Prob answer key

Solutions Manual






A First Course in
PROBABILITY
Seventh Edition





Sheldon Ross

























Prentice Hall, Upper Saddle River NJ 07458

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Table of Contents




Chapter 1 ..............................................................................1

Chapter 2 ..............................................................................10

Chapter 3 ..............................................................................20

Chapter 4 ..............................................................................46

Chapter 5 ..............................................................................64

Chapter 6 ..............................................................................77

Chapter 7 ..............................................................................98

Chapter 8 ..............................................................................133

Chapter 9 ..............................................................................139

Chapter 10 ............................................................................141

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Chapter 1

Problems

1.
(a) By the generalized basic principle of counting there are


26
26 10 10 10 10 10 = 67,600,000

(b)
26
25 10 9 8 7 6 = 19,656,000

2. 64 = 1296

3.
An assignment is a sequence i1, ..., i20 where ij is the job to which person j is assigned. Since
only one person can be assigned to a job, it follows that the sequence is a permutation of the
numbers 1, ..., 20 and so there are 20! different possible assignments.

4.
There are 4! possible arrangements. By assigning instruments to Jay, Jack, John and Jim, in
that order, we see by the generalized basic principle that there are 2 1 2 1 = 4 possibilities.

5.
There were 8 2 9 = 144 possible codes. There were 1 2 9 = 18 that started with a 4.

6.
Each kitten can be identified by a code number i, j, k, l where each of i, j, k, l is any of the
numbers from 1 to 7. The number i represents which wife is carrying the kitten, j then
represents which of that wife's 7 sacks contain the kitten; k represents which of the 7 cats in
sack j of wife i is the mother of the kitten; and l represents the number of the kitten of cat k in
sack j of wife i. By the generalized principle there are thus 7 7 7 7 = 2401 kittens

7.
(a) 6! = 720
(b)
2
3! 3! = 72



(c) 4!3! = 144
(d)
6
3 2 2 1 1 = 72

8.
(a) 5! = 120
7!
(b)
= 1260
2 2
! !
!
11
(c) = 34,650
4 4
! 2
! !
7!
(d)
= 1260
2 2
! !

)
12
(
!
9.
= 27,720
6 4
! !

10.
(a) 8! = 40,320
(b)
2
7! = 10,080

(c) 5!4! = 2,880
(d)
4!24 = 384
Chapter 1
1

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11. (a)
6!
(b)
3!2!3!

(c)
3!4!

12. (a)
305
(b)
30
29 28 27 26

20
13.

2

52
14.

5

1012
15. There
are
possible choices of the 5 men and 5 women. They can then be paired up
5 5
in 5! ways, since if we arbitrarily order the men then the first man can be paired with any of
1012
the 5 women, the next with any of the remaining 4, and so on. Hence, there are
!
5

5 5
possible results.

6 7 4
16. (a)
+ + = 42 possibilities.
2 2 2

(b) There are 6 7 choices of a math and a science book, 6 4 choices of a math and an
economics book, and 7 4 choices of a science and an economics book. Hence, there are
94 possible choices.

17.
The first gift can go to any of the 10 children, the second to any of the remaining 9 children,
and so on. Hence, there are 10 9 8 5 4 = 604,800 possibilities.

56 4
18.
= 600
223

84 824
19. (a)
There
are
+ = 896 possible committees.
33 312
84
824

There
are
that do not contain either of the 2 men, and there are that
33
312
contain exactly 1 of them.

66 266
(b)
There
are
+ = 1000 possible committees.
33 1 23
2
Chapter 1

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75 75 75
75
(c)
There
are
+ + = 910 possible committees. There are in
33 23 32
33
75
which neither feuding party serves; in which the feuding women serves; and
23
75
in which the feuding man serves.
32

6 26 6 6
20.
+ , +
5 14 5 3

7!
21.
= 35. Each path is a linear arrangement of 4 r's and 3 u's (r for right and u for up). For
!
3 4!
instance the arrangement r, r, u, u, r, r, u specifies the path whose first 2 steps are to the right,
next 2 steps are up, next 2 are to the right, and final step is up.

4!
!
3
22. There
are paths from A to the circled point; and
paths from the circled point to B.
2!2!
2 !
1
!
Thus, by the basic principle, there are 18 different paths from A to B that go through the
circled piont.

23. 3!23


52

25.


,
13
,
13
,
13 13

12
12!
27.

=

,
3 ,
4 5


!
3 4 !
5
!

28.
Assuming teachers are distinct.
(a)
48

8

!
8
(b)

=
= 2520.
4
,
2 ,
2 ,
2 2

(2)

29. (a)
(10)!/3!4!2!

3 7!
(b)
3

2
4!2!

30. 2
9! - 228! since 2 9! is the number in which the French and English are next to each other
and 228! the number in which the French and English are next to each other and the U.S. and
Russian are next to each other.

Chapter 1
3

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31.
(a) number of nonnegative integer solutions of x1 + x2 + x3 + x4 = 8.

11

Hence, answer is = 165
3


7

(b) here it is the number of positive solutions--hence answer is = 35
3

32.
(a) number of nonnegative solutions of x1 + ... + x6 = 8
13

answer
=

5


(b) (number of solutions of x1 + ... + x6 = 5) x (number of solutions of x1 + ... + x6 = 3) =
108

5 5

33. (a)
x1 + x2 + x3 + x4 = 20, x1 2, x2 2, x3 3, x4 4

Let
y1 = x1 - 1, y2 = x2 - 1, y3 = x3 - 2, y4 = x4 - 3




y1 + y2 + y3 + y4 = 13, yi > 0

12

Hence,
there
are
= 220 possible strategies.
3

15
(b)
there
are
investments only in 1, 2, 3
2
14

there
are
investments only in 1, 2, 4
2
13

there
are
investments only in 1, 3, 4
2
13

there
are
investments only in 2, 3, 4
2

15
14
13 12


+ +
2
+ = 552 possibilities
2
2
2 3



4
Chapter 1

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Theoretical Exercises

2.
m n
i=1 i

3.
n(n - 1) (n - r + 1) = n!/(n - r)!

