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## P r o b a b i l i t Y

### SAEED GHAHRAMANI

Western New England College

Upper Saddle River, New Jersey 07458

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## C ontents

### 

1 Axioms of Probability 1

1.2 Sample Space and Events 1 1.4 Basic Theorems 2

1.7 Random Selection of Points from Intervals 7 Review Problems 9

### 

2 Combinatorial Methods 13

2.2 Counting Principle 13 2.3 Permutations 16 2.4 Combinations 18 2.5 Stirling’ Formula 31

Review Problems 31

### 

3 Conditional Probability and Independence 35 3.1 Conditional Probability 35

3.2 Law of Multiplication 39 3.3 Law of Total Probability 41 3.4 Bayes’ Formula 46

3.5 Independence 48

3.6 Applications of Probability to Genetics 56 Review Problems 59

### 

4 Distribution Functions and

Discrete Random Variables 63

4.2 Distribution Functions 63 4.3 Discrete Random Variables 66

4.4 Expectations of Discrete Random Variables 71

4.5 Variances and Moments of Discrete Random Variables 77 4.6 Standardized Random Variables 83

Review Problems 83

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### 

5 Special Discrete Distributions 87 5.1 Bernoulli and Binomial Random Variables 87

5.2 Poisson Random Variable 94 5.3 Other Discrete Random Variables 99

Review Problems 106

### 

6 Continuous Random Variables 111

6.1 Probability Density Functions 111

6.2 Density Function of a Function of a Random Variable 113 6.3 Expectations and Variances 116

Review Problems 123

### 

7 Special Continuous Distributions 126

7.1 Uniform Random Variable 126 7.2 Normal Random Variable 131 7.3 Exponential Random Variables 139 7.4 Gamma Distribution 144

7.5 Beta Distribution 147

7.6 Survival Analysis and Hazard Function 152 Review Problems 153

### 

8 Bivariate Distributions 157

8.1 Joint Distribution of Two Random Variables 157 8.2 Independent Random Variables 166

8.3 Conditional Distributions 174

8.4 Transformations of Two Random Variables 183 Review Problems 191

### 

9 Multivariate Distributions 200

9.1 Joint Distribution of n > 2 Random Variables 200 9.2 Order Statistics 210

9.3 Multinomial Distributions 215 Review Problems 218

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Contents v

### 

10 More Expectations and Variances 222

10.1 Expected Values of Sums of Random Variables 222 10.2 Covariance 227

10.3 Correlation 237

10.4 Conditioning on Random Variables 239 10.5 Bivariate Normal Distribution 251

Review Problems 254

### 

11 Sums of Independent Random

Variables and Limit Theorems 261

11.1 Moment-Generating Functions 261

11.2 Sums of Independent Random Variables 269 11.3 Markov and Chebyshev Inequalities 274 11.4 Laws of Large Numbers 278

11.5 Central Limit Theorem 282 Review Problems 287

### 

12 Stochastic Processes 291

12.2 More on Poisson Processes 291 12.3 Markov Chains 296

12.4 Continuous-Time Markov Chains 315 12.5 Brownian Motion 326

Review Problems 331

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## A xioms of P robability

1.2 SAMPLE SPACE AND EVENTS

1. For 1 ≤ i, j ≤ 3, by (i, j) we mean that Vann’s card number is i, and Paul’s card number is j. Clearly, A=

(1, 2), (1, 3), (2, 3)

and B =

(2, 1), (3, 1), (3, 2) . (a) Since A∩ B = ∅, the events A and B are mutually exclusive.

(b) None of (1, 1), (2, 2), (3, 3) belongs to A∪ B. Hence A ∪ B not being the sample space shows that A and B are not complements of one another.

2. S= {RRR, RRB, RBR, RBB, BRR, BRB, BBR, BBB}.

3. {x : 0 < x < 20}; {1, 2, 3, . . . , 19}.

5. EF: One 1 and one even.

EcF: One 1 and one odd.

EcFc: Both even or both belong to{3, 5}.

6. S= {QQ, QN, QP, QD, DN, DP, NP, NN, P P }. (a) {QP }; (b) {DN, DP, NN}; (c) ∅.

7. S=

x: 7 ≤ x ≤ 916

;

x: 7 ≤ x ≤ 714

∪

x: 734 ≤ x ≤ 814

∪

x: 834 ≤ x ≤ 916 . 8. E∪ F ∪ G = G: If E or F occurs, then G occurs.

EF G= G: If G occurs, then E and F occur.

9. For 1 ≤ i ≤ 3, 1 ≤ j ≤ 3, by aibj we mean passenger a gets off at hotel i and passenger b gets off at hotel j . The answers are{aibj: 1 ≤ i ≤ 3, 1 ≤ j ≤ 3} and {a1b1, a2b2, a3b3}, respectively.

10. (a) (E∪ F )(F ∪ G) = (F ∪ E)(F ∪ G) = F ∪ EG.

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(b) Using part (a), we have

(E∪ F )(Ec∪ F )(E ∪ Fc)= (F ∪ EEc)(E∪ Fc)= F (E ∪ Fc)= FE ∪ FFc= FE.

11. (a) ABcCc; (b) A∪ B ∪ C; (c) AcBcCc; (d) ABCc∪ ABcC∪ AcBC; (e) ABcCc∪ AcBcC∪ AcBCc; (f) (A− B) ∪ (B − A) = (A ∪ B) − AB.

12. If B= ∅, the relation is obvious. If the relation is true for every event A, then it is true for S, the sample space, as well. Thus

S= (B ∩ Sc)∪ (Bc∩ S) = ∅ ∪ Bc = Bc, showing that B = ∅.

