Understanding Tree and Venn Diagrams for Probability Analysis
School
University of California, Davis**We aren't endorsed by this school
Course
MISC 114
Subject
Statistics
Date
Dec 10, 2024
Pages
3
Uploaded by CommodoreIronTiger40
pg. 9Tree Diagram A Tree Diagram is a special type of graph used to determine the outcomes of an experiment. It consists of “branches” labeled as frequencies or probabilities. We’ll use a Tree Diagram in our example for Bayes’ Rule. You may view additional examples in the book. Venn Diagram A picture that represents the outcomes of an experiment. Example: Suppose an experiment has outcomes black, white, red, orange, yellow, green, blue, and purple, where each outcome has an equal chance of occurring. Let event C = {green, blue, purple} and event P = {red, yellow, blue}. a)Find ࠵శ࠵? ∩ ࠵ృ࠵?and ࠵శ࠵? ∪ ࠵ృ࠵?. b)Draw a Venn diagram representing this situation. Bayes’ Theorem An extension of the Conditional Probability Rule: C1p=Eblue3Cup=Egreen,blue,purple,red,yellow3SCPgreenredpurple-yellowblack,white,orangeP(AIB)=P(ANB)EconditionalRule3SP(B)AB=PACBIA)EmultiplicationRubP(ANB)+PLA'B)ElawoftotalProbability3=P(A)P(B(A)Bayes'.P(A)P(B(A)+P(A))p(BIA)Emuet.Rule3Rule
pg. 10Example: The probability to have a type of cancer in a population is 0.03. A test for the disease is 99% accurate if the person has the disease; and it is 98% accurate if the person does not have the disease. What is the probability of a person having the disease if the test result is positive (test result indicates the person has the disease)? accuratetest:Ifhavethedisease-o+testIfdonothavedisease-x -testCtCGiven:P(c)=0.03p(+(c)=0.99p)-1C')=0.98Complements:P(C')=0.97P)-(C)=0.01P(+(c)=0.02Question:p(c)+)=?P(c(+)==p(c)p)+(c)Emult.Rule3P(+MC)+P(+MC)&LawoftotalProb-waysofa+testresult3=P(c)P(+(c)P(c)p(+(c)+P(c)P(+(c)=(0.03)(0.99)=0.6049=60.49%(0.03)(0.99)+10.97)(0.02)
ProbabilityTreGiven:p(c)=0.03P(+(c)=0.99p)-1C')=0.98p(c)P(+(c)=P(c1+)=P(+1c)P(c=0.03Ph+12=0.99(0.03)(0.99)=0.0297ps-(c)=0.0P(c)P(-(c)=P((1->10.03)(0.01)=0.0003P(+(c)=0.02P(((>P(+(c)=P(c'n+)(0.97)(0.02)=0.0194p303+0.97ps-est0.sP(c))p(-(C)=P(c'1-)(0.97)(0.98)=0.9506Q:P(c(t)=P(c1+)=0.0297=0.6049=60.49%P(+)0.0297+0.0194