This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Solutions to Homework 03 [1] The number of ways of picking 20 raccoons out of N is N 20 ! The number of ways of picking 5 tagged raccoons out of 10 and 15 untagged raccoons out of N10 is 10 5 ! N 10 15 ! The probability of picking 5 tagged out of 20 is p ( N ) = 10 5 ! N 10 15 ! N 20 ! To find the N that maximizes p ( N ), we need to find the range of N where p ( N ) increases with N. This is equivalent to the range of N where p ( N ) /p ( N 1) ≥ 1. p ( N ) p ( N 1) = N 10 15 ! N 1 20 ! N 20 ! N 11 15 ! = ( N 10)( N 20) ( N 25) N ≥ 1 ( N 10)( N 20) ≥ ( N 25) N ⇒ 40 ≥ N Therefore, when N = 40, p ( N ) has the maximum of p (40) = . 284. [2] (i) P [ A  B ] = P [ A ∩ B ] P [ B ] for P [ B ] > 0. Now ≤ P [ A ∩ B ] , < P [ B ] ⇒ ≤ P [ A  B ] Also, ( A ∩ B ) ⊂ B ⇒ P [ A ∩ B ] ≤ P [ B ] ⇒ P [ A  B ] ≤ 1 Therefore, ≤ P [ A  B ] ≤ 1 ii. First note that B ∩ S = B . Then P [ S  B ] = P [ B ∩ S ] P [ B ] = P [ B ] P [ B ] = 1 1 iii. We’ll use the following result...
View
Full
Document
This note was uploaded on 09/03/2008 for the course ECSE 4510 taught by Professor Woods during the Spring '03 term at Rensselaer Polytechnic Institute.
 Spring '03
 WOODS

Click to edit the document details