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Unformatted text preview: @1993 Springer—Verlag New York. Inc. + Section 2.1 1. a) (I): bv)(1)(1/6)‘(5/613 2. P(2 boys and 2 girls) = (;)(1/2)‘ = (3/2‘I = 0.375 < 0.5. So families with different numbers of
boys and girls are more likely than those having an equal number of boys and girls, and the relative
frequencies are (respectively): 0.625, 0.375. 3. a) P(2 sixes in 5 rolls) = (§)(1/6)’(5/6)“ = 0.160751 b) P(at least 2 sixes in 5 rolls) = 1 — P(a.t most 1 six)
=1—[P(0 ska) + PU. six”
= 1 — (5/6)5 — (f)(1/5)(5/6)‘ = 0.196245
(This is shorter than adding the chances of 2. 3, 4, 5 sins.) c) P(a.t most 2 sixes) = P(0 sixes + PU. six).P(2 sixes)
= (5,/6)5 + (§)(1/6)(5/6)‘ + (3 (1/6)’(5/6)3 = 0.964506 d) The probability that a. single die shows 4 or greater is 3/6 = 1/2.
P(exactly 3 Show 4 or greater) = (30(2)s =: 0.3125 c) P(at least 3 Show 4 or greater) : [(3) + (i) + 6)] (1/?)5 z 0.5
(The binomial (5, 1/2) distribution is symmetric about 2.5) 4.
P(2 sixes in ﬁrst ﬁve rollsi3 sixes in all eight rolls)
“ P(2 sixes in ﬁrst 5, and 3 sixes in all eight)
" P(3 Sim in all eight)
__ P(2 sixes in ﬁrst ﬁve, and 1 six in next three)
 P(3 sixs in all eight)
W2. (:10 (gm/613151615 “ "7:7" = '57 = “'5357“ (1? i3 i3 i3)
5.a)——ﬁ— b)(5a C)1“ +5K—
12 12 33) (f2) ﬁx 6. a) P(exactly 4 hits) = (:)(0.7)‘(0.3)‘ = 0.1361367 b} P(exactly 4 hitslat least 2 hits): Wm“ “41 511:3“? 3‘ ‘3? 2 hi“) ' W ” i __ P exact! 4 hitsi __
_ 1—P(exacty 0 t5)~P(cxac y l t) —0.l363126 c) P(exactly 4 hitsiﬁrst 2 shots hit)
. Hexacti 4 hits 8; first 2 shots hit
“ HHrst 2 shots Ext)
 Ptﬁrst 2 shots hit 8:: exactlz 2 hits in last 6 shots)
” n rst s on t) = P(exactly 2 hits in last 6 shots) (since shots are independent)
= (i)(0~7)2(0.3)‘ == 0.059535 7. The chance that you win in this game is 15/36 =2 5/12 (list all 36 outcomesl), so the chance you win
at least four times in ﬁve plays is s 5 ‘
(4)(5;12)‘(7/12)+ (5)5/12)’ : 0.100459 8. Let n be a positive integer. Using the formula. [or the mode of the binomial (mp) distribution: If up + p < I then the most likely number of successa is in: (up + p) = 0 . Section 2.1 If rip+p = 1 then the most likely number of successes is O or 1 (both equally likely).
