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S olutions M anual

A First Course in

P

ROBABILIT

Y

Seventh Edition

Sheldon Ross

Prentice Hall, Upper Saddle River NJ 07458

Chapter 1

Problems

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

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

4 = 1296

3. An assignment is a sequence i 1 , …, i 20 where i (^) j 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

(b) 2! 2!

(c) 4! 4! 2!

(d) 2! 2!

10. (a) 8! = 40,

(b) 2 ⋅ 7! = 10, (c) 5!4! = 2, (d) 4!

4 = 384

11. (a) 6!

(b) 3!2!3! (c) 3!4!

12. (a) 30

5

(b) 30 ⋅ 29 ⋅ 28 ⋅ 27 ⋅ 26

15. There are  

possible choices of the 5 men and 5 women. They can then be paired up

in 5! ways, since if we arbitrarily order the men then the first man can be paired with any of

the 5 women, the next with any of the remaining 4, and so on. Hence, there are (^)  

possible results.

16. (a) (^)  

= 42 possibilities.

(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.

19. (a) There are (^)  

= 896 possible committees.

There are (^)  

that do not contain either of the 2 men, and there are (^)  

that

contain exactly 1 of them.

(b) There are (^)  

= 1000 possible committees.

31. (a) number of nonnegative integer solutions of x 1 + x 2 + x 3 + x 4 = 8.

Hence, answer is  

(b) here it is the number of positive solutions—hence answer is (^)  

32. (a) number of nonnegative solutions of x 1 + … + x 6 = 8

answer = (^)  

(b) (number of solutions of x 1 + … + x 6 = 5) × (number of solutions of x 1 + … + x 6 = 3) =

33. (a) x 1 + x 2 + x 3 + x 4 = 20, x 1 ≥ 2, x 2 ≥ 2, x 3 ≥ 3, x 4 ≥ 4

Let y 1 = x 1 − 1, y 2 = x 2 − 1, y 3 = x 3 − 2, y 4 = x 4 − 3

y 1 + y 2 + y 3 + y 4 = 13, yi > 0

Hence, there are (^)  

= 220 possible strategies.

(b) there are  

investments only in 1, 2, 3

there are  

investments only in 1, 2, 4

there are (^)  

investments only in 1, 3, 4

there are (^)  

investments only in 2, 3, 4

2 = 552 possibilities

Theoretical Exercises

2. ∑ = i

m i 1 n

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.

5. There are (^)  

j

n different 0 − 1 vectors whose sum is j , since any such vector can be

characterized by a selection of j of the n indices whose values are then set equal to 1. Hence

there are (^)  

∑ = j

n n j k vectors that meet the criterion.

k

n

r

n

r

n

( )!( 1 )!

n r r

n

r n r

n

− r

n

n

r

n

n r

r n r

n

!( )!

8. There are (^)  

r

n m gropus of size r. As there are (^)  

r i

m

i

n groups of size r that consist of i

men and r − i women, we see that

=

 +^

r

i

r i

m

i

n

r

n m

0

= n i

n

i

n

n

n

n

i 0

2

0

=

n

i i

n

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

k

n k = ( )!( 1 )!

− n k k

n

n k k

kn

k

n n k = ( )!( 1 )!

n k k

n

n k k

n k n

k

n n = ( )!( 1 )!

n k k

n

n k k

nn

15. (a) The number of vectors that have xk = j is equal to the number of vectors x 1 ≤ x 2 ≤ … ≤ xk − 1

satisfying 1 ≤ xi ≤ j. That is, the number of vectors is equal to Hk − 1 ( j ), and the result follows.

(b) H 2 (1) = H 1 (1) = 1 H 2 (2) = H 1 (1) + H 1 (2) = 3 H 2 (3) = H 1 (1) + H 1 (2) + H 1 (3) = 6 H 2 (4) = H 1 (1) + H 1 (2) + H 1 (3) + H 1 (4) = 10 H 2 (5) = H 1 (1) + H 1 (2) + H 1 (3) + H 1 (4) + H 1 (5) = 15 H 3 (5) = H 2 (1) + H 2 (2) + H 2 (3) + H 2 (4) + H 2 (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

(b) The number of outcomes in which i players tie for last place is equal to (^)  

i

n , the number

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 ).

