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3.19(b) Before we actually start the problem let’s get a feel for it in the case
n = 5. To avoid confusion assume the boxes are labelled 1;2;3;4;5 and
the balls labelled A;B;C;D;E. An outcome can be represented as a
5▯tuple such as (1;1;3;2;1), and we interpret this as A going to box
1, B going to box 1, C going to box 3, D going to box 2 and E going to
box 1. There are clearly 5 total outcomes, and it is worth noting that
the outcome (1;1;3;2;1) will leave exactly two boxes (boxes 4 and 5)
empty.
So what is an outcome that leaves exactly one box empty going to look
like? Well its going to have to look something like (1;3;1;2;5). We see
that two of the balls have to go in the same box and the other three
have to go in di▯erent boxes. So let’s break the job of writing down
such outcomes into three tasks. First we decide which balls are going
to go into the same box, then we decide which box to put them in,
then we decide on the boxes for the remaining balls. The ▯rst task
amou▯ ▯ to selecting a 2▯element subset of fA;B;C;D;Eg and there
are 5 ways to do this, the second task amounts to choosing one of the
2
available 5 boxes to put them in and there are 5 ways to do this. For
example we might choose fB;Eg in step one and 4 in step two, leading
to a partially completed outcome of ( ;4; ; ;4). To complete this
outcome we need only select the boxes that are going to hold the three
remaining balls, this amounts to selecting a three-digit permutation of
the digits f1;2;3;5g and there are 4 ▯ 3 ▯ 2 = 4! ways to do this. For
example we might select the permutation 532, leading to the complete
outcome (5;4;3;2;4).
▯5▯ ▯5▯
So there are 2 ▯ 5 ▯ 4! = 2 ▯ 5! total outcomes that leave exactly one
box▯em▯ty. In general, ▯i▯h n boxes and n balls, the total number will
be n ▯ n ▯ (n ▯ 1)! = n ▯ n! such outcomes.
2 2
3.39(b) Each state has two senators, say A and B. There are 50 states. So we
may write the complete list of senators as
S = f1 A1 B2 ;A ;:B:;50 ;50Ag : B
1 An outcome here is a 5▯element subset of S such as f4 ;7 A34B;49 A49 g,A B
▯ ▯
and there are 150 total outcomes. In order to write down an outcome
that does not repeat a state let us (i) choose the states that will be
represented and then (ii) choose which senator (A or B) from each
state. Task (i) amounts to selecting a 5▯element subset of f1;2;:::;50g
and task (ii) amounts to selecting a 5▯digit string using only the
letters A,B (repetitions obviously allowed). For example we might
choose f3;9;15;24;42g in step (i) and AABAA in step (ii), leading
▯50
to the committee f3 ;A ;1A ;24B;42 A. ThAre are 5 ways to com-
plete task (i) and 2 ways to complete task (ii), therefore there are
▯ ▯
50 ▯ 2 = 67;800;320 total committees with no state repeated.
5
3.43 We need to ▯ll in the string
▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯
using 1 P, 4 Es, 2 Rs, 1 S and 1 V . put the letters In order to create
such an arrangement we must (i) choose the spot that is going the P,
then (ii) choose the four spots that are going to hold the Es, then :::.
For example if we choose f4g in step (i) and f1;3;7;8g in step (ii) we
get the partially completed outcome
E ▯ ▯ E ▯ P ▯ ▯ ▯ E ▯ E ▯
(6)
3.59(b) The answer is 48 ▯4▯45.
52!
3.101 Suppose for the moment that N = 10 and M = 4. Then an outcome
will be a string of the form nndnnnddnd, consisting of exactly 4 d’s
and 6 n’s. This particular outcome corresponds to the situation where
the third, seventh, eighth and tenth items are defective. In order to
construct such an outcome we need only perform one task, and that is to
select the positions of the 4 d’s. This amounts to selecting a 4-element
subset of f1;2;3;:::;9;10g, and the outcome above corresponds to the
▯10
selection f3;7;8;10g, and there are obviously 4 ways to do this.
2 ▯10
So there are 4 outcomes. In general, for arbitrary M and N (with
M ▯ N) an outcome will be a string of length N, such as ddnd▯▯▯nndn
consisting of exactly M d’s. To construct such an outcome we simply
select a M-element subset of f1;2;3;:::;N ▯1;Ng and put the M d’s
▯N▯
in those positions. So there areM total outcomes.
th
What does an outcome where the k item selected is the last defective
look like? Well for N = 10, M = 4 and k = 6 it would have to look
something like dndnddnnnn. The general form of such an outcome is
▯ ▯ ▯ ▯ ▯ dnnnn. In order to write down such an outcome we
need only choose where, in t▯ ▯▯rst 5 = k ▯ 1 spots, to put the ▯rst
3 = M ▯1 d’s, and there are 5 = k ▯ 1M ▯ 1 ways to do this. So the
▯ 3▯ ▯ ▯
probability of interest isk▯1 = N , assuming of course that k ▯ M
M▯1 M
(if k < M the probability is zero).