13. Parts (a) and (d) are obviously true; part (c) is true by DeMorgan’s law; part (b) is false: throw a four-sided die; let F = {1, 2, 3}, G = {2, 3, 4}, E = {1, 4}.

14. (a)

n=1An; (b)37 n=1An. 15. Straightforward.

16. Straightforward.

17. Straightforward.

18. Let a1, a2, and a3be the ﬁrst, the second, and the third volumes of the dictionary. Let a4, a5, a6, and a7be the remaining books. Let A= {a1, a2, . . . , a7}; the answers are

S=

x1x2x3x4x5x6x7: xi ∈ A, 1 ≤ i ≤ 7, and xi = xjif i = j

and 

x1x2x3x4x5x6x7∈ S : xixi+1xi+2 = a1a2a3for some i, 1≤ i ≤ 5 , respectively.

19. m=1n=mAn.

20. Let B1= A1, B2= A2− A1, B3= A3− (A1∪ A2), . . . , Bn= An−n−1

i=1 Ai, . . . .

1.4 BASIC THEOREMS

1. No; P (sum 11)= 2/36 while P (sum 12) = 1/36.

2. 0.33+ 0.07 = 0.40.

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Section 1.4 Basic Theorems 3 3. Let E be the event that an earthquake will damage the structure next year. Let H be the event that a hurricane will damage the structure next year. We are given that P (E)= 0.015, P (H )= 0.025, and P (EH) = 0.0073. Since

P (E∪ H) = P (E) + P (H ) − P (EH) = 0.015 + 0.025 − 0.0073 = 0.0327, the probability that next year the structure will be damaged by an earthquake and/or a hurricane is 0.0327. The probability that it is not damaged by any of the two natural disasters is 0.9673.

4. Let A be the event of a randomly selected driver having an accident during the next 12 months.

Let B be the event that the person is male. By Theorem 1.7, the desired probability is P (A)= P (AB) + P (ABc)= 0.12 + 0.06 = 0.18.

5. Let A be the event that a randomly selected investor invests in traditional annuities. Let B be the event that he or she invests in the stock market. Then P (A) = 0.75, P (B) = 0.45, and P (A∪ B) = 0.85. Since,

P (AB)= P (A) + P (B) − P (A ∪ B) = 0.75 + 0.45 − 0.85 = 0.35, 35% invest in both stock market and traditional annuities.

6. The probability that the ﬁrst horse wins is 2/7. The probability that the second horse wins is 3/10. Since the events that the ﬁrst horse wins and the second horse wins are mutually exclusive, the probability that either the ﬁrst horse or the second horse will win is

2 7 + 3

10 = 41 70.

7. In point of fact Rockford was right the ﬁrst time. The reporter is assuming that both autopsies are performed by a given doctor. The probability that both autopsies are performed by the same doctor–whichever doctor it may be–is 1/2. Let AB represent the case in which Dr. A performs the ﬁrst autopsy and Dr. B performs the second autopsy, with similar representations for other cases. Then the sample space is S = {AA, AB, BA, BB}. The event that both autopsies are performed by the same doctor is{AA, BB}. Clearly, the probability of this event is 2/4=1/2.

8. Let m be the probability that Marty will be hired. Then m+ (m + 0.2) + m = 1 which gives m= 8/30; so the answer is 8/30 + 2/10 = 7/15.

9. Let s be the probability that the patient selected at random suffers from schizophrenia. Then s+ s/3 + s/2 + s/10 = 1 which gives s = 15/29.

10. P (A∪ B) ≤ 1 implies that P (A) + P (B) − P (AB) ≤ 1.

11. (a) 2/52+ 2/52 = 1/13; (b) 12/52 + 26/52 − 6/53 = 8/13; (c) 1 − (16/52) = 9/13.

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12. (a) False; toss a die and let A= {1, 2}, B = {2, 3}, and C = {1, 3}.

(b) False; toss a die and let A= {1, 2, 3, 4}, B = {1, 2, 3, 4, 5}, C = {1, 2, 3, 4, 5, 6}.

13. A simple Venn diagram shows that the answers are 65% and 10%, respectively.

14. Applying Theorem 1.6 twice, we have

P (A∪ B ∪ C) = P (A ∪ B) + P (C) − P

(A∪ B)C

= P (A) + P (B) − P (AB) + P (C) − P (AC ∪ BC)

= P (A) + P (B) − P (AB) + P (C) − P (AC) − P (BC) + P (ABC)

= P (A) + P (B) + P (C) − P (AB) − P (AC) − P (BC) + P (ABC).

15. Using Theorem 1.5, we have that the desired probability is P (AB− ABC) + P (AC − ABC) + P (BC − ABC)

= P (AB) − P (ABC) + P (AC) − P (ABC) + P (BC) − P (ABC)

= P (AB) + P (AC) + P (BC) − 3P (ABC).

16. 7/11.

17. ni=1pij.

18. Let M and F denote the events that the randomly selected student earned an A on the midterm exam and an A on the ﬁnal exam, respectively. Then

P (MF )= P (M) + P (F ) − P (M ∪ F ), where P (M)= 17/33, P (F ) = 14/33, and by DeMorgan’s law,

P (M∪ F ) = 1 − P (McFc)= 1 −11 33 =22

33. Therefore,

P (MF )= 17 33+14

33−22 33 = 3

11.

19. A Venn diagram shows that the answers are 1/8, 5/24, and 5/24, respectively.

20. The equation has real roots if and only if b2≥ 4c. From the 36 possible outcomes for (b, c), in the following 19 cases we have that b2≥ 4c: (2, 1), (3, 1), (3, 2), (4, 1), . . . , (4, 4), (5, 1), . . ., (5, 6), (6, 1), . . . , (6, 6). Therefore, the answer is 19/36.

