This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 witheutaHWpairinseats #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 nk, 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. HA2A1)=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 + (AK7] = 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. ...
View
Full Document
 Spring '04
 Eisenstein

Click to edit the document details