▯M▯▯ N▯M ▯ ▯N▯
3.109 (a) The probability is k n▯k = n . (b) The probability that we
catch a tagged deer is M=N, and we would expect that the proportion
of tagged deer in our sample would be reasonably close to this number.
Thus we would expect that k=n ▯ M=N and we could estimate N
using N ▯ nM=k = 8▯10=3 ▯ 27. So we might estimate that there are
approximately 27 deer in the forest.
3 5.40 When sampling without replacement there are (N) n N(N ▯ 1)(N ▯
2)▯▯▯(N ▯n+1) possible outcomes. In order to determine the number
th
of outcomes in which the k person selected has the speci▯ed attribute
note that there are M ways to select this person, and (N ▯1) ways
n▯1
to select the remaining people. Therefore the probability that the k
person is selected has the speci▯ed attribute is
M ▯ (N ▯ 1) M ▯ (N ▯ 1) ▯ (N ▯ 2)▯▯▯((N ▯ 1) ▯ (n ▯ 1) + 1)
n▯1 =
(N) n N ▯ (N ▯ 1)▯▯▯(N ▯ n + 1)
M ▯ (N ▯ 1) ▯ (N ▯ 2)▯▯▯(N ▯ n + 1)
= N ▯ (N ▯ 1)▯▯▯(N ▯ n + 1)
M
= :
N
If the sampling is carried out with replacement the total number of
n th
outcomes is N . As before there are M ways to choose the k person
but there are now Nn▯1ways to choose the remaining people. Therefore
the probability the khperson that is selected is
n▯1
M ▯ N = M :
N n N
5.68 Think of the words the students knows as \Type 1." The probability
in part (a) is then
▯ ▯▯ ▯
30;00050;000
4▯ ▯0 = 0:0711937::: :
80;000
100
The probability in part (b) is
40 ▯30;00050;000
X k 100▯k
▯80;000 = 0:4645:::
k=35 100
According to the book’s rule of sum the binomial approximation would
be appropriate since the student knows more than 5% of the words in
the dictionary. Note that if we use the binomial approximation in part
4 (a) we get
▯ ▯▯ ▯
30;00050;000 ▯ 100▯▯ ▯3▯ 40 5 60
4▯ ▯60 ▯ = 0:0711599::: ;
80;000 40 8 8
100
which is quite a good agreement. For (d) I would estimate the size of
the students vocabulary at
38
100 ▯ 80;000 = 30;400 :
5.116 The pmf is
▯ k ▯ 1▯
f(k) = (:67) (:33)▯3 ; k = 3;4;5;:::
2
5 6.6 (a) Start with a probability such P(X = 3;Y = 5). There are two
ways this can occur - we get a three on the ▯rst roll and a ▯ve
on the second roll, or we get a ▯ve on the ▯rst roll and a three
on the second roll. Each of these outcomes has probability 1 , so
36
P(X = 3;Y = 5) = 1 .
18
Now consider a probability such as P(X = 3;Y = 3). The only
way this can happen is if we get a three on both rolls, so P(X =
3;Y = 3) = 1.
36
Finally, consider a probability such as P(X = 3;Y = 2). By
de▯nition the maximum exceeds the minimum, and so there is
now way the minimum can be three when the maximum is two.
Therefore P(X = 3;Y = 2) = 0.
To summarize, the joint probability mass function of (X;Y ) is
8 1
< 36 if 1 ▯ x = y ▯ 6 ;
1
fX;Y (x;y) = P(X = x;Y = y) = 18 if 1 ▯ x < y ▯ 6 ;
:
0 otherwise
(b) In tabular form we have
X
1 2 3 4 5 6 f
Y
1 1 0 0 0 0 0 1
36 36
2 1 1 0 0 0 0 3
18 36 36
Y 3 1 1 1 0 0 0 5
18 18 36 36
4 1 1 1 1 0 0 7
18 18 18 36 36
5 1 1 1 1 1 0 9
18 18 18 18 36 36
6 1 1 1 1 1 1 11
18 18 18 18 18 36 36
fX 11 9 7 5 3 1
36 36 36 36 36 36
(c)-(h) The general idea for problems like this is to identify all points in
the range that satisfy the given condition. The upper left panel
of Figure 1 illustrates the range of (X;Y ), i.e. the set of all points
6 (x;y) for which f (x;y) > 0, while the other three panels il-
X;Y
lustrate (in red) the points in the range that satisfy the given
condition for parts (f), (g) (the others are left to you). Once
we’ve identi▯ed the relevant points in the range we simply sum
the values of the joint pmf evaluated at those points. For example
1
P(X = Y ) = f X;Y (1;1)+f X;Y (2;2)+f X;Y (3;3)+f X;Y (4;4)+f X;Y (5;5)+f X;Y (6;6) = ;
6
and
1
P(Y = 2X) = f (1;2) + f (2;4) + f (3;6) = ;
X;Y X;Y X;Y 6
etc.