o If up + p > 1 then the most. likely number of successes is at least 1. Hence the most likely number of successes is zero if and only if np+p S 1, and the largest p for which zero is the most likely number of successes is p = 31—1 9. a) Most. likely number = int (326 x 1/38) 2 8. P(8)=P(6) 3133‘ all} 325—7+I l 325~8+1 1 :.IO4840———7———§~';’v~——8— a: =.138724 9 10
b) P(10) = P(8) . g3}. m
= .138724 x £591. 31.; . W J— = .1127847 c) P(10 wins in 326 bets) :P(9winsin325 bets)x Elg—fPUO wins in 325 bets)x g;
I 37
=. 3 0 —‘ . ~—
1 2 58x 38+ 1127847x 38 2: .1132919 10. a) k/(n+ 1) b) (n Ic+ l)/(n +1). 11. a.) The tallest bar in} the binomial (15, 0.7) histogram is at int ((n + Up) = int(16 x0.7) = int(11.2) =
11 b) P(11 adults in sample of 15) = if)(o.7)“(o.3)* 5 x 14 13 2
= (145) (o.7)“(o.3)‘ = LZ'QTETEI— = (o.7)“(o.3)‘ = 0.218623 12.. a.) P(makec exactly 8 bets before stopping)
= P(wins at 8th game, and has won 4 out of the previous seven) = (l)(1H3/38)‘(20/38)a 3(18/38) = 0.1216891 b) P(plays at least 9 times)
= P(Wins at most {our bets out of the ﬁrst eight)
= (20/3s)°+(:)(18/33)(2o/3s)7+(g)(18/38)’(2o/3s)°+(3)(18/33)“(20/38)5+(2)(1s/3s)‘(20/3s)‘ H z; 0&92616? »»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»» _ »»»»»»»»»»»»»»»»»»»» r, ,,,,,,,,,,,,,,,,,,,,,,,,,,, , ............................. V eeeeeeeeeeeeeeee 13. a) No! Since the child must get one allele ‘from his/her mother, which will be the dominant. B,
he/she must have brown eyoe. b) Taking one allele from each parent, we see that there are {our (equally likely) allele pairs: Eb,
Bb, bb, bb. Thus the probability of the child having brown eyec is 50% z 0.5. c) Once again there are {our possibilities: Bb. Eb, Bb, bb. Thus there in. a. 75% = 0.75 chance that
the child will have brown eyes. d) Given that the mother is brown eyed and her parents were both Eb, the probability she is Bb
is 2/3 and BB 1/3. Thus P(child browneyed) = P(child brown—eyedlmother Bb) ~ P(mother 8b) +P(child brown~eyedlmother BB)  P(mother BB) 1 2””1 2
"‘ IX=—
3+ 3 3 {D
G) P(child Bblwoman Bb)P(woman Bb) __ (1/2)(2/3) __ . l
P(woman Bblcluld Bb) ._ P(child Bb) 2/3 5 20 . Section 2.1
W 14. a) (Ts, Pw); tall and purple.
Genetic Combinations Probability (TT, PP) tall and purple
(TT, Pw) tall and purple
(TT, ww) tall and white b) (Ts, PP) tall and purple
(Ts, Pw) tall and purple ‘
(Ts, ww) tall and white
(53, PP) short and purple
(as, Pw) short and purple
(55, w) short and white tall and purple tall and white short and purple short and white
Probability 9/16 3/16 3/16 1/16 c) 1— (7/16)10 ~10(9/16)(7/16)’. 15. a.) If 0 < p < 1, then int (np+p) = up, since up is an integer. b) Note that up = in! (up) + [up — int (np)].
If [up  int (1131)] + p 2 1, then in! (np+p) = int (up) + 1, which is the integer above up.
If 0 < [up —— int (np)] +p < I, then int (up +p) = int (up), which is the integer below up. c) Consider the case n = 2,1: = 1/3I where 1 is the closest integer to up and the integer above up,
but 0 is also a. mode. Also 0 is the integer below up, but. 1 is also a. mode. 21 Section 2.2 .
WWW
Section 2.2 1. The number of heads in 400 tosses has binomial (400, 0.5) distribution. Use the normal approximation
with ,1 = 400 x (1/2) = 200 and 0’ = «400 x g x 5‘ = 10. a) P090 3 H g 210) a. <I>(1.05) ~¢I>(—1.05) = 0.7062
b) P(210 5 H g 220) a: 1112.05) — o(0.95) =: 0.1509
c) P(H = 200) z 0(005) w (DO—0.05) = 0.0398
d) P(H =: 210) x 601.05) — 00.95) = 0.0242 2. Now :1 = 204 and a' = 9.998.
a) P(190 5 H g 210) .~. 1:40.65) — <I)(—1.45) = 0.6686
b) P(210 g H g 220) z <I>(1.65) <I>(——0.55) = 0.2417
c) P(H = 200) 7.: o(—0.35)  o(—0.45) = 00368
d) P(H = 210) z <1)(0.55)— 0(055) = 0.0333 3. a) Law of large numbers: the first one.
b) Binomial (100, .5) mean 50, SD 5: 1 — o (5455 50) =1~ @(3) = 1 — .8159 = .1841 Binomial (400, .5) mean 200, SD 10: 219.5 —— 208
1  <1» (
10 4. Let X be the number of patients helped by the treatment. Then E(X) = 100, SD(X) = 8.16 and
P(X > 120) = P(X 2 120.5) s; 1 — 0(251) .. .006. ) =1— 0(1.95)=1—.9744 = .0256 5. Want the chance that you win at least 13 times. The number of times that you win has binomial (25.
