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## Week 2 of Notes - Probability Theory and Statistics

1 review
by: Michelle Schmutz

16

3

5

# Week 2 of Notes - Probability Theory and Statistics 3341

Marketplace > University of Texas at Dallas > General Engineering > 3341 > Week 2 of Notes Probability Theory and Statistics
Michelle Schmutz
UTD
GPA 3.3

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second week of notes
COURSE
Probability Theory and Statistics
PROF.
Dr. Mohammed Saquib
TYPE
Class Notes
PAGES
5
WORDS
CONCEPTS
Math, Statistics, Probability, probability theory, and statistics
KARMA
25 ?

## 3

1 review
"I'm pretty sure these materials are like the Rosetta Stone of note taking. Thanks Michelle!!!"
Dr. Rudolph Hudson

## Popular in General Engineering

This 5 page Class Notes was uploaded by Michelle Schmutz on Saturday January 30, 2016. The Class Notes belongs to 3341 at University of Texas at Dallas taught by Dr. Mohammed Saquib in Winter 2016. Since its upload, it has received 16 views. For similar materials see Probability Theory and Statistics in General Engineering at University of Texas at Dallas.

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## Reviews for Week 2 of Notes - Probability Theory and Statistics

I'm pretty sure these materials are like the Rosetta Stone of note taking. Thanks Michelle!!!

