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[SPAM] Res: [obm-l] Probabilidades e o Primeiro Lema de Kaplansky

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--0-18767870-1206304893=:40862
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Fa=C3=A7amos o seguinte, Ulysses:=0A=0AQueremos que haja pelo menos um par =
de n=C3=BAmeros consecutivos sorteados. Ent=C3=A3o vamos contar os sorteios=
 que n=C3=A3o cont=C3=AAm n=C3=BAmeros consecutivos. Para tanto, considerem=
os seq=C3=BC=C3=AAncias de 60 d=C3=ADgitos formadas por 54 0's e 6 1's, de =
tal maneira que, se o i-=C3=A9simo d=C3=ADgito for 0, ent=C3=A3o o n=C3=BAm=
ero i n=C3=A3o foi sorteado e, caso cotr=C3=A1rio, foi sorteado. Por exempl=
o:=0A=0A000010001000.....001=0A=0ANa seq=C3=BC=C3=AAncia acima, foram sorte=
ados os n=C3=BAmeros 5, 9, 60 etc., pois essas posi=C3=A7=C3=B5es s=C3=A3o =
ocupadas por 1's. Assim, se imaginarmos os 54 0's emparelhados, temos:=0A=
=0A_0_0_0_0_0_..._0_0_0_=0A=0AOnde os 55 tra=C3=A7os _ indicam posi=C3=A7=
=C3=B5es candidatas a serem ocupadas por 6 1's, ou seja, definem os n=C3=BA=
meros sorteados. Logo, podemos selecion=C3=A1-las de C(55,6) maneiras. Como=
 o total de sorteios =C3=A9 C(60,6), segue que a probabilidade de n=C3=A3o =
haver n=C3=BAmeros consecutivos =C3=A9 C(55,6)/C(60,6). Portanto, a probabi=
lidade de haver n=C3=BAmeros consecutivos =C3=A9:=0A=0A1-C(55,6)/C(60,6) qu=
e, ap=C3=B3s algumas manipula=C3=A7=C3=B5es, nos leva =C3=A0 alternativa E.=
=0A=0AUm abra=C3=A7o,=0AEduardo Estrada=0A=0A----- Mensagem original ----=
=0ADe: Ulysses Coelho de Souza Jr. <ucsjr@xxxxxxxxxx>=0APara: obm-l@xxxxxxx=
-rio.br=0AEnviadas: S=C3=A1bado, 22 de Mar=C3=A7o de 2008 20:58:56=0AAssunt=
o: [obm-l] Probabilidades e o Primeiro Lema de Kaplansky=0A=0A  =0AOl=C3=A1=
 a todos,=0A =0AA quest=C3=A3o abaixo =C3=A9 de um vestibular recente. Acre=
dito =0Aque o examinador quis dizer "pelo menos um par" ao inv=C3=A9s de "u=
m =0Apar".=0A =0AComent=C3=A1rios ser=C3=A3o bem-vindos. =0A =0ANo Concurso=
 da Mega-Sena s=C3=A3o sorteados=0A=0A=0A =0A6 n=C3=BAmeros de 01 a 60. Por=
 exemplo, o =0Aconcurso=0A =0A924 teve como n=C3=BAmeros =0Asorteados=0A =
=0A02,20,21,27,51 e 60, ou seja, houve um par =0Ade=0A =0An=C3=BAmeros cons=
ecutivos, 20 e 21. A =0Aprobabilidade=0A =0Ade que no jogo da Mega-Sena haj=
a um par =0Ade=0A =0An=C3=BAmeros consecutivos sorteados =0A=C3=A9:=0A =0A =
=0A =0A(A) =0A54!/60!=0A =0A(B) =0A53!/59!=0A =0A(C) =0A1-(56!55!)/(49!60!)=
=0A =0A(D) =0A1-(54!53!)/(48!60!)=0A =0A(E) =0A1-(55!54!)/(49!60!)=0A =0AUm=
 abra=C3=A7o,=0A =0AUlysses C. de =0ASouza.=0A=0A=0A=0A=0A=0A=0A      Abra =
sua conta no Yahoo! Mail, o =C3=BAnico sem limite de espa=C3=A7o para armaz=
enamento!=0Ahttp://br.mail.yahoo.com/
--0-18767870-1206304893=:40862
Content-Type: text/html; charset=utf-8
