[PDF] Introduction to Probability 2nd Edition Problem Solutions

16685_6prob_solved_2ndedition.pdf

Introduction to Probability

2nd Edition

Problem Solutions

(last updated: 9/29/22) c

Dimitri P. Bertsekas and John N. Tsitsiklis

Massachusetts Institute of TechnologyWWW site for book information and orders http://www.athenasc.comAthena Scientic, Belmont, Massachusetts 1

C H A P T E R 1

Solution to Problem 1.1.We have

A=f2;4;6g; B=f4;5;6g;

soA[B=f2;4;5;6g, and (A[B)c=f1;3g:

On the other hand,

A c\Bc=f1;3;5g \ f1;2;3g=f1;3g:

Similarly, we haveA\B=f4;6g, and

(A\B)c=f1;2;3;5g:

On the other hand,

A c[Bc=f1;3;5g [ f1;2;3g=f1;2;3;5g: Solution to Problem 1.2.(a) By using a Venn diagram it can be seen that for any setsSandT, we have

S= (S\T)[(S\Tc):

(Alternatively, argue that anyxmust belong to eitherTor toTc, soxbelongs toS if and only if it belongs toS\Tor toS\Tc.) Apply this equality withS=Acand

T=B, to obtain the rst relation

A c= (Ac\B)[(Ac\Bc): Interchange the roles ofAandBto obtain the second relation. (b) By De Morgan's law, we have (A\B)c=Ac[Bc; and by using the equalities of part (a), we obtain (A\B)c=(Ac\B)[(Ac\Bc)
[(A\Bc)[(Ac\Bc)
= (Ac\B)[(Ac\Bc)[(A\Bc): (c) We haveA=f1;3;5gandB=f1;2;3g, soA\B=f1;3g. Therefore, (A\B)c=f2;4;5;6g; 2 and A c\B=f2g; Ac\Bc=f4;6g; A\Bc=f5g:

Thus, the equality of part (b) is veried.

Solution to Problem 1.5.LetGandCbe the events that the chosen student is a genius and a chocolate lover, respectively. We haveP(G) = 0:6,P(C) = 0:7, and P(G\C) = 0:4. We are interested inP(Gc\Cc), which is obtained with the following calculation: P(Gc\Cc) = 1P(G[C) = 1P(G)+P(C)P(G\C)
= 1(0:6+0:70:4) = 0:1: Solution to Problem 1.6.We rst determine the probabilities of the six possible outcomes. Leta=P(f1g) =P(f3g) =P(f5g) andb=P(f2g) =P(f4g) =P(f6g). We are given thatb= 2a. By the additivity and normalization axioms, 1 = 3a+ 3b=

3a+ 6a= 9a. Thus,a= 1=9,b= 2=9, andP(f1;2;3g) = 4=9.

Solution to Problem 1.7.The outcome of this experiment can be any nite sequence of the form (a1;a2;:::;an), wherenis an arbitrary positive integer,a1;a2;:::;an1 belong tof1;3g, andanbelongs tof2;4g. In addition, there are possible outcomes in which an even number is never obtained. Such outcomes are innite sequences (a1;a2;:::), with each element in the sequence belonging tof1;3g. The sample space consists of all possible outcomes of the above two types. Solution to Problem 1.8.Letpibe the probability of winning against the opponent played in theith turn. Then, you will win the tournament if you win against the 2nd player (probabilityp2) and also you win against at least one of the two other players [probabilityp1+ (1p1)p3=p1+p3p1p3]. Thus, the probability of winning the tournament is p

2(p1+p3p1p3):

The order (1;2;3) is optimal if and only if the above probability is no less than the probabilities corresponding to the two alternative orders, i.e., p

2(p1+p3p1p3)p1(p2+p3p2p3);

p

2(p1+p3p1p3)p3(p2+p1p2p1):

It can be seen that the rst inequality above is equivalent top2p1, while the second inequality above is equivalent top2p3.

Solution to Problem 1.9.(a) Since

=[ni=1Si, we have A=n[ i=1(A\Si); while the setsA\Siare disjoint. The result follows by using the additivity axiom. (b) The eventsB\Cc,Bc\C,B\C, andBc\Ccform a partition of , so by part (a), we have P(A) =P(A\B\Cc) +P(A\Bc\C) +P(A\B\C) +P(A\Bc\Cc):(1) 3 The eventA\Bcan be written as the union of two disjoint events as follows:

A\B= (A\B\C)[(A\B\Cc);

so that

P(A\B) =P(A\B\C) +P(A\B\Cc):(2)

Similarly,

P(A\C) =P(A\B\C) +P(A\Bc\C):(3)

Combining Eqs. (1)-(3), we obtain the desired result. Solution to Problem 1.10.Since the eventsA\BcandAc\Bare disjoint, we have using the additivity axiom repeatedly, P (A\Bc)[(Ac\B)
=P(A\Bc)+P(Ac\B) =P(A)P(A\B)+P(B)P(A\B): Solution to Problem 1.14.(a) Each possible outcome has probability 1/36. There are 6 possible outcomes that are doubles, so the probability of doubles is 6=36 = 1=6. (b) The conditioning event (sum is 4 or less) consists of the 6 outcomes (1;1);(1;2);(1;3);(2;1);(2;2);(3;1) ;

2 of which are doubles, so the conditional probability of doubles is 2=6 = 1=3.

(c) There are 11 possible outcomes with at least one 6, namely, (6;6), (6;i), and (i;6), fori= 1;2;:::;5. Thus, the probability that at least one die is a 6 is 11/36. (d) There are 30 possible outcomes where the dice land on dierent numbers. Out of these, there are 10 outcomes in which at least one of the rolls is a 6. Thus, the desired conditional probability is 10=30 = 1=3. Solution to Problem 1.15.LetAbe the event that the rst toss is a head and letBbe the event that the second toss is a head. We must compare the conditional probabilitiesP(A\BjA) andP(A\BjA[B). We have

P(A\BjA) =P(A\B)\A
P(A)=P(A\B)P(A);

and

P(A\BjA[B) =P(A\B)\(A[B)
P(A[B)=P(A\B)P(A[B):

SinceP(A[B)P(A), the rst conditional probability above is at least as large, so Alice is right, regardless of whether the coin is fair or not. In the case where the coin is fair, that is, if all four outcomesHH,HT,TH,TTare equally likely, we have

P(A\B)P(A)=1=41=2=12

;P(A\B)P(A[B)=1=43=4=13 : A generalization of Alice's reasoning is that ifA0,B0, andC0are events such thatB0C0andA0\B0=A0\C0(for example ifA0B0C0), then the event 4 A