18/38) distribution. Use the normal approximation with ,u = 11.84,a = 2.50:
P(13 or more wins) z: 1 —— P(12 or fewer wins) z l—~ (I) (W = 1 —— @(026) x 0.3974 G. The number of opposing voters in the sample has the binomial (200, .45) distribution. This giva
p = 90 and a 2 1/200 x .45 x .5 = 7.035. Use the normal approximation: ................................................................. a)TherCQUYredChanceisapproximately . . WWW. . 90.5 —— 90 89.5 — 150
q) ( 7.035 ) _ Q( 7.035 ) z .5283 ~ .4717 = .0567 (about 6%) b) Now the required chance is approximately 100.5  90 I  <1)
7.035 ) z 1 .. (00.49) = 1 ~ .9319 = 0.0681 (about 7%) 7. a) The city A sample has 400 people, the city B sample has 600 people, so city B accuracy is «6/4 = 1.22 times greater.
b) Both have the same accuracy, since the absolute sizes of the two samples are equal. c) The city A sample has 4000 people, the city B sample has 4500 people. so the city B sample is ”4500/4000 = 1.06 times more accurate. even though the percent of population sampled in
City is is smaller than the percent sampled in city A. 8. Use the normal approximation: or = 4an z: 9.1287. Want relative area between 99.5 and 100.5 under the normal curve with mean 600 x (1/6) = 100, and a = 9.1287. That is. want area between
—.055 and .055 under the standard normal curve. Required area : <H.055} — [l »— $00.55)) = .0438. 22 Section 2.2 9. 3.) Think of 324 independent trials, each a “success” (the person shows up) with probability 0_9_ So
the number of arrivals has binomial (324., 0.9) distribution. We want P(rnore than 300 successes in 324 trial;
Use the normal approximation with I‘ = 324 x 0.9 = 291.5 and a = #324 x 0.9 x 0.1 :: 5,4 to
compute the required probability:
o (ES—553%.! = 1  «30.65) 2 0.0495.
b) Increase: in the long run, each group must show up with probability .9. So effectively, traveling
in groups reduces the number of trials, keeping p the same. So the histogram for the proportion
of successes has more mass in the tails, since u is smaller. c) Repeat 3.), with the 300 seats replaced by 150 pairs, and the 324 people replaced by 152
pairs. So the number of pairs that arrive has binomial (162, 0.9) distribution. Use the nor
mal approximation with n = 162 x 0.9 = 145.8 and a' = V162 x 0.9 x 0.1 =: 3.82. We want 4,150.33... 5 = 1 —— 00.23) = 0.1093. 10. We have 30 independent repetitions of a binomial (200, 3' For each of these repetitions, the proba—
bility of getting exactly 100 heads can be gotten by normal approximation where p = (200 x i) = 100 andor: 200x§x§=7.07.