21. The only prime divisors of 63 are 3 and 7. Thus the number selected is relatively prime to 63 if and only if it is neither divisible by 3 nor by 7. Let A and B be the events that the outcome

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Section 1.4 Basic Theorems 5 is divisible by 3 and 7, respectively. The desired quantity is

P (AcBc)= 1 − P (A ∪ B) = 1 − P (A) − P (B) + P (AB)

= 1 −21 63 − 9

63+ 3 63 = 4

7.

22. Let T and F be the events that the number selected is divisible by 3 and 5, respectively.

(a) The desired quantity is the probability of the event T Fc: P (T Fc)= P (T ) − P (T F ) = 333

1000 − 66

1000 = 267 1000. (b) The desired quantity is the probability of the event TcFc:

P (TcFc)= 1 − P (T ∪ F ) = 1 − P (T ) − P (F ) + P (T F )

= 1 − 333

1000− 200

1000 + 66

1000 = 533 1000.

23. (Draw a Venn diagram.) From the data we have that 55% passed all three, 5% passed calculus and physics but not chemistry, and 20% passed calculus and chemistry but not physics. So at least (55+ 5 + 20)% = 80% must have passed calculus. This number is greater than the given 78% for all of the students who passed calculus. Therefore, the data is incorrect.

24. By symmetry the answer is 1/4.

25. Let A, B, and C be the events that the number selected is divisible by 4, 5, and 7, respectively.

We are interested in P (ABcCc). Now ABcCc = A − A(B ∪ C) and A(B ∪ C) ⊆ A. So by Theorem 1.5,

P (ABcCc)= P (A) − P

A(B∪ C)

= P (A) − P (AB ∪ AC)

= P (A) − P (AB) − P (AC) + P (ABC)

= 250

1000 − 50

1000− 35

1000+ 7

1000 = 172 1000. 26. A Venn diagram shows that the answer is 0.36.

27. Let A be the event that the ﬁrst number selected is greater than the second; let B be the event that the second number selected is greater than the ﬁrst; and let C be the event that the two numbers selected are equal. Then P (A)+ P (B) + P (C) = 1, P (A) = P (B), and P (C)= 1/100. These give P (A) = 99/200.

28. Let B1 = A1, and for n ≥ 2, Bn = An −n−1

i=1 Ai. Then {B1, B2, . . .} is a sequence of mutually exclusive events and

i=1Ai =

i=1Bi.Hence

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P

n=1

An

= P

n=1

Bn

= n=1

P (Bn)

n=1

P (An), since Bn⊆ An, n≥ 1.

29. By Boole’s inequality (Exercise 28), P



n=1

An

= 1 − P

n=1

Acn

≥ 1 −

n=1

P (Acn).

30. She is wrong! Consider the next 50 ﬂights. For 1 ≤ i ≤ 50, let Ai be the event that the ith mission will be completed without mishap. Then50

i=1Ai is the event that all of the next 50 missions will be completed successfully. We will show that P 50

i=1Ai

>0. This proves that Mia is wrong. Note that the probability of the simultaneous occurrence of any number of Aci ’s is nonzero. Furthermore, consider any set E consisting of n (n≤ 50) of the Aci’s. It is reasonable to assume that the probability of the simultaneous occurrence of the events of E is strictly less than the probability of the simultaneous occurrence of the events of any subset of E. Using these facts, it is straightforward to conclude from the inclusion–exclusion principle that,

P

 50

i=1

Aci

<

50 i=1

P (Aci)= 50

i=1

1 50 = 1.

Thus, by DeMorgan’s law,

P

 50

i=1

Ai

= 1 − P 50

i=1

Aci

>1− 1 = 0.

31. Qsatisﬁes Axioms 1 and 2, but not necessarily Axiom 3. So it is not, in general, a probability on S. Let S = {1, 2, 3, }. Let P

{1}

= P {2}

= P {3}

= 1/3. Then Q {1}

= Q {2}

= 1/9, whereas Q

{1, 2}

= P {1, 2}2

= 4/9. Therefore, Q

{1, 2, }

= Q {1}

+ Q {2}

.

Ris not a probability on S because it does not satisfy Axiom 2; that is, R(S)= 1.

32. Let BRB mean that a blue hat is placed on the ﬁrst player’s head, a red hat on the second player’s head, and a blue hat on the third player’s head, with similar representations for other cases. The sample space is

S= {BBB, BRB, BBR, BRR, RRR, RRB, RBR, RBB}.

This shows that the probability that two of the players will have hats of the same color and the third player’s hat will be of the opposite color is 6/8= 3/4. The following improvement,

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Section 1.7 Random Selection of Points from Intervals 7 based on this observation, explained by Sara Robinson in Tuesday, April 10, 2001 issue of the New York Times, is due to Professor Elwyn Berlekamp of the University of California at Berkeley.

Three-fourths of the time, two of the players will have hats of the same color and the third player’s hat will be the opposite color. The group can win every time this happens by using the following strategy: Once the game starts, each player looks at the other two players’ hats. If the two hats are different colors, he [or she] passes.

If they are the same color, the player guesses his [or her] own hat is the opposite color. This way, every time the hat colors are distributed two and one, one player will guess correctly and the others will pass, and the group will win the game. When all the hats are the same color, however, all three players will guess incorrectly and the group will lose.

1.7 RANDOM SELECTION OF POINTS FROM INTERVALS 1. 30− 10

30− 0 = 2 3. 2. 0.0635− 0.04

0.12− 0.04 = 0.294.

3. (a) False; in the experiment of choosing a point at random from the interval (0, 1), let A= (0, 1) − {1/2}. A is not the sample space but P (A) = 1.

