engr361hw3

# engr361hw3 - 42 Seats “IEIEEI 2143211 x x x x x...

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Unformatted text preview: 42.. Seats: “IEIEEI 2143211 x x x x x =———=.0667 &Pinl&2 =— Pﬂ ) 6x5><4><3><2><1 15 P(J&Pnexttoeachother) =P{J&Pinl&2)+ ___+P(J&Pin5&6) 1 1 2 SX— 2— 2.333 15 3 P(at least one H next to his W) = l — P1: no H next to his 1W) We enunt the # ufways (1an H next to his W as follows: # ifurderings without aH—W pairin seats #1 and 3 and no H next to his W: 5* x 4 x 1‘I x 2"! x l x 1 = 48 *= pair, #=can’t put the mate of seat #2 here or else a H—W pairweuld be in #5 and IS. #uforderings witheutaH-Wpairinseats #1 and 3,311an Hnexttn his W=6 x 43-: ng 2x 2x 1:192 ‘5: can’t be mate ofperson in seat #1 er#2_ Sn, #efseating arrangements with no H next to W = 48 + 192 = 240 _ 240 1 AndP(noHne)ttehisW]= =—= —,sn 6x5x4x3><2><1 3 _ 1 2 P(at leasteneHnexttohisW)= l - — = — 3 3 J=L=i=[ " ] kIm—k}! (n—k)!k! arr—k The number ofsubsets of size 1: = the number of subsets of size n-k, because to each subset of size I: there corresponds exactly nne subset of size n—k [the n—k ubjeets not in the subset of size 1:}. 48. 59. HA2|A1)=M=E_-50 P(A]) .12 _ .01 b. HA1 n42 n A3 I41): — = .0833 L1. d. .12 We want P[{exactljr one] I (at least one]]. P[at least one) =P(A1 1...! A; U A3] =.l2+ .ﬂ?+.DS — .05 — .03 —.U2 + .01 =.l4 Also notice that the intersection ofthe two events is just the lit event, since “exactly one” is totallyI contained in "at least one." .04 .01 So P[(exactly one] I {at least one)]= ; = .3 571 .14 The pieces of this equation can be found in your answers to exercise 26 (section 2.2): pg; lAlﬂAE)ZWZEZ'833 P{A1 MA!) .05 .4 X .3 = .12 = HA1 as) = £1.41) . P(B| A) .35x.ﬁ=.21=P(A2ﬂB} .25 x .5 :15 = 551.4. ms) in. P{A2 n E): .21 b. 11(3):?{41 nB)+P(A2 n mum3 nB)=.455 1.31.41 r53) .12 — = — = .2154 54(3) .455 HA2I33= % = .462 .P(A3[B)= 1 - 2154—4152:.254 c- HAIIE) = 55. P(sat:is] =51 P(n]ean| sans) = — = .3922 .5] P(rnedian | satis] = .2941 P(rnnde | satin] = .3131Ir Sn Mean (and not Merle!) is the must likely,r author, while Median is least 69. 3. Since the events are independent, thenA' and B' are independent, too. [see paragraph below equation 2.1 P[B'|.A')= . P[B')=1- .T = .3 11- P111 U BFPtAJ + FEB] - P(A}P{B) = -4 + 3 + (AK-7] = 32 P AB ' A 3 AB ' .12 P(AUB) P-(A US) .32 74. P(no errnr on any particular question} = .9, so P(nn error on any of the 10 questinns) ={.9}m = .348? Then Hat least one error) = 1 — £9)“: = .6513. For 1) replacing .1, the two prnhahiljties are {II—p]:11 and 1 —(1—p)n. ...
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