s1 - Stat 5101 (Geyer) 2002 Midterm 1 Problem 1 This is...

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: Stat 5101 (Geyer) 2002 Midterm 1 Problem 1 This is like Problems 1.8.4 and 1.8.10 in DeGroot and Schervish. There are ( 20 5 ) ways to choose 5 light bulbs out of 20 light bulbs. So that is the size of the sample space. The phrase selected at random in the problem statement means all are equally likely. There are ( 17 5 ) ways to choose 5 light bulbs out of the 17 non-defective light bulbs. Hence the probability is ( 17 5 ) ( 20 5 ) = 6188 15504 = 0 . 3991 Problem 2 This is a Bayes rule problem like Problem 2.3.8 in DeGroot and Schervish. Let A be the event tests positive and B be the event uses drugs. Then we are given Pr( A | B ) = 0 . 96 Pr( A c | B ) = 0 . 04 Pr( A | B c ) = 0 . 02 Pr( A c | B c ) = 0 . 98 Pr( B ) = 0 . 05 from which we can derive (by the complement rule) Pr( B c ) = 0 . 95 What we want is Pr( B | A ) = Pr( A | B )Pr( B ) Pr( A | B )Pr( B ) + Pr( A | B c )Pr( B c ) = . 96 . 05 . 96 . 05 + 0 . 02 . 95 = 0 . 7164 Problem 3 This is like Problem 3.2.4 in DeGroot and Schervish....
View Full Document

This note was uploaded on 10/28/2010 for the course STAT 5101 taught by Professor Staff during the Fall '02 term at Minnesota.

Page1 / 3

s1 - Stat 5101 (Geyer) 2002 Midterm 1 Problem 1 This is...

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