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1.4.11. Without the
missing c, there are 50 individuals, 8 of whom have
brown hair and 20 of whom have blue eyes. To satisfy the condition,
we
must have
.Rr+
P[brown hair]
:
;#
=
P[brown hair
I
blue eyes]
'
bu+t
=r
20*z'
Therefore
(8
+r)(20
*r)
:
r(50 +
r)
+
160
*28s
:b0c
+
160
:
22r.
There is no integer value of
r
satisfying the last equation.
3g
/
Nu,
z
I
I
1.4.13. (a)
PlXz: rl
:p(l
p)+
(rp).L:Lp2;
PlX2:3]
:p'p:p2.
(b)
PlX3
:11
:
(1

p)(1

pX1

p)
:
(1

p)3;
P[&:3]
:p(lp)(i
p)+
(1p)p(1
p)+
(1pXtp)p
:3p(r

p)2;
PlX3
:
5l
:
p. p(l

p) +
p(I

p)p + (L

p)p

p
:
3p2
(t

p);
P[X3:7]:p.pp:p3.
1.4.15.
Pflyingltest says lie]
/ret
P(xr=
l
lx,'a)
=
l,,ra{y
('
o*C,r./.z
4E
u4/tM/
49{'
f)r/r,
f(n
t
h)
=
:
_
(.80)(.50)
(.80)(.50)
+ (.10)(.50)
_
(.80)
_pro

(.80) + ('10)

"t
"'
L.4.L7. We assume in this problem that ace counts as 11 for the
dealer.
Of the 4? remaining cards, there are four each of ranks 2, 4,
6,7, 9,
and
ace, there are eleven face cards, and there are three each
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This note was uploaded on 10/27/2010 for the course MATH 351 taught by Professor Moumen,f during the Spring '08 term at George Mason.
 Spring '08
 Moumen,F

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