Content-Transfer-Encoding: quoted-printable

<html><head><style type=3D"text/css"><!-- DIV {margin:0px;} --></style></he=
ad><body><div style=3D"font-family:times new roman,new york,times,serif;fon=
t-size:12pt"><div style=3D"font-family: times new roman,new york,times,seri=
f; font-size: 12pt;">Fa=C3=A7amos o seguinte, Ulysses:<br><br>Queremos que =
haja pelo menos um par de n=C3=BAmeros consecutivos sorteados. Ent=C3=A3o v=
amos contar os sorteios que n=C3=A3o cont=C3=AAm n=C3=BAmeros consecutivos.=
 Para tanto, consideremos seq=C3=BC=C3=AAncias de 60 d=C3=ADgitos formadas =
por 54 0's e 6 1's, de tal maneira que, se o i-=C3=A9simo d=C3=ADgito for 0=
, ent=C3=A3o o n=C3=BAmero i n=C3=A3o foi sorteado e, caso cotr=C3=A1rio, f=
oi sorteado. Por exemplo:<br><br>000010001000.....001<br><br>Na seq=C3=BC=
=C3=AAncia acima, foram sorteados os n=C3=BAmeros 5, 9, 60 etc., pois essas=
 posi=C3=A7=C3=B5es s=C3=A3o ocupadas por 1's. Assim, se imaginarmos os 54 =
0's emparelhados, temos:<br><br>_0_0_0_0_0_..._0_0_0_<br><br>Onde os 55 tra=
=C3=A7os _ indicam posi=C3=A7=C3=B5es candidatas a serem ocupadas por 6 1's=
, ou seja, definem os
 n=C3=BAmeros sorteados. Logo, podemos selecion=C3=A1-las de C(55,6) maneir=
as. Como o total de sorteios =C3=A9 C(60,6), segue que a probabilidade de <=
span style=3D"font-weight: bold;">n=C3=A3o</span> haver n=C3=BAmeros consec=
utivos =C3=A9 C(55,6)/C(60,6). Portanto, a probabilidade de haver n=C3=BAme=
ros consecutivos =C3=A9:<br><br>1-C(55,6)/C(60,6) que, ap=C3=B3s algumas ma=
nipula=C3=A7=C3=B5es, nos leva =C3=A0 alternativa E.<br><br>Um abra=C3=A7o,=
<br>Eduardo Estrada<br><br><div style=3D"font-family: times new roman,new y=
ork,times,serif; font-size: 12pt;">----- Mensagem original ----<br>De: Ulys=
ses Coelho de Souza Jr. &lt;ucsjr@xxxxxxxxxx&gt;<br>Para: obm-l@xxxxxxxxxxx=
.br<br>Enviadas: S=C3=A1bado, 22 de Mar=C3=A7o de 2008 20:58:56<br>Assunto:=
 [obm-l] Probabilidades e o Primeiro Lema de Kaplansky<br><br>=0A=0A =0A =
=0A<style></style>=0A<font color=3D"#000080">=0A</font><p class=3D"MsoNorma=
l" style=3D"margin: 0cm 0cm 0pt; text-align: left;" align=3D"left"><span st=
yle=3D"font-size: 11pt; font-family: 'Arial','sans-serif';"><font color=3D"=
#000080"><font color=3D"#000000" size=3D"2">Ol=C3=A1 a todos,</font></font>=
</span></p>=0A<p class=3D"MsoNormal" style=3D"margin: 0cm 0cm 0pt; text-ali=
gn: left;" align=3D"left"><span style=3D"font-size: 11pt; font-family: 'Ari=
al','sans-serif';"><font color=3D"#000080"><font color=3D"#000000" size=3D"=
2"></font></font></span><font color=3D"#000080">&nbsp;</font></p>=0A<p clas=
s=3D"MsoNormal" style=3D"margin: 0cm 0cm 0pt; text-align: left;" align=3D"l=
eft"><span style=3D"font-size: 11pt; font-family: 'Arial','sans-serif';"><f=
ont color=3D"#000080"><font color=3D"#000000" size=3D"2">A quest=C3=A3o aba=
