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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 HW pair wo uld be in #5 and s. of orderings without a HW 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 nk, 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] PHI1127'" 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(A1rw B.)+P{4.2 r5 B}—P{A3 ('5 B}: .455 121.41 r~.3)_ .12 _ P(B) _ .455—
.21
.455 e. P{A1B}= .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) — MAJMB} = .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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This homework help was uploaded on 01/31/2008 for the course ENGR 361 taught by Professor Eisenstein during the Fall '04 term at Drexel.
 Fall '04
 Eisenstein

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