ECE313.Lecture04

ECE313.Lecture04 - ECE 313 Probability with Engineering...

Info iconThis preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
The Axioms of Probability, Part II Professor Dilip V. Sarwate Department of Electrical and Computer Engineering © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign. All Rights Reserved ECE 313 Probability with Engineering Applications
Background image of page 1

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

View Full DocumentRight Arrow Icon
ECE 313 - Lecture 4 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 2 of 42 Review of Lecture #3 We explored the consequences of the axioms of probability We discussed the notion of a partition and how partitions can be used to calculate the probability of an event We discussed binomial coefficients We looked at a simple combinatorial problem
Background image of page 2
ECE 313 - Lecture 4 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 3 of 42 Drawing a random sample Example: An urn contains 6 identical red balls R 1 , R 2 , R 3 , R 4 , R 5 , R 6 and 3 identical green balls G 1 , G 2 , G 3 . A trial of the experiment consists of simultaneously drawing two balls at random from the urn The outcomes of this experiment are subsets of size 2 of the form {R 1 ,R 5 } or {R 4 ,G 1 } or {G 2 ,G 3 }
Background image of page 3

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

View Full DocumentRight Arrow Icon
ECE 313 - Lecture 4 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 4 of 42 How many subsets of each kind? There are = 36 subsets of size 2 from a set of 9 balls, and the collection of these subsets is the sample space = 15 outcomes consist of two red balls = 3 outcomes have two green balls =18 outcomes have 1 red, 1 green 9 2 6 2 3 2 6 1 3 1
Background image of page 4
ECE 313 - Lecture 4 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 5 of 42 More probabilities A = {R 2 is in the sample drawn} B = {G 2 is in the sample drawn} What is P(A B) ? P(A) = 8/36 P(B) = 8/36 AB = {outcome = {R 2 , G 2 }} is a singleton event, and hence P(AB) = 1/36 P(A B) = P(A) + P(B) – P(AB) = 15/36 Exercise: What is P(A c B c )?
Background image of page 5

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

View Full DocumentRight Arrow Icon
ECE 313 - Lecture 4 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 6 of 42 Sampling without replacement A sample of size k from a set of size n is a subset of size k {R 2 , G 2 } is the same as {G 2 , R 2 } Instead of obtaining the k elements of the subset simultaneously, we could draw them out one at a time, each draw being carried out without replacing the previously drawn elements This is sampling without replacement
Background image of page 6
ECE 313 - Lecture 4 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 7 of 42 Sampling without replacement If a previously drawn element were to be put back, then it could be drawn again This is not what we want! Sampling without replacement results in
Background image of page 7

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

View Full DocumentRight Arrow Icon
Image of page 8
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/29/2009 for the course ECE 123 taught by Professor Mr.pil during the Spring '09 term at University of Iowa.

Page1 / 44

ECE313.Lecture04 - ECE 313 Probability with Engineering...

This preview shows document pages 1 - 8. Sign up to view the full document.

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