Sloution to problem set 3

Probability and Statistics for Engineering and the Sciences (with CD-ROM and InfoTrac )

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Unformatted text preview: 42. 44. l Seats: EIEIE 2x1x4x3x2x1_1 P{J&P in 1&2) = —— =.0667 6x5r4»3x2x1 15 PUe‘IzP next to each other) = P(J'&P in 1&2] + ... + P{J&P in 5&6} 1 1 = 5 X— = — = .3 33 15 3 Pfiat least one H next to his W) = 1 — P( no H next to his- EV} We count the of ways of no H next to his W as follows: I iforderings 1without a H—‘W pair in seats #1 and 3 and no H next to his W = 6* >< 4 >: 1* :4 2" >< l K 1 _= 48 3: pair, ‘5 =c an‘t put the mate of seat #2 here or else a H-W pair wo uld be in #5 and s. of orderings without a H-W pair in seats #1 and 3, and no H next to his W = 6 >< 4 >: 2': X 2 >: :2 :4 1 = 192 #= can’t be mate ofperson in seat #1 ONE. 30: # of seating arrangements 1with no H next to W = 48 + 192 = 240 _ 24D 1 And P(noHnextto h1s W = = —= —__ so 6X5X4X3/2X1 3 1 Pfiat least one H next to his W)=1- — = — 3 3 \ ,» _ 2| =L="_“=‘ ” I ,, kl(n—k]l (H—k)lk! arr—Io. a The number ofsubsets of size 1-: = the number of subsets of size n-k, because to each subset of size k there corresponds exactly one subset of size n—k (the n—k objects not in the subset of size k). 4S. P(.4 .«5 242} _ .06 PTA: I Al} = PfA] PHI-1127'" Ag A1}: = 1'} . a We want P[(exactl}' one) {at least one)]. P(at least one) = H.511 c.- A3 '4.) A3) = .12+.07" + .05 — .06 — .03 — .02 — .01 = .14 Also notice that the intersection ofthe two events is just the l:t event, since “exactly one“ is totallyr contained in “at least one." 1144.01 _ .14 So P[(exact1y one) (at least one}]= .3 57 1 The pieces of this equation can be found in your answers to exercise 26 (section 2.2): . P .4 r~..4. m4“ .05 PM} .4] «12192—1: 1 5 3) = P(A] (5:12) .06 = .833 .4><.3 = .12 = 54.41 .fi.B)=P(.41)-P(B| .4) .35><.6=.21=P{A2 me) a. 1414; m B} = .21 b. PfB}=P(A1-rw B.)+P{4.2 r5 B}—P{A3 ('5 B}: .455 121.41 r~.3)_ .12 _ P(B) _ .455— .21 .455 e. P{A1|B}= .264 H.423): = 452 31:14.43): 1 — 2154—452: .224 69‘. PEaat15)=.51 '3‘ Human saris}: — = .3922 .51 PEmedian sans} = 4941 PEmotle sans) = .313? So Mean {and not Mode!) is the most likely author, while Median i5 least. a. Sinee the events are independent, then A“ and Er are independent. too. (see paragraph below equatton 2.7". P(B'|.'—1.'] = . P[B“] = 1- .7 = .3 13. PG; .J B]=P(A) + P03) — MAJ-MB} = .4 — .F — {.4){.?) = .3: RAF (.4 e133.) = P(AB') = .12 —= . 146 Pita-.13) E. P(AB' A'—...—'B]= P(A B} .32 Pfino error on any particular question) = .9, so P{no error on any ofthe 10 questions} =I:.9jl'fl = .3481 Then Flint least one error} = 1 — [.51]1D = .6513. For preplacing .1. the two probabrlities are (1—5))” and 1 — (1—11)“. ...
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