(b) False; in the same experiment P {1/2}

= 0 while {12} = ∅.

4. P (A∪ B) ≥ P (A) = 1, so P (A ∪ B) = 1. This gives

P (AB)= P (A) + P (B) − P (A ∪ B) = 1 + 1 − 1 = 1.

P

{1, 2, . . . , 1999}

= 1999

i=1

P {i}

= 1999

i=1

0= 0.

6. For i = 0, 1, 2, . . . , 9, the probability that i appears as the ﬁrst digit of the decimal represen- tation of the selected point is the probability that the point falls into the interval

i

10,i+ 1 10

. Therefore, it equals

i+ 1 10 − i

10 1− 0 = 1

10.

This shows that all numerals are equally likely to appear as the ﬁrst digit of the decimal representation of the selected point.

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7. No, it is not. Let S = {w1, w2, . . .}. Suppose that for some p > 0, P {wi}

= p, i = 1, 2, . . .. Then, by Axioms 2 and 3,

i=1p= 1. This is impossible.

8. Use induction. For n = 1, the theorem is trivial. Exercise 4 proves the theorem for n = 2.

Suppose that the theorem is true for n. We show it for n+ 1,

P (A1A2· · · AnAn+1)= P (A1A2· · · An)+ P (An+1)− P (A1A2· · · An∪ An+1)

= 1 + 1 − 1 = 1,

where P (A1A2· · · An)= 1 is true by the induction hypothesis, and P (A1A2· · · An∪ An+1)≥ P (An+1)= 1, implies that P (A1A2· · · An∪ An+1)= 1.

9. (a) Clearly,1 2 ∈

n=1

1 2− 1

2n, 1 2+ 1

2n

. If x n=1

1 2− 1

2n, 1 2+ 1

2n

, then, for all n≥ 1, 1

2− 1

2n < x < 1 2 + 1

2n. Letting n→ ∞, we obtain 1/2 ≤ x ≤ 1/2; thus x = 1/2.

(b) Let Anbe the event that the point selected at random is in

1 2 − 1

2n,1 2+ 1

2n ; then

A1⊇ A2⊇ A3⊇ · · · ⊇ An ⊇ An+1 ⊇ · · · . Since P (An)= 1

n, by the continuity property of the probability function, P

{1/2}

= lim

n→∞P (An)= 0.

10. The set of rational numbers is countable. Let Q = {r1, r2, r3, . . .} be the set of rational numbers in (0, 1). Then

P (Q)= P

{r1, r2, r3, . . .}

=

i=1

P {ri}

= 0.

Let I be the set of irrational numbers in (0, 1); then

P (I)= P (Qc)= 1 − P (Q) = 1.

11. For i = 0, 1, 2, . . . , 9, the probability that i appears as the nth digit of the decimal represen- tation of the selected point is the probability that the point falls into the following subset of (0, 1):

10n−1 −1 m=0

10m+ i

10n , 10m+ i + 1 10n

.

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Chapter 1 Review Problems 9 Since the intervals in this union are mutually exclusive, the probability that the point falls into this subset is

10 n−1−1 m=0

10m+ i + 1

10n10m+ i 10n

1− 0 = 10n−1· 1 10n = 1

10.

This shows that all numerals are equally likely to appear as the nth digit of the decimal representation of the selected point.

12. P (Bm)≤

n=mP (An).Since

n=1P (An)converges,

m→∞lim P (Bm)≤ lim

m→∞

n=m

P (An)= 0.

This gives limm→∞P (Bm)= 0. Therefore,

B1⊇ B2⊇ B3⊇ · · · ⊇ Bm⊇ Bm+1 ⊇ · · · implies that

P



m=1

n=m

An

= P

m=1

Bm

= limm→∞P (Bm)= 0.

13. In the experiment of choosing a random point from (0, 1), let Et = (0, 1)−{t}, for 0 < t < 1.

Then P (Et)= 1 for all t, while

P

t∈(0,1)

Et

= P (∅) = 0.

14. Clearly rn∈ (αn, βn). By the geometric series theorem,

n=1

n− αn)= n=1

ε 2n+1 = ε

1 4 1−1

2

= ε 2 < ε.

REVIEW PROBLEMS FOR CHAPTER 1

1. 3.25− 2

4.3− 2 = 0.54.

2. We have that S=

∅, {1} ,

∅, {2} ,

∅, {1, 2} ,

{1}, {2} ,

{1}, {1, 2} ,

{2}, {1, 2}

.

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The desired events are (a) 

∅, {1} ,

∅, {2} ,

∅, {1, 2} ,

{1}, {2}

; (b) 

∅, {1, 2} ,

{1}, {2}

; (c) 

∅, {1} ,

∅, {2} ,

∅, {1, 2} ,

{1}, {1, 2} ,

{2}, {1, 2}

.

3. Since A⊆ B, we have that Bc⊆ Ac.This implies that (a) is false but (b) is true.

4. In the experiment of tossing a die let A = {1, 3, 5} and B = {5}; then both (a) and (b) are false.

5. We may deﬁne a sample space S as follows.

S=

x1x2· · · xn: n ≥ 1, xi ∈ {H,T}; xi = xi+1, 1≤ i ≤ n − 2; xn−1= xn

. 6. A venn diagram shows that 18 are neither male nor for surgery.

7. We have that ABC⊆ BC, so P (ABC) ≤ P (BC) and hence P (BC) − P (ABC) ≥ 0. This and the following give the result.

P (A∪ B ∪ C) = P (A) + P (B) + P (C) −

P (AB)+ P (AC) + P (BC) − P (ABC)

≤ P (A) + P (B) + P (C).

8. If P (AB)= P (AC) = P (BC) = 0, then P (ABC) = 0 since ABC ⊆ AB. These imply that P (A∪ B ∪ C) = P (A) + P (B) + P (C) − P (AB) − P (AC) − P (BC) + P (ABC)

= P (A) + P (B) + P (C).

Now suppose that

P (A∪ B ∪ C) = P (A) + P (B) + P (C).

This relation implies that

P (AB)+ P (BC) +

P (AC)− P (ABC)

= 0. (1)

Since P (AC)− P (ABC) ≥ 0 we have that the sum of three nonnegative quantities is 0; so each of them is 0. That is,

P (AB)= 0, P (BC) = 0, P (AC) = P (ABC). (2) Now rewriting (1) as

P (AB)+ P (AC) +

P (BC)− P (ABC)

= 0, the same argument implies that

P (AB)= 0, P (AC) = 0, P (BC) = P (ABC). (3) Comparing (2) and (3) we have

P (AB)= P (AC) = P (BC) = 0.

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Chapter 1 Review Problems 11 9. Let W be the event that a randomly selected person from this community drinks or serves white wine. Let R be the event that she or he drinks or serves red wine. We are given that P (W )= 0.40, P (R) = 0.50, and P (W ∪ R) = 0.70. Since

P (W R)= P (W) + P (R) − P (W ∪ R) = 0.40 + 0.50 − 0.70 = 0.20, 20% percent drink or serve both red and white wine.

10. No, it is not right. The probability that the second student chooses the tire the ﬁrst student chose is 1/4.

11. By De Morgan’s second law, P (AcBc)= 1 − P

(AcBc)c

= 1 − P (A ∪ B) = 1 − P (A) − P (B) + P (AB).

12. By Theorem 1.5 and the fact that A− B and B − A are mutually exclusive, P

(A− B) ∪ (B − A)

= P (A − B) + P (B − A) = P (A − AB) + P (B − AB)

= P (A) − P (AB) + P (B) − P (AB) = P (A) + P (B) − 2P (AB).

13. Denote a box of books by ai, if it is received from publisher i, i = 1, 2, 3. The sample space is

S=

x1x2x3x4x5x6: two of the xi’s are a1, two of them are a2, and the remaining two are a3 . The desired event is E=

x1x2x3x4x5x6∈ S : x5= x6

.

14. Let E, F , G, and H be the events that the next baby born in this town has blood type O, A, B, and AB, respectively. Then

P (E)= P (F ), P (G) = 1

10P (F ), P (G)= 2P (H).

These imply

P (E)= P (F ) = 20P (H).

Therefore, from

P (E)+ P (F ) + P (G) + P (H ) = 1, we get

20P (H )+ 20P (H ) + 2P (H) + P (H ) = 1, which gives P (H )= 1/43.

15. Let F , S, and N be the events that the number selected is divisible by 4, 7, and 9, respectively.

We are interested in P (FcScNc)which is equal to 1− P (F ∪ S ∪ N) by DeMorgan’s law.

(18)

Now

P (F ∪ S ∪ N) = P (F ) + P (S) + P (N) − P (FS) − P (F N) − P (SN) + P (FSN)

= 250

1000 + 142

1000+ 111

1000 − 35

1000 − 27

1000− 15

1000+ 3

1000 = 0.429.

So the desired probability is 0.571.

16. The number is relatively prime to 150 if is not divisible by 2, 3, or 5. Let A, B, and C be the events that the number selected is divisible by 2, 3, and 5, respectively. We are interested in P (AcBcCc)= 1 − P (A ∪ B ∪ C). Now

P (A∪ B ∪ C) = P (A) + P (B) + P (C) − P (AB) − P (AC) − P (BC) + P (ABC)

= 75 150+ 50

150 + 30 150− 25

150− 15 150 − 10

150 + 5 150 = 11

15. Therefore, the answer is 1−11

15 = 4 15.

17. (a) UicDci; (b) U1U2· · · Un; (c) (U1cD1c)∪ (U2cDc2)∪ · · · ∪ (UncDnc); (d) (U1D2U3cDc3)∪ (U1U2cD2cD3)∪ (D1U2U3cDc3)∪ (D1U2cD2cU3)

∪(D1cU1cD2U3)∪ (D1cU1cU2D2)∪ (D1cU1cD2cU2cD3cU3c); (e) D1cD2c· · · Dcn.

18. 199− 96 199− 0 = 103

199.

19. We must have b2<4ac. There are 6× 6 × 6 = 216 possible outcomes for a, b, and c. For cases in which a < c, a > c, and a= c, it can be checked that there are 73, 73, and 27 cases in which b2<4ac, respectively. Therefore, the desired probability is

73+ 73 + 27

216 = 173

216.

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## C ombinatorial M ethods

2.2 COUNTING PRINCIPLES

1. The total number of six-digit numbers is 9×10×10×10×10×10 = 9×105since the ﬁrst digit cannot be 0. The number of six-digit numbers without the digit ﬁve is 8× 9 × 9 × 9 × 9 × 9 = 8× 95.Hence there are 9× 105− 8 × 95= 427, 608 six-digit numbers that contain the digit ﬁve.

2. (a) 55= 3125. (b) 53= 125.

3. There are 26× 26 × 26 = 17, 576 distinct sets of initials. Hence in any town with more than 17,576 inhabitants, there are at least two persons with the same initials. The answer to the question is therefore yes.

4. 415 = 1, 073, 741, 824.

5. 2 223 = 1

222 ≈ 0.00000024.

6. (a) 525= 380, 204, 032. (b) 52 × 51 × 50 × 49 × 48 = 311, 875, 200.

7. 6/36= 1/6.

8. (a) 4× 3 × 2 × 2 12× 8 × 8 × 4 = 1

64. (b) 1− 8× 5 × 6 × 2 12× 8 × 8 × 4 = 27

32. 9. 1

415 ≈ 0.00000000093.

10. 26× 25 × 24 × 10 × 9 × 8 = 11, 232, 000.

11. There are 263× 102= 1, 757, 600 such codes; so the answer is positive.

12. 2nm.

13. (2+ 1)(3 + 1)(2 + 1) = 36. (See the solution to Exercise 24.)

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14. There are (26− 1)23= 504 possible sandwiches. So the claim is true.

15. (a) 54= 625. (b) 54− 5 × 4 × 3 × 2 = 505.

16. 212 = 4096.

17. 1−48× 48 × 48 × 48

52× 52 × 52 × 52 = 0.274.

18. 10× 9 × 8 × 7 = 5040. (a) 9× 9 × 8 × 7 = 4536; (b) 5040 − 1 × 1 × 8 × 7 = 4984.

19. 1−(N− 1)n Nn .

20. By Example 2.6, the probability is 0.507 that among Jenny and the next 22 people she meets randomly there are two with the same birthday. However, it is quite possible that one of these two persons is not Jenny. Let n be the minimum number of people Jenny must meet so that the chances are better than even that someone shares her birthday. To ﬁnd n, let A denote the event that among the next n people Jenny meets randomly someone’s birthday is the same as Jenny’s. We have

P (A)= 1 − P (Ac)= 1 −364n 365n. To have P (A) > 1/2, we must ﬁnd the smallest n for which

1−364n 365n > 1

2,

or 364n

365n <1 2. This gives

n >

log1 2 log364

365

= 252.652.

Therefore, for the desired probability to be greater than 0.5, n must be 253. To some this might seem counterintuitive.

21. Draw a tree diagram for the situation in which the salesperson goes from I to B ﬁrst. In this situation, you will ﬁnd that in 7 out of 23 cases, she will end up staying at island I . By symmetry, if she goes from I to H , D, or F ﬁrst, in each of these situations in 7 out of 23 cases she will end up staying at island I . So there are 4× 23 = 92 cases altogether and in 4×7 = 28 of them the salesperson will end up staying at island I. Since 28/92 = 0.3043, the answer is 30.43%. Note that the probability that the salesperson will end up staying at island I is not 0.3043 because not all of the cases are equiprobable.

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Section 2.2 Counting Principle 15 22. He is at 0 ﬁrst, next he goes to 1 or−1. If at 1, then he goes to 0 or 2. If at −1, then he goes to 0 or−2, and so on. Draw a tree diagram. You will ﬁnd that after walking 4 blocks, he is at one of the points 4, 2, 0,−2, or −4. There are 16 possible cases altogether. Of these 6 end up at 0, none at 1, and none at−1. Therefore, the answer to (a) is 6/16 and the answer to (b) is 0.

23. We can think of a number less than 1,000,000 as a six-digit number by allowing it to start with 0 or 0’s. With this convention, it should be clear that there are 96such numbers without the digit ﬁve. Hence the desired probability is 1− (96/106)= 0.469.

24. Divisors of N are of the form pe11pe22· · · pkek,where ei = 0, 1, 2, . . . , ni, 1≤ i ≤ k. Therefore, the answer is (n1+ 1)(n2+ 1) · · · (nk+ 1).

25. There are 64possibilities altogether. In 54of these possibilities there is no 3. In 53of these possibilities only the ﬁrst die lands 3. In 53of these possibilities only the second die lands 3, and so on. Therefore, the answer is

54+ 4 × 53

64 = 0.868.

26. Any subset of the set {salami, turkey, bologna, corned beef, ham, Swiss cheese, American cheese} except the empty set can form a reasonable sandwich. There are 27− 1 possibilities.

To every sandwich a subset of the set {lettuce, tomato, mayonnaise} can also be added. Since there are 3 possibilities for bread, the ﬁnal answer is (27− 1) × 23 × 3 = 3048 and the advertisement is true.

27. 11× 10 × 9 × 8 × 7 × 6 × 5 × 4

118 = 0.031.

28. For i = 1, 2, 3, let Ai be the event that no one departs at stop i. The desired quantity is P (Ac1Ac2Ac3)= 1 − P (A1∪ A2∪ A3).Now

P (A1∪ A2∪ A3)= P (A1)+ P (A2)+ P (A3)

− P (A1A2)− P (A1A3)− P (A2A3)+ P (A1A2A3)

= 26 36+26

36 +26 36− 1

36 − 1 36− 1

36 + 0 = 7 27. Therefore, the desired probability is 1− (7/27) = 20/27.

29. For 0≤ i ≤ 9, the sum of the ﬁrst two digits is i in (i + 1) ways. Therefore, there are (i + 1)2 numbers in the given set with the sum of the ﬁrst two digits equal to the sum of the last two digits and equal to i. For i= 10, there are 92numbers in the given set with the sum of the ﬁrst two digits equal to the sum of the last two digits and equal to 10. For i = 11, the corresponding numbers are 82and so on. Therefore, there are altogether

12+ 22+ · · · + 102+ 92+ 82+ · · · + 12= 670

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numbers with the desired probability and hence the answer is 670/104= 0.067.

30. Let A be the event that the number selected contains at least one 0. Let B be the event that it contains at least one 1 and C be the event that it contains at least one 2. The desired quantity is P (ABC)= 1 − P (Ac∪ Bc∪ Cc), where

P (Ac∪ Bc∪ Cc)= P (Ac)+ P (Bc)+ P (Cc)

− P (AcBc)− P (AcCc)− P (BcCc)+ P (AcBcCc)

= 9r

9× 10r−1 + 8× 9r−1

9× 10r−1 + 8× 9r−1

9× 10r−1 − 8r

9× 10r−1 − 8r 9× 10r−1

− 7× 8r−1

9× 10r−1 + 7r 9× 10r−1.

2.3 PERMUTATIONS

1. The answer is 1 4! = 1

24 ≈ 0.0417.

2. 3! = 6.

3. 8!

3! 5! = 56.

4. The probability that John will arrive right after Jim is 7!/8! (consider Jim and John as one arrival). Therefore, the answer is 1− (7!/8!) = 0.875.

Another Solution: If Jim is the last person, John will not arrive after Jim. Therefore, the remaining seven can arrive in 7! ways. If Jim is not the last person, the total number of possibilities in which John will not arrive right after Jim is 7× 6 × 6!. So the answer is

7! + 7 × 6 × 6!

8! = 0.875.

5. (a) 312= 531, 441. (b) 12!

6! 6! = 924. (c) 12!

3! 4! 5! = 27, 720.

6. 6P2= 30.

7. 20!

4! 3! 5! 8! = 3, 491, 888, 400.

8. (5× 4 × 7) × (4 × 3 × 6) × (3 × 2 × 5)

3! = 50, 400.

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Section 2.3 Permutations 17 9. There are 8! schedule possibilities. By symmetry, in 8!/2 of them Dr. Richman’s lecture precedes Dr. Chollet’s and in 8!/2 ways Dr. Richman’s lecture precedes Dr. Chollet’s. So the answer is 8!/2 = 20, 160.

10. 11!

3! 2! 3! 3! = 92, 400.

11. 1− (6!/66)= 0.985.

12. (a) 11!

4! 4! 2! = 34, 650.

(b) Treating all P ’s as one entity, the answer is 10!

4! 4! = 6300.

(c) Treating all I ’s as one entity, the answer is 8!

4! 2! = 840.

(d) Treating all P ’s as one entity, and all I ’s as another entity, the answer is 7!

4! = 210.

(e) By (a) and (c), The answer is 840/34650= 0.024.

13.  8! 2! 3! 3!



68= 0.000333.

14.  9! 3! 3! 3!



529= 6.043 × 10−13.

15. m! (n+ m)!.

16. Each girl and each boy has the same chance of occupying the 13th chair. So the answer is 12/20= 0.6. This can also be seen from 12× 19!

20! = 12 20 = 0.6.

17. 12!

1212 = 0.000054.

18. Look at the ﬁve math books as one entity. The answer is 5! × 18!

22! = 0.00068.

19. 1−9P7

97 = 0.962.

20. 2× 5! × 5!

10! = 0.0079.

21. n!/nn.

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22. 1− (6!/66)= 0.985.

23. Suppose that A and B are not on speaking terms. 134P4committees can be formed in which neither A serves nor B; 4×134P3committees can be formed in which A serves and B does not.

The same numbers of committees can be formed in which B serves and A does not. Therefore, the answer is134P4+ 2(4 ×134P3)= 326, 998, 056.

24. (a) mn. (b) mPn. (c) n!.

25. 3· 8! 2! 3! 2! 1!



68= 0.003.

26. (a) 20!

39× 37 × 35 × · · · × 5 × 3 × 1 = 7.61 × 10−6.

(b) 1

39× 37 × 35 × · · · × 5 × 3 × 1 = 3.13 × 10−24.

27. Thirty people can sit in 30! ways at a round table. But for each way, if they rotate 30 times (everybody move one chair to the left at a time) no new situations will be created. Thus in 30!/30 = 29! ways 15 married couples can sit at a round table. Think of each married couple as one entity and note that in 15!/15 = 14! ways 15 such entities can sit at a round table. We have that the 15 couples can sit at a round table in (2!)15· 14! different ways because if the couples of each entity change positions between themselves, a new situation will be created.

So the desired probability is

14!(2!)15

29! = 3.23 × 10−16. The answer to the second part is

24!(2!)5

29! = 2.25 × 10−6.

28. In 13! ways the balls can be drawn one after another. The number of those in which the ﬁrst white appears in the second or in the fourth or in the sixth or in the eighth draw is calculated as follows. (These are Jack’s turns.)

8× 5 × 11! + 8 × 7 × 6 × 5 × 9! + 8 × 7 × 6 × 5 × 4 × 5 × 7!

+ 8 × 7 × 6 × 5 × 4 × 3 × 2 × 5 × 5! = 2, 399, 846, 400.

Therefore, the answer is 2, 399, 846, 400/13! = 0.385.

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Section 2.4 Combinations 19 2.4 COMBINATIONS

1.

20 6



= 38, 760.

2.

100 i=51

100 i



= 583, 379, 627, 841, 332, 604, 080, 945, 354, 060 ≈ 5.8 × 1029.

3.

20 6

25 6



= 6, 864, 396, 000.

4.

12 3

40 2



52 5

 = 0.066.

5.

N− 1 n− 1

N n



= n N. 6.

5 3

2 2



= 10.

7.

8 3

5 2

3 3



= 560.

8.

18 6

 +

18 4



= 21, 624.

9.

10 5

12 7



= 0.318.

10. The coefﬁcient of 23x9in the expansion of (2+ x)12is

12 9



. Therefore, the coefﬁcient of x9 is 23

12 9



= 1760.

11. The coefﬁcient of (2x)3(−4y)4in the expansion of (2x− 4y)7is

7 4



. Thus the coefﬁcient of x3y2in this expansion is 23(−4)4

7 4



= 71, 680.

12.

9 3

 6 4

 + 2

6 3

= 4620.

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13. (a)

10 5



210= 0.246; (b) 10

i=5

10 i



210= 0.623.

14. If their minimum is larger than 5, they are all from the set{6, 7, 8, . . . , 20}. Hence the answer is

15 5

20 5



= 0.194.

15. (a)

6 2

28 4



34 6

 = 0.228; (b)

6 6

 +

6 6

 +

10 6

 +

12 6



34 6

 = 0.00084.

16.

50 5

150 45



200 50

 = 0.00206.

17.

n i=0

2i

n i



= n

i=0

n i



2i1n−i = (2 + 1)n= 3n. n

i=0

xi

n i



= n

i=0

n i



xi1n−i = (x + 1)n.

18. 6 2

 54

66= 0.201.

19. 212

24 12



= 0.00151.

20. Royal Flush: 4

52 5

 = 0.0000015.

Straight ﬂush: 36

52 5

 = 0.000014.

Four of a kind:

13× 12

4 1



52 5

 = 0.00024.

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Section 2.4 Combinations 21

Full house:

13

4 3



· 12

4 2



52 5

 = 0.0014.

Flush:

4

13 5



− 40

52 5

 = 0.002.

Straight: 10(4)5− 40 52

5

 = 0.0039.

Three of a kind:

13

4 3



·

12 2

 42

52 5

 = 0.021.

Two pairs:

13 2

4 2

4 2



· 11

4 1



52 5

 = 0.048.

One pair:

13

4 2



·

12 3

 43

52 5

 = 0.42.

None of the above: 1− the sum of all of the above cases = 0.5034445.

21. The desired probability is

12 6

12 6



24 12

 = 0.3157.

22. The answer is the solution of the equation

x 3



= 20. This equation is equivalent to x(x− 1)(x − 2) = 120 and its solution is x = 6.

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23. There are 9×103= 9000 four-digit numbers. From every 4-combination of the set {0, 1, . . . , 9}, exactly one four-digit number can be constructed in which its ones place is less than its tens place, its tens place is less than its hundreds place, and its hundreds place is less than its thousands place. Therefore, the number of such four-digit numbers is

10 4



= 210. Hence the desired probability is 0.023333.

24.

(x+ y + z)2=

n1+n2+n3=2

n!

n1! n2! n3!xn1yn2zn3

= 2!

2! 0! 0!x2y0z0+ 2!

0! 2! 0!x0y2z0+ 2!

0! 0! 2!x0y0z2 + 2!

1! 1! 0!x1y1z0+ 2!

1! 0! 1!x1y0z1+ 2!

0! 1! 1!x0y1z1

= x2+ y2+ z2+ 2xy + 2xz + 2yz.

25. The coefﬁcient of (2x)2(−y)3(3z)2in the expansion of (2x − y + 3z)7is 7!

2! 3! 2!.Thus the coefﬁcient of x2y3z2in this expansion is 22(−1)3(3)2 7!

2! 3! 2! = −7560.

26. The coefﬁcient of (2x)3(−y)7(3)3in the expansion of (2x− y + 3)13 is 13!

3! 7! 3!.Therefore, the coefﬁcient of x3y7in this expansion is 23(−1)7(3)3 13!

3! 7! 3! = −7, 413, 120.

27. In 52!

13! 13! 13! 13! = 52!

(13!)4 ways 52 cards can be dealt among four people. Hence the sample space contains 52!/(13!)4 points. Now in 4! ways the four different suits can be distributed among the players; thus the desired probability is 4!/[52!/(13!)4] ≈ 4.47 × 10−28.

28. The theorem is valid for k = 2; it is the binomial expansion. Suppose that it is true for all integers≤ k − 1. We show it for k. By the binomial expansion,

(x1+ x2+ · · · + xk)n= n n1=0

n n1



x1n1(x2+ · · · + xk)n−n1

= n n1=0

n n1



xn11

n2+n3+···+nk=n−n1

(n− n1)!

n2! n3! · · · nk!x2n2x3n3· · · xknk

=

n1+n2+···+nk=n

n n1

 (n− n1)!

n2! n3! · · · nk!xn11x2n2· · · xknk

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Section 2.4 Combinations 23

=

n1+n2+···+nk=n

n!

n1! n2! · · · nk!x1n1x2n2· · · xknk.

29. We must have 8 steps. Since the distance from M to L is ten 5-centimeter intervals and the ﬁrst step is made at M, there are 9 spots left at which the remaining 7 steps can be made. So the answer is

9 7



= 36.

30. (a)

2 1

98 49

 +

98 48



100 50

 = 0.753; (b) 250100 50



= 1.16 × 10−14.

31. (a) It must be clear that n1=

n 2



n2=

n1 2

 + nn1

n3=

n2

2



+ n2(n+ n1) n4=

n3 2



+ n3(n+ n1+ n2) ...

nk=

nk−1 2



+ nk−1(n+ n1+ · · · + nk−1).

(b) For n= 25, 000, successive calculations of nk’s yield, n1= 312, 487, 500,

n2= 48, 832, 030, 859, 381, 250,

n3= 1, 192, 283, 634, 186, 401, 370, 231, 933, 886, 715, 625, n4= 710, 770, 132, 174, 366, 339, 321, 713, 883, 042, 336, 781, 236,

550, 151, 462, 446, 793, 456, 831, 056, 250.

For n= 25, 000, the total number of all possible hybrids in the ﬁrst four generations, n1+n2+n3+n4, is 710,770,132,174,366,339,321,713,883,042,337,973,520,184,337, 863,865,857,421,889,665,625. This number is approximately 710× 1063.

32. For n= 1, we have the trivial identity x+ y =

1 0



x0y1−0+

1 1

 x1y1−1.

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