(c) ∑

=

n

i

N n i

n

1

=

n

i

N n i n i

n

1

−

=

1

0

n

j

N j j

n

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

(d) N (3) = 1 + 3 N (1) + 3 N (2) = 1 + 3 + 9 = 13 N (4) = 1 + 4 N (1) + 6 N (2) + 4 N (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 n 1 , n 2 , …, n (^) r are to be made up from

elements 1, 2, …, n where n = ∑

=

r

i

ni 1

. As 

i −^ nr

n n n ,...

1 ,..., 1 represents the number of

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

19. By induction:

( x 1 + x 2 + … + xr )

n

1

1 (^2 ... )

0 1

n i r

i

n

i

x x x i

n (^) −

=

∑ by the Binomial theorem

1

1 0 1

i

n

i

x i

n

=

i (^) 2 ,..., i r

1 2

1 i r

i

r

x x i i

n i  

i (^) 2 +... + ir = n − i 1

i 1 ,..., i r

∑∑ ...∑ ir

r

i

r

x x i i

n ... ,...,

1 1 1

i 1 + i 2 +...+ ir = n

where the second equality follows from the induction hypothesis and the last from the

identity (^)  

n i i r

n

i i

n i

i

n

2 ,...,^1 ,...,

1

1

20. The number of integer solutions of

x 1 + … + xr = n , xi ≥ mi

is the same as the number of nonnegative solutions of

y 1 + … + yr = n − ∑

r mi 1

, yi ≥ 0.

Proposition 6.2 gives the result 

1 r

n m r

r

i (^).

21. There are (^)  

k

r choices of the k of the x ’s to equal 0. Given this choice the other r − k of the

x ’s must be positive and sum to n.

By Proposition 6.1, there are  

n r k

n

r k

n 1

1

such solutions.

Hence the result follows.

n

n r by Proposition 6.2.

Chapter 2

Problems

1. (a) S = {( r , r ), ( r , g ), ( r , b ), ( g , r ), ( g , g ), ( g , b ), ( b , r ), b , g ), ( b , b )}

(b) S = {( r , g ), ( r , b ), ( g , r ), ( g , b ), ( b , r ), ( b , g )}

2. S = {( n , x 1 , …, xn − 1 ), n ≥ 1, xi ≠ 6, i = 1, …, n − 1}, with the interpretation that the outcome is

( n , x 1 , …, xn − 1 ) if the first 6 appears on roll n , and xi appears on roll, i , i = 1, …, n − 1. The

event

c ( (^) n 1 En )

∞ ∪ (^) = is the event that 6 never appears.

3. EF = {(1, 2), (1, 4), (1, 6), (2, 1), (4, 1), (6, 1)}.

E ∪ F occurs if the sum is odd or if at least one of the dice lands on 1. FG = {(1, 4), (4, 1)}. EF

c is the event that neither of the dice lands on 1 and the sum is odd. EFG = FG.

4. A = {1,0001,0000001, …} B = {01, 00001, 00000001, …}

( A ∪ B )

c = {00000 …, 001, 000001, …}

5. (a) 2

5 = 32 (b) W = {(1, 1, 1, 1, 1), (1, 1, 1, 1, 0), (1, 1, 1, 0, 1), (1, 1, 0, 1, 1), (1, 1, 1, 0, 0), (1, 1, 0, 1, 0) (1, 1, 0, 0, 1), (1, 1, 0, 0, 0), (1, 0, 1, 1, 1), (0, 1, 1, 1, 1), (1, 0, 1, 1, 0), (0, 1, 1, 1, 0), (0, 0, 1, 1, 1) (0, 0, 1, 1, 0), (1, 0, 1, 0, 1)}

(c) 8 (d) AW = {(1, 1, 1, 0, 0), (1, 1, 0, 0, 0)}

6. (a) S = {(1, g ), (0, g ), (1, f ), (0, f ), (1, s ), (0, s )}

(b) A = {(1, s ), (0, s )} (c) B = {(0, g ), (0, f ), (0, s )} (d) {(1, s ), (0, s ), (1, g ), (1, f )}

7. (a) 6

15

(b) 6

15 − 3

15

(c) 4

15

8. (a).

(b). (c) 0

9. Choose a customer at random. Let A denote the event that this customer carries an American

Express card and V the event that he or she carries a VISA card.

P ( A ∪ V ) = P ( A ) + P ( V ) − P ( AV ) = .24 + .61 − .11 = .74.

Therefore, 74 percent of the establishment’s customers carry at least one of the two types of credit cards that it accepts.

10. Let R and N denote the events, respectively, that the student wears a ring and wears a

necklace.