ixo =C3=A9 de um vestibular recente.&nbsp;Acredito =0Aque o examinador quis=
 dizer "pelo menos um par" ao inv=C3=A9s de "um =0Apar".</font></font></spa=
n></p>=0A<p class=3D"MsoNormal" style=3D"margin: 0cm 0cm 0pt; text-align: l=
eft;" align=3D"left"><span style=3D"font-size: 11pt; font-family: 'Arial','=
sans-serif';"></span><font color=3D"#000080">&nbsp;</font></p>=0A<p class=
=3D"MsoNormal" style=3D"margin: 0cm 0cm 0pt; text-align: left;" align=3D"le=
ft"><span style=3D"font-size: 11pt; font-family: 'Arial','sans-serif';"><fo=
nt color=3D"#000080"><font color=3D"#000000" size=3D"2">Coment=C3=A1rios se=
r=C3=A3o bem-vindos.</font>&nbsp;</font></span></p>=0A<p class=3D"MsoNormal=
" style=3D"margin: 0cm 0cm 0pt; text-align: left;" align=3D"left"><span sty=
le=3D"font-size: 11pt; font-family: 'Arial','sans-serif';"><font color=3D"#=
000080"><font color=3D"#000000" size=3D"2"></font></font></span><font color=
=3D"#000080">&nbsp;</font></p>=0A<p class=3D"MsoNormal" style=3D"margin: 0c=
m 0cm 0pt; text-align: left;" align=3D"left"><span style=3D"font-size: 11pt=
; font-family: 'Arial','sans-serif';"><font color=3D"#000080"><font color=
=3D"#000000"><font size=3D"2">No Concurso da Mega-Sena s=C3=A3o sorteados=
=0A=0A</font></font></font></span></p><font color=3D"#000080"> =0A</font><p=
 class=3D"MsoNormal" style=3D"margin: 0cm 0cm 0pt; text-align: left;" align=
=3D"left"><span style=3D"font-size: 11pt; font-family: 'Arial','sans-serif'=
;"><font color=3D"#000080"><font color=3D"#000000"><font size=3D"2">6 n=C3=
=BAmeros de 01 a 60. Por exemplo, o =0Aconcurso</font></font></font></span>=
</p><font color=3D"#000080"> =0A</font><p class=3D"MsoNormal" style=3D"marg=
in: 0cm 0cm 0pt; text-align: left;" align=3D"left"><span style=3D"font-size=
: 11pt; font-family: 'Arial','sans-serif';"><font color=3D"#000080"><font c=
olor=3D"#000000"><font size=3D"2">924 teve como n=C3=BAmeros =0Asorteados</=
font></font></font></span></p><font color=3D"#000080"> =0A</font><p class=
=3D"MsoNormal" style=3D"margin: 0cm 0cm 0pt; text-align: left;" align=3D"le=
ft"><span style=3D"font-size: 11pt; font-family: 'Arial','sans-serif';"><fo=
nt color=3D"#000080"><font color=3D"#000000"><font size=3D"2">02,20,21,27,5=
1 e 60, ou seja, houve um par =0Ade</font></font></font></span></p><font co=
lor=3D"#000080"> =0A</font><p class=3D"MsoNormal" style=3D"margin: 0cm 0cm =
0pt; text-align: left;" align=3D"left"><span style=3D"font-size: 11pt; font=
-family: 'Arial','sans-serif';"><font color=3D"#000080"><font color=3D"#000=
000"><font size=3D"2">n=C3=BAmeros consecutivos, 20 e 21. A =0Aprobabilidad=
e</font></font></font></span></p><font color=3D"#000080"> =0A</font><p clas=
s=3D"MsoNormal" style=3D"margin: 0cm 0cm 0pt; text-align: left;" align=3D"l=