The chance of exactly 100 heads can be approximated by _ (100 + .5) —— 100 __ (100 — .5) — 100 ~ _ _
P000 heads ) _ o ( 7‘07 0 M707 ~ .5282 .4718 _ .0564 Since the students are independent, the probability that all 30 students do not get exactly 100 heads
is (1  P(100 heads ”30 and the normal approximation gives us (1  P000 heads n” z (1 — .0564)” = 0.175 11. The number of hits in the next 100 times at bat has binomial (100, 0.3) distribution. Use the normal
approximation with u = 30 and a = V100 x 0.3 x 0. = 4.58. a) P(> 31 hits) 1— o (”f—z—m = 1  0(011) = 0.4.562
b) H? 33 hits) ~ 1 — o (32 54°) = 1 — 0(0. 545) = 0.2929
c) P(_<_ 27 hits)~ o (’———7f'3°)= (40. 545): 0.2929 d) No, independence would be lost, because if the player has been doing well the last few times at
bat, he’s more likely to do well the next time. Similarly, if he’s been doing badly, then he’s more
likely to continue doing badly. This will increase all the chances above. c) From part b), we see that the player has about a 29% chance of such a performance, if his form
stayed the some (long run average .300) and if the hits were independent. So it is reasonable to
conclude that this performance is “due to chance,” since 29% is quite a large probability. ................................ 12.. Fornmltiﬁeﬁindcpendenttrialswithsuccess prbablllbypgl/g, .. ,, ., u = up = $000 and 0’ = anq = 50 Hence, by the normal approximation, the probability of between 5000 — m and 5000 +17: successes is approximately
m+1l2 _ «m—1/2 _ m+1/2 _~
“’( so ) @l .. )“’( .. )1 This equals 2/3 if and only if (5((m + l/2)/50) = 5/6 which implies that (m + l/2)/50 = 0.97 and m
= 48. In other words, there is about a 2/3 chance that the number of heads in 10,000 tosses of a {air
coin is within about one standard deviation of the mean. 13. First ﬁnd 2 such that <I>(—z,z) = 95%. This means (Hz) :— 0.9750 and by the normal table. .2 = 1.96. .5
959’=P(" irssﬂ<1> “ mom
0 pmp n .. ‘plnpd: ﬂ We want l.96(.5/m S 1%, so '2
l.96(.5)
> =
n__( ‘01 ) 9604 23 Section 2.2 14. 15. 16. 17. a.) # working devices in box has binomial (400, .95) distribution. This is approximately normal with
p = 380, a z 4.36. Required percent 7:: 1 — MW) = 1 — <I>(2.18) = 0.0146 (This normal
approximation is pretty rough due to the skewness of the distribution. The exact probability
is 0.0094 correct to 4 decimal places The skewnormal approao'mation is (3.0099 which is much
better.) —————*"i‘:;3‘°) 2 0.95 so 5 :3: = > b) Using the normal approximation, want largest 1: so that 1  (I)
«1.65, so k = 373. a) ¢’(z) = 7;:(z)e"’" = ~z¢<z) 5) ¢"(Z) = ¢5(Z)  z(Z¢'(Z)) = (12 ~ 1)¢(2) c) Sketch: Outside (—4, 4), they are close to zero.
d) Lef f(I) = iﬂii‘il Want f"(£) mm) = $45”? “‘) = if”)? —1)¢(: ‘ " 1 a o' :7 «0'2 e) For n — a' < 1: < :1 + a', the second derivative in d) is negative. So the curve is concave in this
region. For 1: > u + a or 2: < u  a. The second derivative is positive, so curve is convex. a) Using the results of Exercise 15,
¢m(z) = 2z¢(z) — (z2 — 1)z¢(z) = («z2 + 3z)¢(z) b) Fundamental theorem of calculus; ¢” vanishes at ~00. The ﬁrst equality of integrals is because
W" is an odd function. 2¢(\/§) = ((s/i)2 — 1)¢(~\/§) = ¢”(J5) ‘5 V3
A ¢m(z)dz) 3/ ¢'"(z)dz __ [0 ¢"’(z)dz) 3 2¢(\/§) "‘ ("45(0» c) ¢'" switches signs at «J5 and v’i. {5(0) + Mix/5 is the biggest area possible over an interval
because areas are negative when the curve is negative. a) 1—o(z) = 1]; ¢(::)dz =1: amp]; ¢(z)dz=£°° $(z)dx. b) Over the range of the integral 2 < z, so 1:]: is greater than I. Multiplying a positive integrand
by a number greater than 1 gives an integral with a larger value. 63. a , o . 3
1 C"? :cdz;
tz 2 1—<I>(z)</;  [0° 1 e‘“du [u ~ it: (in  rdz]
ég V21rz 3 ' .. I.. C 2:: w
1 gn “2) e \IZ‘xz Z 24 Section 2.3
WM
Section 2.3 1. Let p =‘P(0), then Pm = 120;) .120:  1)  . 12(1),).