0is at least as likely if we know thatB0has occurred than if we know thatC0has

occurred. Alice's reasoning corresponds to the special case whereA0=A[B,B0=A, andC0=A[B. Solution to Problem 1.16.In this problem, there is a tendency to reason that since the opposite face is either heads or tails, the desired probability is 1/2. This is, however, wrong, because given that heads came up, it is more likely that the two-headed coin was chosen. The correct reasoning is to calculate the conditional probability p=P(two-headed coin was chosenjheads came up) = P(two-headed coin was chosen and heads came up)P(heads came up):

We have

P(two-headed coin was chosen and heads came up) =13 ;

P(heads came up) =12

; so by taking the ratio of the above two probabilities, we obtainp= 2=3. Thus, the probability that the opposite face is tails is 1p= 1=3. Solution to Problem 1.17.LetAbe the event that the batch will be accepted. ThenA=A1\A2\A3\A4, whereAi,i= 1;:::;4, is the event that theith item is not defective. Using the multiplication rule, we have P(A) =P(A1)P(A2jA1)P(A3jA1\A2)P(A4jA1\A2\A3) =95100
9499
9398
9297 = 0:812: Solution to Problem 1.18.Using the denition of conditional probabilities, we have

P(A\BjB) =P(A\B\B)P(B)=P(A\B)P(B)=P(AjB):

Solution to Problem 1.19.LetAbe the event that Alice does not nd her paper in draweri. Since the paper is in draweriwith probabilitypi, and her search is successful with probabilitydi, the multiplication rule yieldsP(Ac) =pidi, so that P(A) = 1pidi. LetBbe the event that the paper is in drawerj. Ifj6=i, then

A\B=B,P(A\B) =P(B), and we have

P(BjA) =P(A\B)P(A)=P(B)P(A)=pj1pidi:

Similarly, ifi=j, we have

P(BjA) =P(A\B)P(A)=P(B)P(AjB)P(A)=pi(1di)1pidi:

Solution to Problem 1.20.(a) Figure 1.1 provides a sequential description for the three dierent strategies. Here we assume 1 point for a win, 0 for a loss, and 1/2 point 5 0 - 0

Timid play

p w

Bold play

(a) (b) (c)

Bold play

Bold play

Bold play

Bold play

Timid play

Timid play

Timid play

0 - 0 1 - 0 2 - 0 1 - 1 1 - 1 0 - 1 0 - 2 0 - 2 0 - 1 1 - 1

0.5-0.5

0.5-1.5

0.5-1.5

0 - 0 1 - 0 1 - 1 1 - 1 0 - 1 0 - 2

1.5-0.5

p w p w p w p w p d p d p d 1-p d 1-p d 1-p d p d 1-p d 1-p w 1-p w 1-p w 1-p w 1-p wFigure 1.1:Sequential descriptions of the chess match histories under strategies (i), (ii), and (iii). for a draw. In the case of a tied 1-1 score, we go to sudden death in the next game, and Boris wins the match (probabilitypw), or loses the match (probability 1pw). (i) Using the total probability theorem and the sequential description of Fig. 1.1(a), we have

P(Boris wins) =p2w+ 2pw(1pw)pw:

The termp2wcorresponds to the win-win outcome, and the term 2pw(1pw)pwcorre- sponds to the win-lose-win and the lose-win-win outcomes. (ii) Using Fig. 1.1(b), we have

P(Boris wins) =p2dpw;

corresponding to the draw-draw-win outcome. (iii) Using Fig. 1.1(c), we have

P(Boris wins) =pwpd+pw(1pd)pw+ (1pw)p2w:

6 The termpwpdcorresponds to the win-draw outcome, the termpw(1pd)pwcorre- sponds to the win-lose-win outcome, and the term (1pw)p2wcorresponds to lose-win- win outcome. (b) Ifpw<1=2, Boris has a greater probability of losing rather than winning any one game, regardless of the type of play he uses. Despite this, the probability of winning the match with strategy (iii) can be greater than 1/2, provided thatpwis close enough to 1/2 andpdis close enough to 1. As an example, ifpw= 0:45 andpd= 0:9, with strategy (iii) we have P(Boris wins) = 0:45
0:9 + 0:452
(10:9) + (10:45)
0:4520:54: With strategies (i) and (ii), the corresponding probabilities of a win can be calculated to be approximately 0.43 and 0.36, respectively. What is happening here is that with strategy (iii), Boris is allowed to select a playing styleafterseeing the result of the rst game, while his opponent is not. Thus, by being able to dictate the playing style in each game after receiving partial information about the match's outcome, Boris gains an advantage. Solution to Problem 1.21.Letp(m;k) be the probability that the starting player wins when the jar initially containsmwhite andkblack balls. We have, using the total probability theorem, p(m;k) =mm+k+km+k1p(m;k1)
= 1km+kp(m;k1): The probabilitiesp(m;1);p(m;2);:::;p(m;n) can be calculated sequentially using this formula, starting with the initial conditionp(m;0) = 1. Solution to Problem 1.22.We derive a recursion for the probabilitypithat a white ball is chosen from theith jar. We have, using the total probability theorem, p i+1=m+ 1m+n+ 1pi+mm+n+ 1(1pi) =1m+n+ 1pi+mm+n+ 1; starting with the initial conditionp1=m=(m+n). Thus, we have p

2=1m+n+ 1
mm+n+mm+n+ 1=mm+n:

More generally, this calculation shows that ifpi1=m=(m+n), thenpi=m=(m+n).

Thus, we obtainpi=m=(m+n) for alli.

Solution to Problem 1.23.Letpi;ni(k) denote the probability that afterkex- changes, a jar will containiballs that started in that jar andniballs that started in the other jar. We want to ndpn;0(4). We argue recursively, using the total probability 7 theorem. We have p n;0(4) =1n
1n
pn1;1(3); p n1;1(3) =pn;0(2) + 2
n1n
1n
pn1;1(2) +2n
2n
pn2;2(2); p n;0(2) =1n
1n
pn1;1(1); p n1;1(2) = 2
n1n
1n
pn1;1(1); p n2;2(2) =n1n
n1n
pn1;1(1); p n1;1(1) = 1:

Combining these equations, we obtain

p n;0(4) =1n 2 1n

2+4(n1)2n

4+4(n1)2n

4 =1n 2 1n

2+8(n1)2n

4 : Solution to Problem 1.24.Intuitively, there is something wrong with this rationale. The reason is that it is not based on a correctly specied probabilistic model. In particular, the event where both of the other prisoners are to be released is not properly accounted in the calculation of the posterior probability of release. To be precise, let A, B, and C be the prisoners, and let A be the one who considers asking the guard. Suppose that all prisoners are a priori equally likely to be released. Suppose also that if B and C are to be released, then the guard chooses B or C with equal probability to reveal to A. Then, there are four possible outcomes: (1) A and B are to be released, and the guard says B (probability 1/3). (2) A and C are to be released, and the guard says C (probability 1/3). (3) B and C are to be released, and the guard says B (probability 1/6). (4) B and C are to be released, and the guard says C (probability 1/6). Thus, P(A is to be releasedjguard says B) =P(A is to be released and guard says B)P(guard says B) =

1=31=3 + 1=6=23

:

Similarly,

P(A is to be releasedjguard says C) =23

: Thus, regardless of the identity revealed by the guard, the probability that A is released is equal to 2/3, the a priori probability of being released. Solution to Problem 1.25.Letmandmbe the larger and the smaller of the two amounts, respectively. Consider the three events

A=fX < m); B=fm< X 8 LetA(orBorC) be the event thatA(orBorC, respectively) occursandyou rst select the envelope containing the larger amountm. LetA(orBorC) be the event thatA(orBorC, respectively) occursandyou rst select the envelope containing the smaller amountm. Finally, consider the event

W=fyou end up with the envelope containingmg:

We want to determineP(W) and check whether it is larger than 1/2 or not.

By the total probability theorem, we have

P(WjA) =12

P(WjA) +P(WjA)
=12 (1 + 0) =12 ;

P(WjB) =12

P(WjB) +P(WjB)
=12 (1 + 1) = 1;

P(WjC) =12

P(WjC) +P(WjC)
=12 (0 + 1) =12 : Using these relations together with the total probability theorem, we obtain

P(W) =P(A)P(WjA) +P(B)P(WjB) +P(C)P(WjC)

= 12 P(A) +P(B) +P(C)
+12 P(B) = 12 +12 P(B): SinceP(B)>0 by assumption, it follows thatP(W)>1=2, so your friend is correct.

Solution to Problem 1.26.(a) We use the formula

P(AjB) =P(A\B)P(B)=P(A)P(BjA)P(B):

Since all crows are black, we haveP(B) = 1q. Furthermore,P(A) =p. Finally, P(BjA) = 1q=P(B), since the probability of observing a (black) crow is not aected by the truth of our hypothesis. We conclude thatP(AjB) =P(A) =p. Thus, the new evidence, while compatible with the hypothesis \all cows are white," does not change our beliefs about its truth. (b) Once more,

P(AjC) =P(A\C)P(C)=P(A)P(CjA)P(C):

Given the eventA, a cow is observed with probabilityq, and it must be white. Thus, P(CjA) =q. Given the eventAc, a cow is observed with probabilityq, and it is white with probability 1=2. Thus,P(CjAc) =q=2. Using the total probability theorem,

P(C) =P(A)P(CjA) +P(Ac)P(CjAc) =pq+ (1p)q2

:

Hence,

P(AjC) =pqpq+ (1p)q2

=

2p1 +p> p:

9 Thus, the observation of a white cow makes the hypothesis \all cows are white" more likely to be true. Solution to Problem 1.27.Since Bob tosses one more coin that Alice, it is im- possible that they toss both the same number of heads and the same number of tails. So Bob tosses either more heads than Alice or more tails than Alice (but not both). Since the coins are fair, these events are equally likely by symmetry, so both events have probability 1/2. An alternative solution is to argue that if Alice and Bob are tied after 2ntosses, they are equally likely to win. If they are not tied, then their scores dier by at least 2, and toss 2n+1 will not change the nal outcome. This argument may also be expressed algebraically by using the total probability theorem. LetBbe the event that Bob tosses more heads. LetXbe the event that after each has tossednof their coins, Bob has more heads than Alice, letYbe the event that under the same conditions, Alice has more heads than Bob, and letZbe the event that they have the same number of heads. Since the coins are fair, we haveP(X) =P(Y), and alsoP(Z) = 1P(X)P(Y).

Furthermore, we see that

P(BjX) = 1;P(BjY) = 0;P(BjZ) =12

:

Now we have, using the total probability theorem,

P(B) =P(X)
P(BjX) +P(Y)
P(BjY) +P(Z)
P(BjZ)

=P(X) +12
P(Z) = 12
P(X) +P(Y) +P(Z)
= 12 : as required. Solution to Problem 1.30.Consider the sample space for the hunter's strategy.

The events that lead to the correct path are:

(1) Both dogs agree on the correct path (probabilityp2, by independence). (2) The dogs disagree, dog 1 chooses the correct path, and hunter follows dog 1 [probabilityp(1p)=2]. (3) The dogs disagree, dog 2 chooses the correct path, and hunter follows dog 2 [probabilityp(1p)=2]. The above events are disjoint, so we can add the probabilities to nd that under the hunter's strategy, the probability that he chooses the correct path is p 2+12 p(1p) +12 p(1p) =p: On the other hand, if the hunter lets one dog choose the path, this dog will also choose the correct path with probabilityp. Thus, the two strategies are equally eective. 10 Solution to Problem 1.31.(a) LetAbe the event that a 0 is transmitted. Using the total probability theorem, the desired probability is P(A)(10) +1P(A)
(11) =p(10) + (1p)(11): (b) By independence, the probability that the string 1011 is received correctly is (10)(11)3: (c) In order for a 0 to be decoded correctly, the received string must be 000, 001, 010, or 100. Given that the string transmitted was 000, the probability of receiving 000 is (10)3, and the probability of each of the strings 001, 010, and 100 is0(10)2.

Thus, the probability of correct decoding is

30(10)2+ (10)3:

(d) When the symbol is 0, the probabilities of correct decoding with and without the scheme of part (c) are 30(10)2+ (10)3and 10, respectively. Thus, the probability is improved with the scheme of part (c) if

30(10)2+ (10)3>(10);

or (10)(1 + 20)>1; which is equivalent to 0< 0<1=2. (e) Using Bayes' rule, we have

P(0j101) =P(0)P(101j0)P(0)P(101j0) +P(1)P(101j1):

The probabilities needed in the above formula are

P(0) =p;P(1) = 1p;P(101j0) =20(10);P(101j1) =1(11)2: Solution to Problem 1.32.The answer to this problem is not unique and depends on the assumptions we make on the reproductive strategy of the king's parents. Suppose that the king's parents had decided to have exactly two children and then stopped. There are four possible and equally likely outcomes, namely BB, GG, BG, and GB (B stands for \boy" and G stands for \girl"). Given that at least one child was a boy (the king), the outcome GG is eliminated and we are left with three equally likely outcomes (BB, BG, and GB). The probability that the sibling is male (the conditional probability of BB) is 1/3 . Suppose on the other hand that the king's parents had decided to have children until they would have a male child. In that case, the king is the second child, and the sibling is female, with certainty. 11 Solution to Problem 1.33.Flip the coin twice. If the outcome is heads-tails, choose the opera. if the outcome is tails-heads, choose the movies. Otherwise, repeat the process, until a decision can be made. LetAkbe the event that a decision was made at thekth round. Conditional on the eventAk, the two choices are equally likely, and we have

P(opera) =1X

k=1P(operajAk)P(Ak) =1X k=112

P(Ak) =12

:

We have used here the property

P1 k=0P(Ak) = 1, which is true as long asP(heads)>0 andP(tails)>0. Solution to Problem 1.34.The system may be viewed as a series connection of three subsystems, denoted 1, 2, and 3 in Fig. 1.19 in the text. The probability that the entire system is operational isp1p2p3, wherepiis the probability that subsystemiis operational. Using the formulas for the probability of success of a series or a parallel system given in Example 1.24, we have p

1=p; p3= 1(1p)2;

and p

2= 1(1p)1p1(1p)3

:

Solution to Problem 1.35.LetAibe the event that exactlyicomponents are operational. The probability that the system is operational is the probability of the union[ni=kAi, and since theAiare disjoint, it is equal to n X i=kP(Ai) =nX i=kp(i); wherep(i) are the binomial probabilities. Thus, the probability of an operational system is nX i=k n i p i(1p)ni: Solution to Problem 1.36.(a) LetAdenote the event that the city experiences a black-out. Since the power plants fail independent of each other, we have

P(A) =nY

i=1p i: (b) There will be a black-out if either allnor anyn1 power plants fail. These two events are disjoint, so we can calculate the probabilityP(A) of a black-out by adding their probabilities:

P(A) =nY

i=1p i+nX i=1 (1pi)Y j6=ip j! : 12

Here, (1pi)Q

j6=ipjis the probability thatn1 plants have failed and plantiis the one that has not failed. Solution to Problem 1.37.The probability thatk1voice users andk2data users simultaneously need to be connected isp1(k1)p2(k2), wherep1(k1) andp2(k2) are the corresponding binomial probabilities, given by p i(ki) =n i k i p kii(1pi)niki; i= 1;2: The probability that more users want to use the system than the system can accommodate is the sum of all productsp1(k1)p2(k2) ask1andk2range over all possible values whose total bit rate requirementk1r1+k2r2exceeds the capacitycof the system.

Thus, the desired probability is

X f(k1;k2)jk1r1+k2r2>c; k1n1; k2n2gp

1(k1)p2(k2):

Solution to Problem 1.38.We have

p T=P(at least 6 out of the 8 remaining holes are won by Telis); p W=P(at least 4 out of the 8 remaining holes are won by Wendy):

Using the binomial formulas,

p T=8X k=6 8 k p k(1p)8k; pW=8X k=4 8 k (1p)kp8k: The amount of money that Telis should get is 10
pT=(pT+pW) dollars. Solution to Problem 1.39.Let the eventAbe the event that the professor teaches her class, and letBbe the event that the weather is bad. We have

P(A) =P(B)P(AjB) +P(Bc)P(AjBc);

and

P(AjB) =nX

i=k n i p ib(1pb)ni;

P(AjBc) =nX

i=k n i p ig(1pg)ni:

Therefore,

P(A) =P(B)nX

i=k n i p ib(1pb)ni+1P(B)
nX i=k n i p ig(1pg)ni: 13 Solution to Problem 1.40.LetAbe the event that the rstn1 tosses produce an even number of heads, and letEbe the event that thenth toss is a head. We can obtain an even number of heads inntosses in two distinct ways: 1) there is an even number of heads in the rstn1 tosses, and thenth toss results in tails: this is the eventA\Ec; 2) there is an odd number of heads in the rstn1 tosses, and thenth toss results in heads: this is the eventAc\E. Using also the independence ofAand E, q n=P(A\Ec)[(Ac\E)
=P(A\Ec) +P(Ac\E) =P(A)P(Ec) +P(Ac)P(E) = (1p)qn1+p(1qn1): We now use induction. Forn= 0, we haveq0= 1, which agrees with the given formula forqn. Assume, that the formula holds withnreplaced byn1, i.e., q n1=1 + (12p)n12 :

Using this equation, we have

q n=p(1qn1) + (1p)qn1 =p+ (12p)qn1 =p+ (12p)1 + (12p)n12 =

1 + (12p)n2

; so the given formula holds for alln.

Solution to Problem 1.41.We have

P(N=n) =P(A1;n1\An;n) =P(A1;n1)P(An;njA1;n1);

where forij,Ai;jis the event that contestanti's number is the smallest of the numbers of contestants 1;:::;j. We also have

P(A1;n1) =1n1:

We claim that

P(An;njA1;n1) =P(An;n) =1n

:

The reason is that by symmetry, we have

P(An;njAi;n1) =P(An;njA1;n1); i= 1;:::;n1;

while by the total probability theorem,

P(An;n) =n1X

i=1P(Ai;n1)P(An;njAi;n1) =P(An;njA1;n1)n1X i=1P(Ai;n1) =P(An;njA1;n1): 14 Hence

P(N=n) =1n1
1n

: An alternative solution is also possible, using the counting methods developed in Section 1.6. Let us x a particular choice ofn. Think of an outcome of the experiment as an ordering of the values of thencontestants, so that there aren! equally likely outcomes. The eventfN=ngoccurs if and only if the rst contestant's number is smallest among the rstn1 contestants, and contestantn's number is the smallest among the rstncontestants. This event can occur in (n2)! dierent ways, namely, all the possible ways of ordering contestants 2;:::;n1. Thus, the probability of this event is (n2)!=n! = 1=(n(n1)), in agreement with the previous solution. Solution to Problem 1.49.A sum of 11 is obtained with the following 6 combina- tions: (6;4;1) (6;3;2) (5;5;1) (5;4;2) (5;3;3) (4;4;3): A sum of 12 is obtained with the following 6 combinations: (6;5;1) (6;4;2) (6;3;3) (5;5;2) (5;4;3) (4;4;4): Each combination of 3 distinct numbers corresponds to 6 permutations, while each combination of 3 numbers, two of which are equal, corresponds to 3 permutations. Counting the number of permutations in the 6 combinations corresponding to a sum of 11, we obtain 6 + 6 + 3 + 6 + 3 + 3 = 27 permutations. Counting the number of permutations in the 6 combinations corresponding to a sum of 12, we obtain 6 + 6 +

3+3+6+1 = 25 permutations. Since all permutations are equally likely, a sum of 11

is more likely than a sum of 12.

Note also that the sample space has 6

3= 216 elements, so we haveP(11) =

27=216,P(12) = 25=216.

Solution to Problem 1.50.The sample space consists of all possible choices for the birthday of each person. Since there arenpersons, and each has 365 choices for their birthday, the sample space has 365 nelements. Let us now consider those choices of birthdays for which no two persons have the same birthday. Assuming that n365, there are 365 choices for the rst person, 364 for the second, etc., for a total of 365
364


(365n+ 1). Thus, P(no two birthdays coincide) =365
364


(365n+ 1)365 n: It is interesting to note that fornas small as 23, the probability that there are two persons with the same birthday is larger than 1=2. Solution to Problem 1.51.(a) We number the red balls from 1 tom, and the white balls fromm+ 1 tom+n. One possible sample space consists of all pairs of integers (i;j) with 1i;jm+nandi6=j. The total number of possible outcomes is (m+n)(m+n1). The number of outcomes corresponding to red-white selection, (i.e.,i2 f1;:::;mgandj2 fm+ 1;:::;m+ng) ismn. The number of outcomes corresponding to white-red selection, (i.e.,i2 fm+ 1;:::;m+ngandj2 f1;:::;mg) is alsomn. Thus, the desired probability that the balls are of dierent color is

2mn(m+n)(m+n1):

15 Another possible sample space consists of all the possible ordered color pairs, i.e., fRR;RW;WR;WWg. We then have to calculate the probability of the eventfRW;WRg. We consider a sequential description of the experiment, i.e., we rst select the rst ball and then the second. In the rst stage, the probability of a red ball ism=(m+n). In the second stage, the probability of a red ball is eitherm=(m+n1) or (m1)=(m+n1) depending on whether the rst ball was white or red, respectively. Therefore, using the multiplication rule, we have

P(RR) =mm+n
m1m1 +n;P(RW) =mm+n
nm1 +n;

P(WR) =nm+n
mm+n1;P(WW) =nm+n
n1m+n1:

The desired probability is

P fRW;WRg
=P(RW) +P(WR) = mm+n
nm1 +n+nm+n
mm+n1 =

2mn(m+n)(m+n1):

(b) We calculate the conditional probability of all balls being red, given any of the possible values ofk. We haveP(Rjk= 1) =m=(m+n) and, as found in part (a), P(RRjk= 2) =m(m1)=(m+n)(m1 +n). Arguing sequentially as in part (a), we also haveP(RRRjk= 3) =m(m1)(m2)=(m+n)(m1 +n)(m2 +n). According to the total probability theorem, the desired answer is 13  mm+n+m(m1)(m+n)(m1 +n)+m(m1)(m2)(m+n)(m1 +n)(m2 +n) : Solution to Problem 1.52.The probability that the 13th card is the rst king to be dealt is the probability that out of the rst 13 cards to be dealt, exactly one was a king, and that the king was dealt last. Now, given that exactly one king was dealt in the rst 13 cards, the probability that the king was dealt last is just 1=13, since each \position" is equally likely. Thus, it remains to calculate the probability that there was exactly one king in the rst 13 cards dealt. To calculate this probability we count the \favorable" outcomes and divide by the total number of possible outcomes. We rst count the favorable outcomes, namely those with exactly one king in the rst 13 cards dealt. We can choose a particular king in 4 ways, and we can choose the other

12 cards in48

12
ways, therefore there are 4
48

12
favorable outcomes. There are52

13
total outcomes, so the desired probability is 113

4
48 12 52
13 : For an alternative solution, we argue as in Example 1.10. The probability that the rst card is not a king is 48=52. Given that, the probability that the second is 16 not a king is 47=51. We continue similarly until the 12th card. The probability that the 12th card is not a king, given that none of the preceding 11 was a king, is 37/41. (There are 5211 = 41 cards left, and 4811 = 37 of them are not kings.) Finally, the conditional probability that the 13th card is a king is 4=40. The desired probability is

48
47


37
452
51


41
40:

Solution to Problem 1.53.Suppose we label the classesA,B, andC. The proba- bility that Joe and Jane will both be in classAis the number of possible combinations for classAthat involve both Joe and Jane, divided by the total number of combinations for classA. Therefore, this probability is 88 28
90
30
: Since there are three classes, the probability that Joe and Jane end up in the same class is 3
 88
28
90
30
: A much simpler solution is as follows. We place Joe in one class. Regarding Jane, there are 89 possible \slots", and only 29 of them place her in the same class as Joe. Thus, the answer is 29/89, which turns out to agree with the answer obtained earlier. Solution to Problem 1.54.(a) Since the cars are all distinct, there are 20! ways to line them up. (b) To nd the probability that the cars will be parked so that they alternate, we count the number of \favorable" outcomes, and divide by the total number of possible outcomes found in part (a). We count in the following manner. We rst arrange the US cars in an ordered sequence (permutation). We can do this in 10! ways, since there are 10 distinct cars. Similarly, arrange the foreign cars in an ordered sequence, which can also be done in 10! ways. Finally, interleave the two sequences. This can be done in two dierent ways, since we can let the rst car be either US-made or foreign. Thus, we have a total of 2
10!
10! possibilities, and the desired probability is

2
10!
10!20!

: Note that we could have solved the second part of the problem by neglecting the fact that the cars are distinct. Suppose the foreign cars are indistinguishable, and also that the US cars are indistinguishable. Out of the 20 available spaces, we need to choose

10 spaces in which to place the US cars, and thus there are20

10
possible outcomes.

Out of these outcomes, there are only two in which the cars alternate, depending on 17 whether we start with a US or a foreign car. Thus, the desired probability is 2=20 10
, which coincides with our earlier answer. Solution to Problem 1.55.We count the number of ways in which we can safely place 8 distinguishable rooks, and then divide this by the total number of possibilities. First we count the number of favorable positions for the rooks. We will place the rooks one by one on the 88 chessboard. For the rst rook, there are no constraints, so we have 64 choices. Placing this rook, however, eliminates one row and one column. Thus, for the second rook, we can imagine that the illegal column and row have been removed, thus leaving us with a 77 chessboard, and with 49 choices. Similarly, for the third rook we have 36 choices, for the fourth 25, etc. In the absence of any restrictions, there are 64
63


57 = 64!=56! ways we can place 8 rooks, so the desired probability is64
49
36
25
16
9
464! 56!
:

Solution to Problem 1.56.(a) There are8

4
ways to pick 4 lower level classes, and10

3
ways to choose 3 higher level classes, so there are

 8 4 10 3 valid curricula. (b) This part is more involved. We need to consider several dierent cases: (i) Suppose we do not chooseL1. Then bothL2andL3must be chosen; otherwise no higher level courses would be allowed. Thus, we need to choose 2 more lower level classes out of the remaining 5, and 3 higher level classes from the available

5. We then obtain5

2
5

3
valid curricula.

(ii) If we chooseL1but choose neitherL2norL3, we have5 3
5

3
choices.

(iii) If we chooseL1and choose one ofL2orL3, we have 2
5 2
5

3
choices. This is

because there are two ways of choosing betweenL2andL3,5

2
ways of choosing

2 lower level classes fromL4;:::;L8, and5

3
ways of choosing 3 higher level

classes fromH1;:::;H5. (iv) Finally, if we chooseL1,L2, andL3, we have5 1
10

3
choices.

Note that we are not double counting, because there is no overlap in the cases we are considering, and furthermore we have considered every possible choice. The total is obtained by adding the counts for the above four cases. Solution to Problem 1.57.Let us x the order in which letters appear in the sentence. There are 26! choices, corresponding to the possible permutations of the 26- letter alphabet. Having xed the order of the letters, we need to separate them into words. To obtain 6 words, we need to place 5 separators (\blanks") between the letters. With 26 letters, there are 25 possible positions for these blanks, and the number of choices is25

5
. Thus, the desired number of sentences is 26!25

5
. Generalizing, the

number of sentences consisting ofwnonempty words using exactly once each letter 18 from al-letter alphabet is equal to l!l1 w1 : Solution to Problem 1.58.(a) The sample space consists of all ways of drawing 7 elements out of a 52-element set, so it contains52

7
possible outcomes. Let us count

those outcomes that involve exactly 3 aces. We are free to select any 3 out of the 4 aces, and any 4 out of the 48 remaining cards, for a total of4 3
48

4
choices. Thus,

P(7 cards include exactly 3 aces) =

4 3 48
4 52
7 : (b) Proceeding similar to part (a), we obtain

P(7 cards include exactly 2 kings) =

4 2 48
5 52
7 : (c) IfAandBstand for the events in parts (a) and (b), respectively, we are looking forP(A[B) =P(A) +P(B)P(A\B). The eventA\B(having exactly 3 aces and exactly 2 kings) can occur by choosing 3 out of the 4 available aces, 2 out of the 4 available kings, and 2 more cards out of the remaining 44. Thus, this event consists of4 3
4 2
44

2
distinct outcomes. Hence,

P(7 cards include 3 aces and/or 2 kings) =

4 3 48
4 +4 2 48
5 4 3 4 2 44
2 52
7 : Solution to Problem 1.59.Clearly ifn > m, orn > k, ormn >100k, the probability must be zero. Ifnm,nk, andmn100k, then we can nd the probability that the test drive foundnof the 100 cars defective by counting the total number of sizemsubsets, and then the number of sizemsubsets that containn lemons. Clearly, there are100 m
dierent subsets of sizem. To count the number of size msubsets withnlemons, we rst choosenlemons from thekavailable lemons, and then choosemngood cars from the 100kavailable good cars. Thus, the number of ways to choose a subset of sizemfrom 100 cars, and getnlemons, is k n 100k
mn ; 19 and the desired probability is k n 100k
mn 100
m : Solution to Problem 1.60.The size of the sample space is the number of dierent ways that 52 objects can be divided in 4 groups of 13, and is given by the multinomial formula52!13!13!13!13! : There are 4! dierent ways of distributing the 4 aces to the 4 players, and there are

48!12!12!12!12!

dierent ways of dividing the remaining 48 cards into 4 groups of 12. Thus, the desired probability is 4!

48!12!12!12!12!

52!

13!13!13!13!

: An alternative solution can be obtained by considering a dierent, but proba- bilistically equivalent method of dealing the cards. Each player has 13 slots, each one of which is to receive one card. Instead of shuing the deck, we place the 4 aces at the top, and start dealing the cards one at a time, with each free slot being equally likely to receive the next card. For the event of interest to occur, the rst ace can go anywhere; the second can go to any one of the 39 slots (out of the 51 available) that correspond to players that do not yet have an ace; the third can go to any one of the

26 slots (out of the 50 available) that correspond to the two players that do not yet

have an ace; and nally, the fourth, can go to any one of the 13 slots (out of the 49 available) that correspond to the only player who does not yet have an ace. Thus, the desired probability is39
26
1351
50
49: By simplifying our previous answer, it can be checked that it is the same as the one obtained here, thus corroborating the intuitive fact that the two dierent ways of dealing the cards are probabilistically equivalent. 20

C H A P T E R 2

Solution to Problem 2.1.LetXbe the number of points the MIT team earns over the weekend. We have

P(X= 0) = 0:6
0:3 = 0:18;

P(X= 1) = 0:4
0:5
0:3 + 0:6
0:5
0:7 = 0:27;

P(X= 2) = 0:4
0:5
0:3 + 0:6
0:5
0:7 + 0:4
0:5
0:7
0:5 = 0:34; P(X= 3) = 0:4
0:5
0:7
0:5 + 0:4
0:5
0:7
0:5 = 0:14;

P(X= 4) = 0:4
0:5
0:7
0:5 = 0:07;

P(X >4) = 0:

Solution to Problem 2.2.The number of guests that have the same birthday as you is binomial withp= 1=365 andn= 499. Thus the probability that exactly one other guest has the same birthday is 499 1 1365


364365

 498
0:3486: Let=np= 499=3651:367. The Poisson approximation ise=e1:367
1:367

0:3483, which closely agrees with the correct probability based on the binomial.

Solution to Problem 2.3.(a) LetLbe the duration of the match. If Fischer wins a match consisting ofLgames, thenL1 draws must rst occur before he wins.

Summing over all possible lengths, we obtain

P(Fischer wins) =10X

l=1(0:3)l1(0:4) = 0:571425: (b) The match has lengthLwithL <10, if and only if (L1) draws occur, followed by a win by either player. The match has lengthL= 10 if and only if 9 draws occur. The probability of a win by either player is 0.7. Thus p

L(l) =P(L=l) =(

(0:3)l1(0:7); l= 1;:::;9, (0:3)9; l= 10,

0;otherwise.

Solution to Problem 2.4.(a) LetXbe the number of modems in use. Fork <50, the probability thatX=kis the same as the probability thatkout of 1000 customers need a connection: p

X(k) =1000

k (0:01)k(0:99)1000k; k= 0;1;:::;49: 21
The probability thatX= 50, is the same as the probability that 50 or more out of

1000 customers need a connection:

p

X(50) =1000X

k=50 1000
k (0:01)k(0:99)1000k: (b) By approximating the binomial with a Poisson with parameter= 1000
0:01 = 10, we have p

X(k) =e1010kk!; k= 0;1;:::;49;

p

X(50) =1000X

k=50e 1010kk!: (c) LetAbe the event that there are more customers needing a connection than there are modems. Then,

P(A) =1000X

k=51 1000
k (0:01)k(0:99)1000k: With the Poisson approximation,P(A) is estimated by 1000
X k=51e 1010kk!: Solution to Problem 2.5.(a) LetXbe the number of packets stored at the end of the rst slot. Fork < b, the probability thatX=kis the same as the probability that kpackets are generated by the source: p

X(k) =ekk!; k= 0;1;:::;b1;

while p

X(b) =1X

k=be kk!= 1b1X k=0e kk!: LetYbe the number of number of packets stored at the end of the second slot. Since minfX;cgis the number of packets transmitted in the second slot, we have

Y=XminfX;cg. Thus,

p

Y(0) =cX

k=0p

X(k) =cX

k=0e kk!; p

Y(k) =pX(k+c) =ek+c(k+c)!; k= 1;:::;bc1;

22
p

Y(bc) =pX(b) = 1b1X

k=0e kk!: (b) The probability that some packets get discarded during the rst slot is the same as the probability that more thanbpackets are generated by the source, so it is equal to 1 X k=b+1e kk!; or 1bX k=0e kk!: Solution to Problem 2.6.We consider the general case of part (b), and we show thatp >1=2 is a necessary and sucient condition forn= 2k+ 1 games to be better thann= 2k1 games. To prove this, letNbe the number of Celtics' wins in the rst 2k1 games. IfAdenotes the event that the Celtics win withn= 2k+1, andB denotes the event that the Celtics win withn= 2k1, then

P(A) =P(Nk+ 1) +P(N=k)
1(1p)2
+P(N=k1)
p2;

P(B) =P(Nk) =P(N=k) +P(Nk+ 1);

and therefore

P(A)P(B) =P(N=k1)
p2P(N=k)
(1p)2

= 2k1 k1 p k1(1p)kp22k1 k (1p)2pk(1p)k1 = (2k1)!(k1)!k!pk(1p)k(2p1):

It follows thatP(A)>P(B) if and only ifp >12

. Thus, a longer series is better for the better team. Solution to Problem 2.7.Let random variableXbe the number of trials you need to open the door, and letKibe the event that theith key selected opens the door. (a) In case (1), we have p

X(1) =P(K1) =15

; p

X(2) =P(Kc1)P(K2jKc1) =45

14 =15 ; p

X(3) =P(Kc1)P(Kc2jKc1)P(K3jKc1\Kc2) =45

34
13 =15 :

Proceeding similarly, we see that the PMF ofXis

p

X(x) =15

; x= 1;2;3;4;5: 23
We can also view the problem as ordering the keys in advance and then trying them in succession, in which case the probability of any of the ve keys being correct is 1=5. In case (2),Xis a geometric random variable withp= 1=5, and its PMF is p

X(k) =15

45  k1 ; k1: (b) In case (1), we have p

X(1) =P(K1) =210

; p

X(2) =P(Kc1)P(K2jKc1) =810

29 ; p

X(3) =P(Kc1)P(Kc2jKc1)P(K3jKc1\Kc2) =810

79
28 =710
29 :

Proceeding similarly, we see that the PMF ofXis

p

X(x) =2
(10x)90

; x= 1;2;:::;10: Consider now an alternative line of reasoning to derive the PMF ofX. If we view the problem as ordering the keys in advance and then trying them in succession, the probability that the number of trials required isxis the probability that the rst x1 keys do not contain either of the two correct keys and thexth key is one of the correct keys. We can count the number of ways for this to happen and divide by the total number of ways to order the keys to determinepX(x). The total number of ways to order the keys is 10! For thexth key to be the rst correct key, the other key must be among the last 10xkeys, so there are 10xspots in which it can be located. There are 8! ways in which the other 8 keys can be in the other 8 locations. We must then multiply by two since either of the two correct keys could be in thexth position. We therefore have 2
10x
8! ways for thexth key to be the rst correct one and p

X(x) =2
(10x)8!10!

=2
(10x)90 ; x= 1;2;:::;10; as before. In case (2),Xis again a geometric random variable withp= 1=5. Solution to Problem 2.8.Fork= 0;1;:::;n1, we have p

X(k+ 1)p

X(k)=

n k+ 1 p k+1(1p)nk1 n k p k(1p)nk=p1p
nkk+ 1:

Solution to Problem 2.9.Fork= 1;:::;n, we have

p X(k)p

X(k1)=

n k p k(1p)nk n k1 p k1(1p)nk+1=(nk+ 1)pk(1p)=(n+ 1)pkpkkp: 24
Ifkk, thenk(n+1)p, or equivalentlykkp(n+1)pkp, so that the above ratio is greater than or equal to 1. It follows thatpX(k) is monotonically nondecreasing. If k > k , the ratio is less than one, andpX(k) is monotonically decreasing, as required. Solution to Problem 2.10.Using the expression for the Poisson PMF, we have, for k1, p X(k)p

X(k1)=k
ek!
(k1)!

k1
e=k : Thus ifkthe ratio is greater or equal to 1, and it follows thatpX(k) is monotonically increasing. Otherwise, the ratio is less than one, andpX(k) is monotonically decreasing, as required. Solution to Problem 2.13.We will use the PMF for the number of girls among the natural children together with the formula for the PMF of a function of a random variable. LetNbe the number of natural children that are girls. ThenNhas a binomial PMF p

N(k) =8

< : 5 k
12  5 ;if 0k5,

0;otherwise.

LetGbe the number of girls out of the 7 children, so thatG=N+ 2. By applying the formula for the PMF of a function of a random variable, we have p

G(g) =X

fnjn+2=ggp

N(n) =pN(g2):

Thus p

G(g) =8

< : 5 g2
12  5 ;if 2g7,

0;otherwise.

Solution to Problem 2.14.(a) Using the formulapY(y) =P fxjxmod(3)=ygpX(x), we obtain p

Y(0) =pX(0) +pX(3) +pX(6) +pX(9) = 4=10;

p

Y(1) =pX(1) +pX(4) +pX(7) = 3=10;

p

Y(2) =pX(2) +pX(5) +pX(8) = 3=10;

p

Y(y) = 0;ify62 f0;1;2g:

(b) Similarly, using the formulapY(y) =P fxj5 mod(x+1)=ygpX(x), we obtain p

Y(y) =8

>>< > >:2=10;ify= 0,

2=10;ify= 1,

1=10;ify= 2,

5=10;ify= 5,

0;otherwise.

25
Solution to Problem 2.15.The random variableYtakes the valuesklna, where k= 1;:::;n, if and only ifX=akorX=ak. Furthermore,Ytakes the value 0, if and only ifX= 1. Thus, we have p

Y(y) =8

>>>< > >>:22n+ 1;ify= lna;2lna;:::;nlna,

12n+ 1;ify= 0,

0;otherwise.