eft"><span style=3D"font-size: 11pt; font-family: 'Arial','sans-serif';"><f=
ont color=3D"#000080"><font color=3D"#000000"><font size=3D"2">de que no jo=
go da Mega-Sena haja um par =0Ade</font></font></font></span></p><font colo=
r=3D"#000080"> =0A</font><p class=3D"MsoNormal" style=3D"margin: 0cm 0cm 0p=
t; text-align: left;" align=3D"left"><span style=3D"font-size: 11pt; font-f=
amily: 'Arial','sans-serif';"><font color=3D"#000080"><font color=3D"#00000=
0"><font size=3D"2">n=C3=BAmeros consecutivos sorteados =0A=C3=A9:</font></=
font></font></span></p><font color=3D"#000080"> =0A</font><p class=3D"MsoNo=
rmal" style=3D"margin: 0cm 0cm 0pt; text-align: left;" align=3D"left"><span=
 style=3D"font-size: 11pt; font-family: 'Arial','sans-serif';"><font color=
=3D"#000080"><font color=3D"#000000" size=3D"2">&nbsp;</font></font></span>=
</p><font color=3D"#000080"> =0A</font><p class=3D"MsoNormal" style=3D"marg=
in: 0cm 0cm 0pt; text-align: left;" align=3D"left"><font color=3D"#000080">=
<font color=3D"#000000"><span><font size=3D"2">(A) =0A54!/60!</font></span>=
</font></font></p><font color=3D"#000080"> =0A</font><p class=3D"MsoNormal"=
 style=3D"margin: 0cm 0cm 0pt; text-align: left;" align=3D"left"><font colo=
r=3D"#000080"><font color=3D"#000000"><span><font size=3D"2">(B) =0A53!/59!=
</font></span></font></font></p><font color=3D"#000080"> =0A</font><p class=
=3D"MsoNormal" style=3D"margin: 0cm 0cm 0pt; text-align: left;" align=3D"le=
ft"><font color=3D"#000080"><font color=3D"#000000"><span><font size=3D"2">=
(C) =0A1-(56!55!)/(49!60!)</font></span></font></font></p><font color=3D"#0=
00080"> =0A</font><p class=3D"MsoNormal" style=3D"margin: 0cm 0cm 0pt; text=
-align: left;" align=3D"left"><font color=3D"#000080"><font color=3D"#00000=
0"><span><font size=3D"2">(D) =0A1-(54!53!)/(48!60!)</font></span></font></=
font></p><font color=3D"#000080"> =0A</font><div><font color=3D"#000080"><f=
ont color=3D"#000000" size=3D"2"><span>(E)<strong> =0A</strong>1-(55!54!)/(=
49!60!)</span></font></font></div>=0A<div><font color=3D"#000080"><font col=
or=3D"#000000" size=3D"2"><span></span></font>&nbsp;</font></div>=0A<div><f=
ont color=3D"#000080"><font color=3D"#000000" size=3D"2"><span>Um abra=C3=
=A7o,</span></font></font></div>=0A<div><font color=3D"#000080"><font color=
=3D"#000000" size=3D"2"><span></span></font>&nbsp;</font></div>=0A<div><fon=
t color=3D"#000080"><font color=3D"#000000" size=3D"2"><span>Ulysses C. de =
=0ASouza.</span></font></font></div></div><br></div></div><br>=0A=0A=0A    =
  <hr size=3D1>Abra sua conta no <a href=3D"http://br.rd.yahoo.com/mail/tag=
lines/mail/*http://br.mail.yahoo.com/";>Yahoo! Mail</a>, o =C3=BAnico sem li=
mite de espa=C3=A7o para armazenamento! =0A</body></html>
--0-18767870-1206304893=:40862--
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