Use :21 P(k) = l —— p to get the value of p. 2. Put 1: = 10,000, p = 0.5, I: = 5000 in Formula (3). Then p = 5000, a = 50, and the deS'ued probability is approximately ‘ ——————— = 0.008 z 0.01
21 x 50 ' 3. a) Use P(m+1in 2m) = P(m in 2111)  R(m + 1) where R(m + 1): W = (1.. 3&3.) m+1
Similarly, P(m — 1 in 2m) 2 P(m in 2m)/R(m) where R(m) == 23,331 (This could also have
been deduced by symmetry.) 5)
P(m+l in 2m+2)
= P(m — 1 in 2m, then HH) + P(m in 2m, then HT or TH) + P(m +1 in 2m, then T’I‘) = 41mm —1 in 2m)+ —:P(m in 2m) + %P(m +1 in 2m) .J... n+1 c) Substitute P(m — l in 2172) = P(m in 2m) (1 — ﬁf) and P(m +1 in 2m) 2 P(m in 2m) (1 
into (b), and simplify. This can also be checked by cancelling factorials. d) Write ’
. _ P(m in 2m) P(m — 1 in 2m — 2) P(2 in 4) .
mm” 2m)“ P(m~ 1 in 2m—2) P(m2 in 2m~4) P(1 in 2) P“ m 2) 15%).(cm).....(12;2).(1_2;1). e) Use the fact that 1 —— I S c“ with equality iff r = 0 (see diagram). Then the inequalities follow easily: 0 < P(m in 2m) < cé(f+%+§+~+ 9:} < c—éiogm) z 1
Wm (For the last inequality, draw a. graph of 1/1: and remember that the area under this graph from
1 to m is log(m)). I) By part, (C), am ._ l I
am”; .. 2m.
Square this, substitute, and simplify:
m+;— a", 2_2m+l(2m~1)2_(2m1)(2m+1)_l 1
m ~1+§5 any”; _ 2m ~l 2m — (2m)? — 4m“ 25 Section 2.3 (since (0 + '15)O'§ = 1/2) The result. follows from factoring (1’41; =(1‘_) (1+2k) 1
21m 11) am = P(m in 2m) :: P(mode) ~ with a' = §V2m. So am ~ 5% where K = 71;. But this means 2 . 2m+12m—12m1 2m—3 31
~=2K =2 hm “=2 m: . .—._._.._.._.__......
1r muoo(m)a mh&(m+% )" mhfm 2m 2m. 2(m ~1) 2(m 1) 2 2° 26 Section 2.4 MW Section 2.4 1. 3.) Approximately Poisson(1).
b) Approximately Poisson(2).
c) Approximately Poisson(0.3284). d) This is the distribution of the number of successes in 1000 independent trials, Where the success
probability is p = .998. The distribution of the number of failures has binomial (1000, .002)
distribution, which is approximately Poissou(2). Since #successes + #iailura = 1000, it follows
that the histogram for the number of successa (i.e., the desired histogram) looks like the left~to~ right mirror image of a Poissona) histogram, with the block at 0 being sent to 1000, the block
at 1 being sent to 999, etc. 2. a) The number of successes in 500 independent trials with success probability .02 has binomial
(500, .02) distribution with mean u z 10. By the Poisson approximation,
[‘1
PO. success) a: 6'“? = pe'“ = .000454. b) P(2 or fewer successes) == PU) or 1 or 2 successes) 0 1 2
~ ufi_ WE... 'HL
”5 01+“ 1: +3 2: 2 = e‘“(1 + u + ’12—) = .002769
c) P(4 or more successes) = 1 , P(3 or fewer successes) = ,989664.
8. The number of times you see 25 or more sixes has binomial distribution with p = 365 x .022 = 8.03. a) P(at least once) = 1  P(exactly 0 times) 2: 1  c"“ 2: .999674.
b) P(a.t least twice) 2 l — P(exactly 0 times)  P(exactiy 1 time) 2: 1  e‘“  ”6'" = .997060. 4. Here 11 = 365 x 0.00068 = 0.2482, and a) 1  6'“ = .219796;
b) 1 —— e'” — ne‘“ = 0.026150. 5. The number of wins has binomial (52, 1/100) distribution With mean p. = 52/100, and by the i’oisso‘ﬁ ““““““ '
approximation PU: wins) 2 c‘”u"/H.
I: = 0 : P(0 wins) 2: e‘“ 2: .594521;
I: = 1:?(1 win) z pe‘“ = .309151;
I: = 2 ; P(2 wins) z L336" = .080379. 6. a.) The number of black balls seen in a series of 100 draws with replacement has binomial (1000.