(a) P ( R ∪ N ) = 1 − .6 =.

(b) .4 = P ( R ∪ N ) = P ( R ) + P ( N ) − P ( RN ) = .2 + .3 − P ( RN ) Thus, P ( RN ) =.

11. Let A be the event that a randomly chosen person is a cigarette smoker and let B be the event

that she or he is a cigar smoker.

(a) 1 − P ( A ∪ B ) = 1 − (.07 + .28 − .05) = .7. Hence, 70 percent smoke neither.

(b) P ( A

c B ) = P ( B ) − P ( AB ) = .07 − .05 = .02. Hence, 2 percent smoke cigars but not cigarettes.

12. (a) P ( S ∪ F ∪ G ) = (28 + 26 + 16 − 12 − 4 − 6 + 2)/100 = 1/

The desired probability is 1 − 1/2 = 1/2.

(b) Use the Venn diagram below to obtain the answer 32/100.

S F

G

(c) since 50 students are not taking any of the courses, the probability that neither one is

taking a course is (^)  

= 49/198 and so the probability that at least one is taking a

course is 149/198.

13. (a) 20,

(b) 12, (c) 11, (d) 68, (e) 10,

I II

III

21. (a) p 1 = 4/20, p 2 = 8/20, p 3 = 5/20, p 4 = 2/20, p 5 = 1/

(b) There are a total of 4 ⋅ 1 + 8 ⋅ 2 + 5 ⋅ 3 + 2 ⋅ 4 + 1 ⋅ 5 = 48 children. Hence,

q 1 = 4/48, q 2 = 16/48, q 3 = 15/48, q 4 = 8/48, q 5 = 5/

22. The ordering will be unchanged if for some k , 0 ≤ k ≤ n , the first k coin tosses land heads and

the last n − k land tails. Hence, the desired probability is ( n + 1/

n

23. The answer is 5/12, which can be seen as follows:

1 = P {first higher} + P {second higher} + p {same} = 2 P {second higher} + p {same} = 2 P {second higher} + 1/

Another way of solving is to list all the outcomes for which the second is higher. There is 1 outcome when the second die lands on two, 2 when it lands on three, 3 when it lands on four, 4 when it lands on five, and 5 when it lands on six. Hence, the probability is (1 + 2 + 3 + 4 + 5)/36 = 5/12.

25. P ( En ) = 5

1

1

^ = 

∞

=

−

n

n

n PE

27. Imagine that all 10 balls are withdrawn

P ( A ) =

28. P {same} =

 

P {different} = (^)  

If sampling is with replacement

P {same} = (^3)

3 3 3

P {different} = P ( RBG ) + P { BRG ) + P (RGB) + … + P ( GBR )

29. (a) ( )( 1 )

n m n m

nn mm

(b) Putting all terms over the common denominator ( n + m )

2 ( n + m − 1) shows that we must prove that

n

2 ( n + m − 1) + m

2 ( n + m − 1) ≥ n ( n − 1)( n + m ) + m ( m − 1)( n + m )

which is immediate upon multiplying through and simplifying.

30. (a)

4! 4

(b)

 

(c)

 

31. P ({complete} = 9

P {same} = 9

b g

g

b g

gb g

45. 1/ n if discard, (^) k

k

n

n

1 ( 1 )

− − if do not discard

46. If n in the room,

P {all different} = 1212 12

⋅ ⋅ ⋅ − n

When n = 5 this falls below 1/2. (Its value when n = 5 is .3819)

12

20 4 4 (^12 ) ( 3 !)( 2 !)

n m n n N m

n ( 1 ) /

−  − 

52. (a) 2019181716151413

(b) 2019181716151413

53. Let A (^) i be the event that couple i sit next to each other. Then

2 3 4 4 1

P ∪ (^) i = Ai =

and the desired probability is 1 minus the preceding.

54. P ( S ∪ H ∪ D ∪ C ) = P ( S ) + P ( H ) + P ( D ) + P (C) − P ( SH ) − … − P ( SHDC )

55. (a) P ( S ∪ H ∪ D ∪ C ) = P ( S ) + … − P ( SHDC )

3 4

(b) P (1 ∪ 2 ∪ … ∪ 13) =

 

56. Player B. If Player A chooses spinner (a) then B can choose spinner (c). If A chooses (b)

then B chooses (a). If A chooses (c) then B chooses (b). In each case B wins probability 5/9.