IEOR 172: Probability and Risk Analysis for Engineers, Fall 2007
Homework 3 Solution
Chapter 3
Question 58
(a)
P(head) = P(head

HH
)P(
HH
) + P(head

HT
)P(
HT
)
+P(head

TH
)P(
TH
) + P(head

TT
)P(
TT
)
= P(head)
p
2
+ 0
·
p
(1

p
) + 1
·
(1

p
)
p
+ P(head)(1

p
)
2
then
P(head) =
p
(1

p
)
1

p
2

(1

p
)
2
=
1
2
,
and P(tail) = 1

P(head) =
.
5.
(b) If the ﬁrst ﬂip is
H
, then the procedure continues until we get a
T
, and the result is
T
; simi
larly, if the ﬁrst ﬂip is
T
, then the result is
H
. Therefore, P(head) = P(
T
) = 1

p
and P(tail) =
p
,
so the result is not equally likely to be either heads or tail.
Theoretical exercise
Question 21
(a) For
n
= 2,
m
= 1, voting order
AAB
:
P
2
,
1
=
1
3
.
For
n
= 3,
m
= 1, voting order
AAAB
,
AABA
:
P
3
,
1
=
2
4
=
1
2
.
For
n
= 3,
m
= 1, voting order
AAAB
,
AABA
:
P
3
,
1
=
2
4
=
1
2
.
For
n
= 3,
m
= 2, voting order
AAABB
,
AABAB
:
P
3
,
2
=
2
10
=
1
5
.
For