This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Solutions  Homework #3 1 Chapter 2 1.1 Problems 1. Problem 17 : If 8 castles (rooks) are randomly placed on a chessboard, compute the probability that none of the rooks can capture any of the others. That is, compute the probability that no row or file contains more than one rook. Since there are 64 squares on a chessboard with 8 rows and 8 columns, there are: ˆ 64 8 ! ways to place 8 rooks in 64 squares (assuming no two are in the same spot). With 8 rows & 8 columns, every row & every column must have exactly 1 rook. So choosing where to place a rook is equivalent to choosing a row/column pair. How many such pairs are possible? Choose row 1. There are 8 ways to choose a column as its mate. Choose row 2. There are 7 remaining ways to choose its column mate, etc. ⇒ 8 · 7 ··· 2 · 1 = 8! ways to choose the pairs. ⇒ P (No row/column with more than 1 rook) = 8! ˆ 64 8 ! = 9 . 109 × 10 6 . 2. Problem 37 : An instructor gives her class a set of 10 problems with the information that the final exam will consist of a random selection of 5 of them. If a student has figured out how to do 7 of the problems, what is the probability that he or she will answer correctly: (a) All 5 problems? There are: ˆ 10 5 ! total ways to have a random selection of 5 out of 10 problems, ˆ 7 5 ! ways to have 5 of the 7 “workable” problems selected, and ˆ 3 ! = 1 way to have none of the “unworkable” problems selected....
View
Full
Document
This note was uploaded on 04/15/2008 for the course STAT 425 taught by Professor Pembridge during the Fall '07 term at University of Michigan.
 Fall '07
 Pembridge
 Probability

Click to edit the document details