36225 Chapter Notes - Chapter 2: Disjoint Sets, Bayes Estimator, List Of Poker Hand Categories

127 views23 pages
School
Department
Course
Professor

Document Summary

A = {ff}, b = {mm}, c = {mf, fm, mm}. Then, a b = 0/ , b c = {mm}, C = {(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)} A b = {(2,2), (4,2), (6,2), (2,4), (4,4), (6,4), (2,6), (4,6), (6,6)} Ba = {(1,2), (3,2), (5,2), (1,4), (3,4), (5,4), (1,6), (3,6), (5,6)} Ba = everything but {(1,2), (1,4), (1,6), (3,2), (3,4), (3,6), (5,2), (5,4), (5,6)} A = {two males} = {m1, m2), (m1,m3), (m2,m3)} B = {at least one female} = {(m1,w1), (m2,w1), (m3,w1), (m1,w2), (m2,w2), (m3,w2), B = {no females} = a: 36 + 6 = 42. S = {a+, b+, ab+, o+, a-, b-, ab-, o-: p({a}) = 0. 41, p({b}) = 0. 10, p({ab}) = 0. 04, p({o}) = 0. 45, p({a} or {b}) = p({a}) + p({b}) = 0. 51, since the events are mutually exclusive. Since p(s) = p(e1) + + p(e5) = 1, 1 = . 15 + . 15 + . 40 + 3p(e5).

Get access

Grade+20% off
$8 USD/m$10 USD/m
Billed $96 USD annually
Grade+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
40 Verified Answers
Class+
$8 USD/m
Billed $96 USD annually
Class+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
30 Verified Answers

Related Documents