Lecture Notes 1-20

# Lecture Notes 1-20 - BMW 1-1 An W ﬁg 3 maﬁa-amaﬁ \$5...

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Unformatted text preview: BMW 1-1: An W ﬁg 3. maﬁa-amaﬁ \$5 an mam. A 32: aim 4317 againy 1513:1552; with: Wig 5mm: £9 (335% an mm; 0 w=WogW Qeﬁﬁiﬁw 1.2. Armﬁaéé‘ﬁgzsm i535. ﬂaw-{agile ﬁﬁm} wheres} 35 “mph spam g is a {mum 9% yams ﬁrm the mph spate, am? \$9: \$3 a. mm 2 Wm Mammyﬁmﬁ 3mm: 11M: fnﬁmmgaawﬁwg mama wt 3 g 35 méthiza ﬂ; gr mm m 9" :4? \$2; Miuajj 4.9131325? ﬁﬁﬁammm¢ at 2 1 ark-mi? mm mamm‘ The four condition is actually a result of the ﬁrst three: AU A” = Q A f] A“ ¢ Q5 Pr{A U A“} = Pr{Q} =1 Pr{A U A”} = Pr{A} + Pr{Ac} =1 Pr{A“} = 1- Pr{A} ﬂgﬁaﬁiﬁim 3.3- L2; he a mama? WE wig: 35a mwﬁ 35121 .§ mm m3] g: as, mﬁﬁmmﬁz‘ﬁﬂiﬁ’ um \$213.35 damn: Rigiij is 335 La \$13“; :‘Fiiﬁlgi = - ﬂmﬁa Li A Esilepilmﬁﬁ mwﬁsximag imam? maﬁa mm mam 3% mix: 13km mmbum at” m {game E39132 ﬁg a: ham}. a qualiﬁr {mﬁmmm‘zimmm watts :3 Wﬁms gmnrﬂa‘ ma'gsgeﬂfpkm Quake hm: mpﬁami mm the 133m mi mark the 1113;331:152" were {iﬁfmﬂu The mph spam is ﬂ m“ 5‘33 1:5 E315 33!. \$33 “£333.33? gag 43\$ £3 £55: afaﬁ 535mg 3mm m There an my ﬁgﬁme muggng aim maﬁa! 5w asmﬁztad wéﬁx 31mm. ﬁlm meﬁﬁm-‘ia Mir: \$21} 7: Mi , wmwﬂw em Frﬁn: Iii}-m-W?, mm 1;} 2 @333 ﬁ WEE 22!}m-ﬁam a mﬁmj} m mm s: m; a :és- gelazm‘ mag {mm méﬁas lama Eét‘mﬂﬁi at] is: giaw 12323? "3‘ -- a s: at :tieaet'me we is ﬁsheries in the box maturing seam: moses: ﬁeﬁae the went a! te be ﬁe set { it; 33% lit; E), in, 6:35, (3;-5a E and the re be {t};- ﬂf; t a}, ﬁrﬁﬂeliryteﬂeemorﬁe, .«f is the event nfhassizsgg a! reset we ﬁeﬁcﬂee phase, see? 5‘ is: the was! of a hex efanﬁiie glam 3‘13? gamete: {are nwhe wﬁtteu ea , _ _§r§§ﬂ3§_ E‘ﬁ'ggﬁjiﬁrgljg} “ass Wig“ we; meaIi—Elaéiliiblriaﬁjrﬁ = #3:??1‘ A key aSpect of using the probability rules is when non-disjoint events are encountered. If A n B at Q , then show that Pr{AU B} = Pr{A} + Pr{B} — Pr{A ﬂB}. Proof: The approach is to separate these areas into disjoint sets (events) and then utilize the property that the probabilities of disjoint events add. 9 ‘ B Since B=(AnB)U(A“nB) Pr{B} = Pr{(A f) 3)} + Pr{(A“ f] 3)} and A=(AﬂB)U(AﬂB”) Pr{2"1'}= Pr{(A ﬂ 3)} + Pr{(A f? B”)} Now A U B as disjoints sets is AUB=(AnBC)U(AnB)U(Ac/73) PPM U B} = PIKAH 36)} + PF{(A/7 B)}+ PIKAC f7 3)} Note that Pr{(A” n 3)} = Pr{B} -— Pr{(A f7 3)} and Pr{(A [7 BC)} = PIM} - Pr{(A n 3)} So Pr{(AUB)} = Pr{A}+Pr{B}—Pr{(AﬂB)}. Q.E.D. Do homework 1.1, 1.2, (pages 36-37 in the textbook) and derive the formula Pr{A U BU C} = Pr{A} + Pr{B}+ Pr{C} — Pr{A {7 B} — Pr{A n C} —Pr{B f} C} + Pr{A n B {7 C} Turn in on Monday 1-25, at the beginning of class! ...
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Lecture Notes 1-20 - BMW 1-1 An W ﬁg 3 maﬁa-amaﬁ \$5...

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