4.
Each arrangement is determined by the choice of the r positions where the black balls are
situated.

n
5. There
are
different 0 - 1 vectors whose sum is j, since any such vector can be
j
characterized by a selection of j of the n indices whose values are then set equal to 1. Hence
n
n
there are
vectors that meet the criterion.
j =k

j

n
6.

k

n -1 n -1
(n - )
1 !
(n - )
1 !
7.

+
=
+

r r -1
r (
! n -1- r)! (n - r) (
! r - )
1 !
n!
n - r r n




=
+
=



r (
! n - r)! n
n r

n + m
n
m
8. There
are

gropus of size r. As there are

groups of size r that consist of i
r
i r -

i
men and r - i women, we see that

n + m
r n
m



=
.
r
i r -


i
i=0




2n
n n
n
2
n n
9.

=
=
n
i n i
i
i=0
-


i=0


10.
Parts (a), (b), (c), and (d) are immediate. For part (e), we have the following:

n
k!n!
n!


k =
=

k
(n - k)!k!
(n - k) (
! k - )
1 !
n
(n - k + )
1 n!
n!

(n - k + )
1
=
=

k -
1
(n - k + )
1 (
! k - )
1 !
(n - k) (
! k - )
1 !
n -1
n(n - )
1 !
n!


n
=
=

k -1
(n - k) (
! k - )
1 !
(n - k) (
! k - )
1 !


Chapter 1
5

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11.
The number of subsets of size k that have i as their highest numbered member is equal to
i -1

, the number of ways of choosing k - 1 of the numbers 1, ..., i - 1. Summing over i
k -1
yields the number of subsets of size k.

n
12.
Number of possible selections of a committee of size k and a chairperson is k and so
k
n n
k represents the desired number. On the other hand, the chairperson can be anyone of
k
k =1

the n persons and then each of the other n - 1 can either be on or off the committee. Hence,
n2n - 1 also represents the desired quantity.

n
(i) 2
k
k

(ii)
n2n - 1 since there are n possible choices for the combined chairperson and secretary and
then each of the other n - 1 can either be on or off the committee.
(iii)
n(n - 1)2n - 2


(c) From a set of n we want to choose a committee, its chairperson its secretary and its
treasurer (possibly the same). The result follows since


(a)
there
are
n2n - 1 selections in which the chair, secretary and treasurer are the same
person.


(b) there are 3n(n - 1)2n - 2 selection in which the chair, secretary and treasurer jobs are
held by 2 people.

(c)
there
are
n(n - 1)(n - 2)2n - 3 selections in which the chair, secretary and treasurer are
all different.
n

(d)
there
are
3
k selections in which the committee is of size k.
k


n n
13. (1
- 1)n =
n-
(-
1
)
1

i
i=0


n
j n
n - i
14. (a)

=


j
i i j -

i

n n
j
n
n
n - i n
(b)
From
(a),
=
n-i

= 2
j i
i
j -1
i
j =i


j=i


n n
j
n
n
n - i
(c)
n- j
(- )1 =
n- j

(- )
1

j i
i
j -1
j =i


j=i

n-i
n
n - i




=
n-i-k

(- )
1
= 0
i
k
k=0


6
Chapter 1

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15.
(a) The number of vectors that have xk = j is equal to the number of vectors x1 x2 ... xk-1
satisfying 1 xi j. That is, the number of vectors is equal to Hk-1(j), and the result follows.

(b)




H2(1) = H1(1) = 1



H2(2) = H1(1) + H1(2) = 3



H2(3) = H1(1) + H1(2) + H1(3) = 6



H2(4) = H1(1) + H1(2) + H1(3) + H1(4) = 10



H2(5) = H1(1) + H1(2) + H1(3) + H1(4) + H1(5) = 15



H3(5) = H2(1) + H2(2) + H2(3) + H2(4) + H2(5) = 35

16.
(a) 1 < 2 < 3, 1 < 3 < 2, 2 < 1 < 3, 2 < 3 < 1, 3 < 1 < 2, 3 < 2 < 1,


1 = 2 < 3, 1 = 3 < 2, 2 = 3 < 1, 1 < 2 = 3, 2 < 1 = 3, 3 < 1 = 2, 1 = 2 = 3
n

(b) The number of outcomes in which i players tie for last place is equal to , the number
i
of ways to choose these i players, multiplied by the number of outcomes of the remaining
n - i players, which is clearly equal to N(n - i).

n n
n n
(c)
N(n- )1 = N(n-i)
i
n - i
i=1

i=1

n 1 n




= - N( j)
j
j =0




where the final equality followed by letting j = n - i.

(d)
N(3) = 1 + 3N(1) + 3N(2) = 1 + 3 + 9 = 13


N(4) = 1 + 4N(1) + 6N(2) + 4N(3) = 75

17.
A choice of r elements from a set of n elements is equivalent to breaking these elements into
two subsets, one of size r (equal to the elements selected) and the other of size n - r (equal to
the elements not selected).

18. Suppose
that
r labelled subsets of respective sizes n1, n2, ..., nr are to be made up from
r

n-1

elements 1, 2, ..., n where n = n . As n ,...,n -1 n
,... represents the number of
i
1
i
r
i=1


possibilities when person n is put in subset i, the result follows.

Chapter 1
7

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