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Chapter 5 Notes (4)

# Chapter 5 Notes (4) - mil-fetch“ CandD‘ th =(6 CMD=CUD...

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Unformatted text preview: mil-fetch“. CandD‘ th = (6) CMD=CUD =(2.3.4.6) 5} (1) I CmD Candi) Consider the use nf losing lwa wins A-[e'tllnl at least one heads I! —getting at least one tails S = {(HVH)-(H:T)v(T-H).(T.T)) A = {(Hr”):(H-T).(T.H)} B = ((HlT)‘(T»H)r(T- T» A and!) =A ns = ((H,T).(T,H)) A ur a , A u s = ((M.H). (”.1“). (T.H),(‘I‘,1‘)) = : Amt-l AandB “WWW For the union mm events, PM at a) = m) + P(B) - pm and B). ”h! EEBSE 11: g ﬂung menP(A and B) = 0. so P(A or H) = P(A) + P(B) PM or B) ., A :an Ii UlM‘lHCLI Imu- ELIE; Out 0' a class of 50, the number of students who passed exam 1 is 30, number of sludenu wha pamd exam 2 ls 204 and the number 0' students who wassed both exams are 15. Find the pmbabiliﬁv of a randomly selected sllmem cm 0' the class ta have passed mess: one exam. A - selected student has passed exam 1 B — selected sludenl has passed exam 2 The event a! selected student having passed at least one exam = A nr D From the abave formula, P(A 079) = PM) + P(B) — P(A mm a) 20_ 50' A and H v is the event 0' the selected swden! having passed both exams. There'ore, NA) = 3 Pa?) = 15 P(AnndB) = E?» Therefore, 30 20 15 35 Pm or a) = —+ ————— =o.7 II'i ‘ t For the Intersection of (Wu lndeplndelll events, A and B. P(A and a) : 9(a) x ma) P( :\ and B,- ...
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