Figure 1:
Range of (X,Y) Points in Range where X=2X in Red
7 7
6 6
5 5
4 4
y y
3 3
2 2
1 1
00 2 4 6 00 2 4 6
x x
7 9.26 The desired probability is
Z 3=4Z 1 Z 3=4▯Z 1 ▯
(x + y)dy dx = (x + y)dy dx
1=4 1=2 1=4 1=2
Z 3=4▯ 2▯y=1
y
= xy + 2 dx
1=4 y=1=2
Z 3=4▯ ▯
= x + 3 dx
2 8
1=4
7
= :
16
9.172 The domain of the pair (X;Y ) is the rectangle [0;1]▯[0;2]. The region
where the total loss exceeds one is the portion of this rectangle that
lies above the line y = 1 ▯ x (why?). Thus the desired probability is
Z 1Z 2
(2x + 2 ▯ y)=4dy dx :
0 1▯x
9.178 (c) gives the desired probability. This is because, in order that the
device function, both variables must exceed 0.5.
9.190 (a) The marginal pdf of X is
Z x
f (x) = 8xy dy = 4x ; 0 < x < 1 :
X
0
(b) The marginal pdf of Y is
Z 1
2
fY(y) = 8xy dx = 4y(1 ▯ y ) ; 0 < y < 1 :
y
6.48 Using the table from the solution to 6.6 we get the following table.
The ▯rst row corresponds to the conditional mass function f (x;1) =
XjY
fX;Y(x;1)=fY(1) and gives the conditional distribution P(X = xjY =
1). The second rwo contians f XjY(x;2) = f X;Y(x;2)=fY(2) and gives
8 the conditional distribution P(X = xjY = 2), etc.
X
1 2 3 4 5 6
1 1 0 0 0 0 0
2 2 1 0 0 0 0
2 2 1
Y 3 5 5 5 0 0 0
2 2 2 1
4 7 7 7 7 0 0
2 2 2 2 1
5 9 9 9 9 9 0
2 2 2 2 2 1
6 11 11 11 11 11 11
A similar tanble can be constructed for conditional distributions of Y ,
given X, in which case each column would correspond to a conditional
pmf.
9.64 In each case the probability of interest is
Z
1
fXjY (x;0:8)dx :
0:9
(a) Here we have f (y) = 1 for 0 < y < 1 (i.e. the marginal distribu-
Y
tion of Y is uniform) and so f (x;0:8) = 1 (i.e. the conditional
XjY
distribution of X, given Y = 0:8, is uniform) the probability of
interest is 0:1.
R 1
(b) Here the marignal density of Y is f (y)Y= 0 (x + y)dx = 0:5 + y
x+0:8
and so f XjY (x;0:8) = [x + 0:8]=[0:5 + 0:8] = 1:3. The proba-
1▯(0:9) +(0:8)(0:1)
bility of interest is therefore 1:3 = 0:207:::. This is
larger than what we got in (a) because the conditional distribu-
tion of X here is more heavily weighted towards larges values of
X.
1 3 3
(c) Here the marginal density of Y is f (y)Y= 2 + 2 , leading to the
conditional density
3x + 3(0:8) 2
fXjY (x;0:8) = ;
1 + 3(0:8) 2
9 and the probability of interest is
1 ▯ (0:9) + 3(0:8) (0:1)
2 = 0:158:::
1 + 3(0:8)
9.69 Ignore part (a), hard to see why there is a di▯erence until we work
through the example. You are encouraged to sketch T, noting that it
is a triangle bounded below by the horizontal axis, above by the line
y = x and on the right by the vertical line at x = 1.
(b) In both cases the joint density is positive for all (x;y) 2 T, i.e. all
(x;y) such that 0 < y < x < 1. For Method I the joint density is
simply
1 1
fX;Y(x;y) = = = 2:
Area(T) 1=2
For Method II the joint density is obtained as follows
1 1
fX;Y(x;y) = fY jX(y;x) ▯Xf (x) = ▯ 1 = :
x x
(c) For Method I we have
Z x
fX(x) = 2dy = 2x ;
Z0
1
fY(y) = 2dx = 2(1 ▯ y) ;
y
and for Method II we have
Z x
f (x) = 1 dy = 1 ;
X 0 x
Z 1
1
fY(y) = x dx = ▯ln(y) :
y
(d) For Method I the relevant conditional pdfs are
2 1
fXjY(x;y) = 2(1 ▯ y) = 1 ▯ y ; y < x < 1 ;
10 so that the conditional distribution of X, given Y = y, is uniform
on (y;1), and
f (y;x) = 2 = 1 ; 0 < y < x ;
Y jX 2x x
so that the conditional distribution of Y , given X = x, is uniform
on (0;x).
For Method II we have
1=x 1
fXjY (x;y) = ▯ln(y) = ▯ xln(y) ; y < x < 1 ;
and
f (y;x) = 1=x = 1 ; 0 < y < x ;
Y jX 1 x
so that the conditional distribution of Y , given X = x, is uniform
on (0;x).
In both cases the conditional distribution of Y , given X, is uni-

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