# MATH 5010–001 SUMMER 2003 SOLUTIONS TO ASSIGNMENT 5 Problems from pp. 171–180 28.

```MATH 5010–001 SUMMER 2003
SOLUTIONS TO ASSIGNMENT 5
Problems from pp. 171–180
28. Let X denote the number of defectives in the sample. The possible values of X are
0, 1, 2, 3. The probabilities are:
16
P {X = 0} =
P {X = 1} =
P {X = 2} =
P {X = 3} =
Therefore,
E(X) = 1 &middot;
16 4
2
1
20
3
3
20
3
16 4
2
1
20
3 16 4
1
2
20
3
4
3
20 .
3
+2&middot;
16 4
1
2
20
3
+3&middot;
4
3
20 .
3
30. Let Hj denote the event that the jth toss results in heads, and Tj the event that the
jth toss results in tails. Then,
n
1
n
P {X = 2 } = P {H1 ∩ &middot; &middot; &middot; ∩ Hn−1 ∩ Tn } =
= 2−n .
2
Therefore,
E(X) =
∞
X
2n P {X = 2n } =
n=1
n
X
2n 2−n = +∞,
n=1
as asserted.
46. Let G denote the event that the randomly-selected defendant is guilty; we know that
P (G) = 0.65. Conditionally on G, the number X of jurors who vote “innocent” is
binomial with parameters n = 12 and p1 = 0.2. Conditionally on G c , X is binomial
with parameters n = 12 and p2 = 0.9. We are after P {X ≥ 9}.
P {X ≥ 9} = P {X ≥ 9 | G}P (G) + P {X ≥ 9 | G c }P (G c )
= P {X ≥ 9 | G}0.65 + P {X ≥ 9 | G c }0.35
12 12 X
X
12
12
j
12−j
= 0.65
0.2 0.8
0.9j 0.112−j .
+ 0.35
j
j
j=9
j=9
1
Theoretical Problem from p. 183
32. There are 2n − 1 nonempty subsets. On the other hand, for any j = 1, . . . , n, the
number of subsets that have exactly j elements is nj . Therefore,
P {X = j} =
Thus,
n
j
2n − 1
,
n
j
n
X
j = 1, . . . , n.
n
1 X
n!
E(X) =
= n
j&middot; n
j&middot;
2 −1
2 − 1 j=1
j!(n − j)!
j=1
n
1 X
n!
h
i
= n
2 − 1 j=1 (j − 1)! (n − 1) − (j − 1) !
n n−1 n X n−1
n X n−1
= n
= n
k
2 − 1 j=1 j − 1
2 −1
k=0
=
n2n−1
,
2n − 1
Pn−1
= (1 + 1)n−1 = 2n−1 . Divide
because the binomial theorem asserts that k=0 n−1
k
the numerator and the denominator by 2n−1 to obtain the first statement.
Similarly,
n
n
n
X
1 X
n!
j
E(X 2 ) =
= n
j2 &middot; n
j&middot;
2 −1
2 −1
(j − 1)!(n − j)!
j=1
=
2n
1
−1
1
= n
2 −1
j=1
n
X
(j − 1 + 1) &middot;
j=1
n
X
n!
(j − 1)!(n − j)!
n
1 X
n!
n!
+ n
(j − 1) &middot;
(j − 1)!(n − j)! 2 − 1 j=1 (j − 1)!(n − j)!
j=1
n
=
n
1 X
(n − 1)!
n X
n!
h
i
+
(j
−
1)
&middot;
2n − 1 j=2
(j − 1)!(n − j)! 2n − 1 j=1 (j − 1)! (n − 1) − (j − 1) !
n
n n(n − 1) X
(n − 2)!
n X n−1
h
i + n
= n
j −1
2 −1
(j − 2)! (n − 2)(j − 2) ! 2 − 1
j=2
j=1
n n−1 n(n − 1) X n − 2
n X n−1
= n
+ n
k
2 − 1 j=2 j − 2
2 −1
=
n(n − 1)
2n − 1
n−2
X
`=0
k=0
n−2
n2n−1
n(n − 1)2n−2 + n2n−1
+ n
=
.
`
2 −1
2n − 1
2
Now, n(n − 1)2n−2 + n2n−1 = n2 2n−2 + n(2n−1 − 2n−2 ) = n2 2n−2 + n2n−2 (2 − 1) =
(n2 + n)2n−2 . Because n2 + n = n(n + 1), this yields,
n(n + 1)2n−2
Var(X) =
−
2n − 1
n2n−1
2n − 1
2
=
n(n + 1)2n−2 (2n − 1) − n2 22n−2
.
(2n − 1)2
The numerator can be simplified a bit by first observing that n(n + 1)2n−2 (2n − 1) = n(n +
1)22n−2 −n(n+1)2n−2 . Therefore, n(n+1)2n−2 (2n −1)−n2 22n−2 = n22n−2 −n(n+1)2n−2 ,
and so
n22n−2 − n(n + 1)2n−2
Var(X) =
,
(2n − 1)2
as claimed.
To finish, note that as n → ∞, (2n − 1)2 ∼ 22n . Therefore,
n(n + 1)2n−2
= 0.
n→∞
(2n − 1)2
lim
Consequently,
Var(X) ∼
n2n−2
n
= n2−2 = .
2n
2
4
3
```