.21 e. The conjecture of (c) is true when n + m = 1 ( n = 1 ,m = 0). Assume by induction that P n,m = n-m n + m when n + m = k . Now suppose that n + m = k + 1. By (d) and the induction hypothesis, we have that P n,m = n n + m n-1-m n-1 + m + m n + m n-m + 1 n + m-1 = n-m n + m (1) which completes the proof. Theoretical Exercise 3.22 If the weather is dry on January 1st, then P = P (dry on Jan 1st) = 1 and, conditioning 3...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern