Introduction_to_Probability-Grinstead-Snell-solutions.pdf

Charles M. Grinstead and J. Laurie Snell:

INTRODUCTION to PROBABILITY

Published by AMS

Solutions to the exercises

SECTION 1.1

1. As n increases, the proportion of heads gets closer to 1/2, but the di ! erence between the number

of heads and half the number of ﬂips tends to increase (although it will occasionally be 0).

3. (b) If one simulates a su!ciently large number of rolls, one should be able to conclude that the

gamblers were correct.

5. The smallest n should be about 150.

7. The graph of winnings for betting on a color is much smoother (i.e. has smaller ﬂuctuations) than

the graph for betting on a number.

9. Each time you win, you either win an amount that you have already lost or one of the original

numbers 1,2,3,4, and hence your net winning is just the sum of these four numbers. This is not a

foolproof system, since you may reach a point where you have to bet more money than you have.

If you and the bank had unlimited resources it would be foolproof.

11. For two tosses, the probabilities that Peter wins 0 and 2 are 1/2 and 1/4, respectively. For four

tosses, the probabilities that Peter wins 0, 2, and 4 are 3/8, 1/4, and 1/16, respectively.

13. Your simulation should result in about 25 days in a year having more than 60 percent boys in the

large hospital and about 55 days in a year having more than 60 percent boys in the small hospital.

15. In about 25 percent of the games the player will have a streak of ﬁve.

SECTION 1.2

P ({a, b, c}) = 1

P ({a}) = 1/2

P ({a, b}) = 5/6

P ({b}) = 1/3

P ({b, c}) = 1/2

P ({c}) = 1/6

P ({a, c}) = 2/3

P (!) = 0

3. (b), (d)

5. (a) 1/2

(b) 1/4

(d) 7/8

7. 11/12

9. 3/4,

11. 1 : 12, 1 : 3, 1 : 35

13. 11:4

15. Let the sample space be:

" 1 = {A, A}

" 4 = {B, A}

" 7 = {C, A}

" 2 = {A, B} " 5 = {B, B} " 8 = {C, B}

" 3 = {A, C} " 6 = {B, C} " 9 = {C, C}

where the ﬁrst grade is John’s and the second is Mary’s. You are given that

P (" 4 ) + P (" 5 ) + P (" 6 ) = .3,

P (" 2 ) + P (" 5 ) + P (" 8 ) = .4,

P (" 5 ) + P (" 6 ) + P (" 8 ) = .1.

Adding the ﬁrst two equations and subtracting the third, we obtain the desired probability as

P (" 2 ) + P (" 4 ) + P (" 5 ) = .6.

17. The sample space for a sequence of m experiments is the set of m-tuples of S’s and F ’s, where S

represents a success and F a failure. The probability assigned to a sample point with k successes

and m − k failures is

n − 1

m−k

(a) Let k = 0 in the above expression.

(b) If m = n log 2, then

# !

1 − 1

log 2

lim

n"#

= lim

n"#

1 − 1

log 2

n"# (

lim

log 2

−1

= 1

19. The left-side is the sum of the probabilities of all elements in one of the three sets. For the right

side, if an outcome is in all three sets its probability is added three times, then subtracted three

times, then added once, so in the ﬁnal sum it is counted just once. An element that is in exactly

two sets is added twice, then subtracted once, and so it is counted correctly. Finally, an element in

exactly one set is counted only once by the right side.

21. 7/2 12

23. We have

1 − (1 − r) = 1 .

m(" n ) =

r(1 − r) n =

n=0

25. They call it a fallacy because if the subjects are thinking about probabilities they should realize

that

P (Linda is bank teller and in feminist movement) # P (Linda is bank teller).

One explanation is that the subjects are not thinking about probability as a measure of likelihood.

For another explanation see Exercise 52 of Section 4.1.

27.

number of male survivors at age x

100, 000

P x = P (male lives to age x) =

Q x = P (female lives to age x) =

number of female survivors at age x

100, 000

29. (Solution by Richard Beigel)

(a) In order to emerge from the interchange going west, the car must go straight at the ﬁrst point

of decision, then make 4n + 1 right turns, and ﬁnally go straight a second time. The probability

P (r) of this occurring is

(1 − r) 2 r 4n+1 = r(1 − r) 2

1 − r 4

1 + r 2

− 1

1 + r

P (r) =

n=0

if 0 # r < 1, but P (1) = 0. So P (1/2) = 2/15.

(b) Using standard methods from calculus, one can show that P (r) attains a maximum at the value

r = 1 +

1 +

−

" .346 .

At this value of r, P (r) " .15.

31. (a) Assuming that each student gives any given tire as an answer with probability 1/4, then prob-

ability that they both give the same answer is 1/4.

(b) In this case, they will both answer ‘right front’ with probability (.58) 2 , etc. Thus, the probability

that they both give the same answer is 39.8%.

SECTION 2.1

The problems in this section are all computer programs.

SECTION 2.2

1. (a) f(") = 1/8 on [2, 10]

(b) P ([a, b]) = b − a

3. (a) C = 1

log 5

" .621

(b)

P ([a, b]) = (.621) log(b/a)

(c)

P (x > 5) = log 2

log 5

" .431

P (x < 7) = log(7/2)

log 5

" .778

P (x 2 − 12x + 35 > 0) = log(25/7)

log 5

" .791 .

5. (a) 1 − e 1 " .632

(b) 1 − e 3 " .950

(d) 1

7. (a) 1/3, (b) 1/2, (c) 1/2, (d) 1/3

13. 2 log 2 − 1.

15. Yes.

SECTION 3.1

1. 24

3. 2 32

5. 9, 6.

5 5 .

1000 .

13. (a) 26 3 × 10 3

(b)

3n − 2

n 3

27 ,

’

(

× 26 3 × 10 3

’

(

× (2 n − 2)

3 n

15.

17. 1 − 12 · 11 · . . . · (12 − n + 1)

12 n

, if n # 12, and 1, if n > 12.

21. They are the same.

23. (a)

(b) She will get the best candidate if the second best candidate is in the ﬁrst half and the best

candidate is in the secon half. The probability that this happens is greater than 1/4.

n ,

SECTION 3.2

1. (a) 20

(b) .0064

(d) 1

(e) .0256

(f) 15

(g) 10

= 36

5. .998, .965, .729

11.

b(n, p, j)

b(n, p, j − 1) =

p j q n−j

j!(n − j)!

(n − j + 1)!(j − 1)!

j − 1

p j−1 q n−j+1

= (n − j + 1)

& 1 if and only if j # p(n + 1), and so j = [p(n + 1)] gives b(n, p, j) its largest

value. If p(n + 1) is an integer there will be two possible values of j, namely j = p(n + 1) and

j = p(n + 1) − 1.

(n − j + 1)

9. n = 15, r = 7

11. Eight pieces of each kind of pie.

13. The number of subsets of 2n objects of size j is

= 2n − i + 1

& 1 ’ i # n + 1

2 .

i − 1

Thus i = n makes

maximum.

15. .3443, .441, .181, .027.

’

(

’

(

ways of putting

b di"erent objects into the 2nd and then one way to put the remaining objects into the 3rd box.

Thus the total number of ways is

ways of putting a di"erent objects into the 1st box, and then

n−a

n − a

a!b!(n − a − b)! .

19. (a)

= 7.23 × 10 −8 .

(b)

= .044.

(c)

= .315.

21. 3(2 5 ) − 3 = 93 (We subtract 3 because the three pure colors are each counted twice.)

23. To make the boxes, you need n + 1 bars, 2 on the ends and n − 1 for the divisions. The n − 1 bars

and the r objects occupy n−1 + r places. You can choose any n−1 of these n−1 + r places for the

bars and use the remaining r places for the objects. Thus the number of ways this can be done is

n − 1 + r

n − 1

n − 1 + r

But

17. There are

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