-Dr. Rudolph Hudson

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Date Created: 01/30/16
Weekly Notes #2) Week 2 Notes mutually Theorems f exclusive ) • PCA , U Az ) =p (A Dtp CAD A A 2 The Union Ai and Az is to the addition of ' Ai Az of both which equal the universal . and , of equal set • If A = A U U for then Az U .. . Am and Ain Aj sum¥0tion itj , Ptt ) =§s PGD If the sets are collectively exhaustive and exclusive then the mutually , probability set A is to the all the of equal 1 at of of parts of A , starting at and ending M . • : PE ] ( probability measure ) Satisfies - P[0]=O There are 0 elements within an empty Sample - P[A '] = - P [A ] space ASetd its complement makeup the entire A sample AC space , whether there are other sets or not . If Something takes the entire it to 1 up sample space , is equal , A+Ac=1=s since we are talking about parts a whole . of - - to For any A and B , P[AUB]=P[A]tP[B ] P [ AAB ] ( doesn't have be mutually exclusive ) When finding the probability of a union , to - avoid what twice subtract counting overlaps , the intersection ⾨ 6 M probability of rolling a die on an even number or on a ¥={ multiple 063 . 2.4.63 n={ 3,6 } - If ACB ,then P (A) EP (B) * If A is B then A ) within , the property of B MBHKAUBAC is less than or to the A = PCAHPCBA 9>-0 equal property of B. µBA4=P( B) • The an event B= { sgsz .. . Sm } is the sum the probability of , , of probabilities the outcomes contained in the event . of PCB )=¥sP[{ si} ] an event is the sum its outcomes . of • For an experiment with spaces = { 5 , ... Sn } where each outcome si , sample , , is equally lively. unless restrictions are stated, all the outcomes have an chance p ( si) =£ I < i En . efud of occurring . P(dia)= to , 's BayeTheorem isplaying scavengerntgamewithherfrienMoria.Outq 9 it,ms considthedwinnTh.yarebothery gprobabilitygin,sotheyrebotequa?llosewithnortlower Gisheneedsatleaststobecoreofttwhigher Find:PH¥Pp{{}U{8 }U,9}) PN UsesO}.. } EDNIHA Hu/+1110+1110) =PYa..tkP({ = 8.BBAAF=PCBPCAABTTheconditiprobability PCA Occuranuqtheeventb ) *PABnlAUBD=PHB) 㱺 PCA- I conditprobability PCBA)CBHA e)InAmericapopulation inearsighted ,2s%qthe -0.20=201100opubtionislqthanded.20Yoqthepopulationwereeithennear-sightedorlefthanded.a)WhatistheprobabilityP(NL)thatanAmericanisbothnear-sightedW)audislyt Bhathecpoftheunionqndl P(N)=g) # I P( of㱺PwL)=Ffo=7s,othe410 ( NLAPNU) - NLACNULD 'tS occumnaqthe(NL # P(YN)=P(NL) interseQNLon㱺5p( 't >P( PTNJ ND - PCNHPCL N L 㱺6P(N4=PW)+p(f ls㱺P( p¥u =¥ot÷o=¥k ABN*p(B)=PHB) Dependent Independent b) What is the conditional probability that an American is near.sighted that they given are left handed ? )=PfY¥-=YsYfo㱺Fz×l0s¥=s±o - KHH PKPN ¥P(4=÷o P ( L ) t PCLDN ) ȼ the occurrence near -sightedness could the Occurrence handedness . change of left Therefore ,left handedness depends on nearsightedness . q * : the the this Dependent property of one event changing ( ijust an Occurrence another event. example 1 made a.,3{Y• PCAHP of this hotfatf (At B) up ; is & Independent :Occurrence one event doesn't change / influence the likelihood another event .his also means must intersect , of they ., ; , PCAKPCAIB ) - Pfaff 㱺 MAB ) =PLAPIB ) * 㱺 Are A Bc , ' and inde'endent ? ¥thIB P C ) = Iftwo events PCAIPCBD AB CA ) 1- PCB are { } Ⱦ Bc } =P ( )) independent, allof AB { A, % , =P (A) -PLA) MB ) its are 9 INDEPENDENT t =P (A) - PCAB alsoindependent { A' B } - { A ' Bc} ) , , ED Jemma is Leo some screenshots her conversation with Will .However sending to of tend be reScreenshots of order. She also sends some regular text that will have tmessagese issue. Jemma has bad Jemma observes two sent her If ⊥ " things being from phone . wif¥phots send one goes another , then itshould be labeled ( SS ) .hesetwo set are independentfrom each Othe,but the probabilitof tub beingsent in the orders is . right 0.9 Identity set I by Ci Ⱦ G=T G=S an of , :* " .ee initiate "sej¥÷ Ⱦ Msz1}ȾP(Mv=2 0.1×0.1=0.01 Ms=2 , Y wmiw independent 9) { and )=p (s )Pk)= 0.1×0.1=0.01 } { ,H=P¢ , Ⱦ P(Mv 21) =P (S,SZUSTZUISD = 0.01+0.09+0.09=0.19 exclusive PCT ,Tz) = 1-0.81=0.19 - = - mutually 㱺T We Know that PLAB ) = PCA ) PCB ) needs to be the for this case to be independent PCSSDE P(Mv=z)P( Mvzl ) 0.01£ 0.01 × 0.19 0.01=10.0019 Ⱦ not independent Loofotalobability and G=T b){ Cz=S} { } PKED =P (,Sz )UtszSfs)+ PC,SD = 0.01+0.09=0.10 PCTSTUTTD P(C,=D = =P ( IN tP(TTz)= 0.09+0.81=0.90 s PCAB)=PCA)PCB) itPkzasnc ,=T)±P(Cz㱺)PART) It{p( TSD 0.10×0.90 0.09±0.09 ȼ independent Theorem A • Forapartition= (i,Bz,.)and event in thesamplespac,letCs=AAB .For it,theevents CiandCj are mutuallyxclusived .. *PAD=PcB)PCApB B' BeB BaBs MHE;Up;U£nhE , \ Abthtktnjpbttlktfbpfbnpaibn • ForGU ==*Pp{ ) ȼ Lately a a a a a Plty event ,andpartition,Bz,.,Bm), )=§aP( AABD ciaan E Ci e)Custom(SzaorimorehantwoSs.Thecustomeranaveit,evanillaortwo SChocolate.ManypeopfindVanillatobla,dstheymakup q customers Fioutthe z 5 PC,)= PCS) =p(Ss) y CS CszVS} PA) =1/3 45 , 03 PCC Valen'sdayrecentpassedd manycustomersvehadthefiofchocolate forawhil713of>3/5ustomerslchoosonescooofchocolatehere will choosetable5oopof vanillalouthetableiththisdditionalorma.ion S, Sz 5 i tIt )=V YS %%Y4 31/6 6 /S C 1/6 Yz Yz B

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