Download HW #6 Solutions

Math 4610/5810 - Fall 2009
HW #6 Solutions
Exercise 3.2.10
a) Let X denote the random variable (IA + IB )2 . Then the possible values
for X are 0, 1, 4 with corresponding probabilities:
p(0) = P (IA
p(1) = P (IA
= P (IA
p(4) = P (IA
= 0 and IB = 0) = P (IA = 0)P (IB = 0) = (1 − P (A))(1 − P (B))
= 1 and IB = 0) + P (IA = 0 and IB = 1)
= 1)P (IB = 0) + P (IA = 0)P (IB = 1) = P (A) − 2P (A)P (B)) + P (B)
= 1 and IB = 1) = P (IA = 1)P (IB = 1) = P (A)P (B).
b)
E(IA + IB )2 = E(IA2 + 2IA IB + IB2 ) = E(IA ) + 2E(IA )E(IB ) + E(IB )
= P (A) + 2P (A)P (B) + P (B)
Exercise 3.2.13
a) Let Xi denote the number showing up in the ith roll. Then
E(
10
X
7
Xi ) = 10E(X1) = 10( ) = 35.
2
i=1
b). Let T be the sum of the numbers in the first three rolls, S be the sum
of the largest two numbers in the first three rolls, M be the minimum of the
numbers in the first three rolls. Then
6
X
6−j +1 3
7
P (M ≥ j) = 3( ) − (
)
E(S) = E(T ) − E(M) = 3E(X1 ) −
2
6
j=1
j=1
=
203
≈ 8.4583
24
6
X
Exercise 3.4.1
a) It follows from the binomial distribution that
!
9 5
p (1 − p)
P ( exactly 5 heads appear in the first 9 tosses ) =
5
b) It follows from the geometric distribution that
P ( the first head appears on the 7th toss ) = (1 − p)6 p.
c) It follows from the negative binomial distribution that
!
11 4
p (1 − p)7 p.
P ( the fifth head appears on the 12th toss ) =
4
d) Let X and Y be the number of heads appear in the first 8 tosses
and in the next 5 tosses, respectively. Then both X and Y have binomial
distributions. By independence of the tosses, we have
P (X = Y ) =
X
P (X = x, Y = y) =
x=y
5
X
x=0
!
!
5 x
8 x
p (1 − p)5−x
p (1 − p)8−x
x
x
Exercise 3.4.3
Since X has the geometric distribution with the probability of success
1
p = 12
, E(X) = p1 = 12.
Exercise 3.4.6
a) Let Tr denote the number if trials until the rth success in Bernoulli (p)
trials. Then T1 is the number of trials until the 1st success so that T1 − 1
is the number of failures before the first success. Since T1 has geometric
distribution, we have
P (T1 − 1 = k) = P (T1 = k + 1) = q k p, for k = 0, 1, 2, · · ·
In other words, T1 − 1 has the same distribution has W .
b)
P (W > k) =
∞
X
i=k+1
qip = p
∞
X
qi = p
i=k+1
q k+1
= q k+1
1−q
c) It follows from the expected value of the geometric distribution that
E(W ) = E(T1 − 1) = E(T1 ) − 1 =
q
1
−1= .
p
p
d) It follows from the variance of the geometric distribution as derived in
the lecture that
V ar(W ) = V ar(T1 ) =
q
.
p2
Problem # 6
Proof: By definition of conditional probability, we have
P (X = n + k)
P (X = n + k and X > n)
=
P (X > n)
P (X > n)
n+k−1
q
p
=
= q k−1 p = P (X = k)
n
q
P (X = n + k|X > n) =