set3

First Course in Probability, A (7th Edition)

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Third Homework Set — Solutions Chapter 2 Problem 37 (a) There are ( 10 5 ) selections for the final exam. The number of selec- tions that allow the student to solve all problems is ( 7 5 ) , so that the desired probability is ( 7 5 ) ( 10 5 ) = 0 . 08333. (b) There are ( 7 4 ) · ( 3 1 ) selections that’ll let the student solve exactly four problems, so that the probability of solving at least four prob- lems is ( 7 5 ) + ( 7 4 ) · ( 3 1 ) ( 10 5 ) = 1 2 . Problem 43 (a) There are n ! ways to arrange n people in a line. There are 2( n- 1)! ways to arrange them such that A and B are next to each other. Hence, the probability of A and B being next to each other is 2( n- 1)! n ! = 2 n . (b) If n = 2, then A and B will always be next to each other. Now, assume that n > 3. After A picks a seat, there are n- 1 seats left, two of which are next to A , so that the desired probability is 2 n- 1 . Problem 50 The probability that you have five spades and your partner has the remaining eight spades is ( 13 5 )( 39 8 )( 8 8 )( 31 5 ) ( 52 13 , 13 , 26 ) = 2 . 6084 · 10- 6 ....
View Full Document

This document was uploaded on 04/23/2010.

Page1 / 4

set3 - Third Homework Set — Solutions Chapter 2 Problem...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online