2/1000) distribution with mean u = 2. By the Poisson approximation, 1 0
P(fewer than 2 black balls) 2 (“z—T + {'2’}; : rm + u) = .405005, 2
P(exactly 2 black balls) = e_“—%~ = .270671. Calculate th probability of more than '3. black balls by subtractionrconciude that getting fewer
than 2 black balls is most likely. 27 Section 2.4
b) P(both series see same number of black balls) = 2:0 P(both series see it black Hang) °° k ist:
z E: e”"E—c“" —— by independence 1:! kl
— k=0
. °° 4"
= e Z W = 0.207002. 53:0 7. a.) 2 b) .2659 c) .2475 d) .2565 e) m = 250. Normal approximation J.0266 f) m = 2.
Poisson approximation: .2565 8. k
_ 1t
P(k)=e “7;? k4
.. z ”49......
PO: 1) e (ls—l)! so 11(17):: P(k)/P(k— 1) is given by
120:) = 1;. which is decreasing as 1: increases. The maximimum probability will occur at the largest value of I:
for which 111(k) Z 1; after this PUc) will decrease. Thus the maximum occurs at 1.11101). There is a
double maximum if and only if R(Ic) = 1 {or some k. This can only occur if p is an integer. The two
values of k that maximize are then p. and p — 1. There can never be a triple maximum since this would imply that R(}:) = 1 and RU:  1) z 1 for some is. 9. Assume that each box of cereal has a prize with chance .95, independently of all others. The number
of prizes collected by the family in 52 weeks has the binomial distribution with parameters 52 and .95. This gives .u = 52 x .95 = 49.4, and a = «52 x .95 x .05 = 1.57. Since 0’ is very small (less than
3), the normal approximation is not good. Use the Poisson instead. The number of ”dud” boxes has
the Poisson distribution with paramter 52 x .05 = 2.6. We want the chance of 46 or more prizes, that is, 6 or lac duds. This is
5'” {1 + 2.6 + 2.s’/2 + 2.63/3: + 2.6‘/4! + 2.65/5! + 2.65/62} = .982830. [For comparison, the exact binomial probability is .985515. The normal approximation gives 393459.] 10. Distribution of the number of successes is ............ , _ ’ ‘ ,, V binomial (n,1/N)z Poisson (n/N) z Poisson (5/3). . Pm least two) =1  P(o)  P(I) z1 (5’30 + 5/3) = 1 . {5’3  g ~ 0.49633 z 0.5 28 Section 25 www
Section 2.5 30 30
1 a) b) (‘2)(2/5)‘(3/5)‘
1°
2. a.) §.§%,§% b)3x a) c)1——~§%a§%c§3—
4 GI (3 l. ( 4‘
3. a) b) 3 c) ‘ d) 8
u 13 u ’ x:
22m (”'°°°)(‘°'°°°
4. The exact chance is k‘“ (“0:300 100—1:
100 For the approximation, use the normal curve with u = 40. e = 4.9. Chance is approximstdy 1  ov—Lﬁﬂ) = 0.1788
"'°5 < —2.326, then u _>_ 537. Se SOlVC m __ u (0 M
e. a) u b) ii’l — gal c) (a) + 4(b).
13 I3 I;
7. Denote by B; the event (ith ball is black), similarly for Re. a.) P(81323334)= P(B:)P(Ba181)P(B:!BzBa)P(B‘$3232le = 5%71“? =~
b) This is {our time; P(B1BzBa&), so by (1) equals .3716.
c) 13(313253R4) 3 P(Bl)P(BﬁlBI)P(BSlBIBi)P(R‘lBIB333) = no 79 a 17 Mement subsets from the set {1, ...,100}, so that, e.g., the 8. Let the outcome space be the set of all
3set {23, 21, 1} means that the winning tickets were tickets #1, #21, and #23. Each of the (120) such Saets is equally likely. a.
to
Q
o
It
4
E
03 e) The event (one person gets all three winning tickets) is the disjoint union of the 10 equally likely
events (person i gets all three tickets). The event (person 3‘ gets all three tickets) corresponds
to all 3sets consisting entirely of tickets bought by person i. There are (1:) such 3‘sets, so the desired probability is
10
10 x (3) = .007421. (“3") """"" ” "‘ ”‘syrueeaear(mergers>mree~aiﬁereue~wimmyeum
(the winning tickets were bought by persons ...
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