Vinh Nguyen
Stat 120A(Fall 2007): HW5 Solutions
November 28, 2007
Text:
Probabilty and Statistics
, by Degroot and Schervish (3
rd
ed)
Problems:
5.2.3, 5.2.4, 5.3.6, 5.3.7, 5.5.2, 5.4.1, 5.4.2, and Additional Problems
5.2.3 Let
X
be the number of heads in 10 tosses. We have
X
∼
Bin
(10
,
1
/
2). In order to get more heads than
tails, we must have at least six heads. Thus,
P
(
X
≥
6) = 1

P
(
X
≤
5) = 1

5
X
x
=0
±
10
x
²
(1
/
2)
x
(1
/
2)
10

x
=
0
.
3770
.
5.2.4 We have
X
∼
Bin
(15
,
0
.
4). Thus,
P
(6
≤
X
≤
9) =
9
X
x
=6
±
15
x
²
(0
.
4)
x
(0
.
6)
15

x
=
0
.
5630
.
5.3.6 Suppose
X
1
∼
Bin
(
n
1
,p
) and
X
2
∼
Bin
(
n
2
,p
), where
X
1
and
X
2
are independent. Note that
X
1
+
X
2
∼
Bin
(
n
1
+
n
2
,p
). We would like to compute the conditional distribution of
X
1
given
X
1
+
X
2
=
k
, for
k
= 1
,...,n
1
+
n
2
. In order to do so, we apply the deﬁnition of conditional probability:
P
(
X
1
=
x

X
1
+
X
2
=
k
) =
P
(
X
1
=
x,X
1
+
X
2
=
k
)
P
(
X
1
+
X
2
=
k
)
(Deﬁnition)
=
P
(
X
1
=
x,X
2
=
k

x
)
P
(
X
1
+
X
2
=
k
)
=
P
(
X
1
=
x
)
P