Solution to Problem 2.16.(a) The scalaramust satisfy 1 = X x p

X(x) =1a

3 X x=3x 2; so a=3X x=3x

2= (3)2+ (2)2+ (1)2+ 12+ 22+ 32= 28:

We also haveE[X] = 0 because the PMF is symmetric around 0. (b) Ifz2 f1;4;9g, then p

Z(z) =pX(pz) +pX(pz) =z28

+z28 =z14 :

OtherwisepZ(z) = 0.

(c) var(X) =E[Z] =X z zp

Z(z) =X

z2f1;4;9gz 214
= 7. (d) We have var(X) =X x (xE[X])2pX(x) = 1 2
p

X(1) +pX(1)
+ 22
p

X(2) +pX(2)
+ 32
p

X(3) +pX(3)

= 2
128 + 8
428 + 18
928 = 7: Solution to Problem 2.17.IfXis the temperature in Celsius, the temperature in

Fahrenheit isY= 32 + 9X=5. Therefore,

E[Y] = 32 + 9E[X]=5 = 32 + 18 = 50:

Also var(Y) = (9=5)2var(X); 26
where var(X), the square of the given standard deviation ofX, is equal to 100. Thus, the standard deviation ofYis (9=5)
10 = 18. Hence a normal day in Fahrenheit is one for which the temperature is in the range [32;68].

Solution to Problem 2.18.We have

p

X(x) =1=(ba+ 1);ifx= 2k, whereakb,kinteger,

0;otherwise,

and

E[X] =bX

k=a1ba+ 12k=2aba+ 1(1 + 2 +


+ 2ba) =2b+12aba+ 1:

Similarly,

E[X2] =bX

k=a1ba+ 1(2k)2=4b+14a3(ba+ 1); and nally var(X) =4b+14a3(ba+ 1)2b+12aba+ 1 2 : Solution to Problem 2.19.We will nd the expected gain for each strategy, by computing the expected number of questions until we nd the prize. (a) With this strategy, the probability of nding the location of the prize withiques- tions, wherei= 1;:::;8, is 1=10. The probability of nding the location with 9 questions is 2/10. Therefore, the expected number of questions is 210

9 +110 8 X i=1i= 5:4: (b) It can be checked that for 4 of the 10 possible box numbers, exactly 4 questions will be needed, whereas for 6 of the 10 numbers, 3 questions will be needed. Therefore, with this strategy, the expected number of questions is 410

4 +610
3 = 3:4: Solution to Problem 2.20.The numberCof candy bars you need to eat is a geometric random variable with parameterp. Thus the mean isE[C] = 1=p;and the variance is var(C) = (1p)=p2. Solution to Problem 2.21.The expected value of the gain for a single game is innite since ifXis your gain, then

E[X] =1X

k=12 k
2k=1X k=11 =1: 27
Thus if you are faced with the choice of playing for given feefor not playing at all, and your objective is to make the choice that maximizes your expected net gain, you would be willing to pay any value off. However, this is in strong disagreement with the behavior of individuals. In fact experiments have shown that most people are willing to pay only about $20 to $30 to play the game. The discrepancy is due to a presumption that the amount one is willing to pay is determined by the expected gain. However, expected gain does not take into account a person's attitude towards risk taking. Solution to Problem 2.22.(a) LetXbe the number of tosses until the game is over. Noting thatXis geometric with probability of success P fHT;THg
=p(1q) +q(1p); we obtain p

X(k) =1p(1q)q(1p)

k1p(1q) +q(1p)
; k= 1;2;:::

Therefore

E[X] =1p(1q) +q(1p)

and var(X) =pq+ (1p)(1q) p(1q) +q(1p)
2: (b) The probability that the last toss of the rst coin is a head is P HTjfHT;THg
=p(1q)p(1q) + (1q)p: Solution to Problem 2.23.LetXbe the total number of tosses. (a) For each toss after the rst one, there is probability 1/2 that the result is the same as in the preceding toss. Thus, the random variableXis of the formX=Y+1, where Yis a geometric random variable with parameterp= 1=2. It follows that p

X(k) =

(1=2)k1;ifk2,

0;otherwise,

and

E[X] =E[Y] + 1 =1p

+ 1 = 3:

We also have

var(X) = var(Y) =1pp 2= 2: (b) Ifk >2, there arek1 sequences that lead to the eventfX=kg. One such sequence isH


HT, wherek1 heads are followed by a tail. The otherk2 possible sequences are of the formT


TH


HT, for various lengths of the initialT


T 28
segment. For the case wherek= 2, there is only one (hencek1) possible sequence that leads to the eventfX=kg, namely the sequenceHT. Therefore, for anyk2,

P(X=k) = (k1)(1=2)k:

It follows that

p

X(k) =

(k1)(1=2)k;ifk2,

0;otherwise,

and

E[X] =1X

k=2k(k1)(1=2)k=1X k=1k(k1)(1=2)k=1X k=1k

2(1=2)k1X

k=1k(1=2)k= 62 = 4:

We have used here the equalities

1 X k=1k(1=2)k=E[Y] = 2; and 1X k=1k

2(1=2)k=E[Y2] = var(Y) +E[Y]

2= 2 + 22= 6;

whereYis a geometric random variable with parameterp= 1=2. Solution to Problem 2.24.(a) There are 21 integer pairs (x;y) in the region

R=(x;y)j 2x4;1yx1 ;

so that the joint PMF ofXandYis p

X;Y(x;y) =n1=21;if (x;y) is inR,

0;otherwise.

For eachxin the range [2;4], there are three possible values ofY. Thus, we have p

X(x) =n3=21;ifx=2;1;0;1;2;3;4,

0;otherwise.

The mean ofXis the midpoint of the range [2;4]:

E[X] = 1:

The marginal PMF ofYis obtained by using the tabular method. We have p

Y(y) =8

>>>>< > >>>:1=21;ify=3,

2=21;ify=2,

3=21;ify=1;0;1;2;3,

2=21;ify= 4,

1=21;ify= 5,

0;otherwise.

29

The mean ofYis

E[Y] =121

(3 + 5) +221
(2 + 4) +321
(1 + 1 + 2 + 3) = 1: (b) The prot is given by

P= 100X+ 200Y;

so that

E[P] = 100
E[X] + 200
E[Y] = 100
1 + 200
1 = 300:

Solution to Problem 2.25.(a) Since all possible values of (I;J) are equally likely, we have p

I;J(i;j) =(

1P n k=1mk;ifjmi,

0;otherwise.

The marginal PMFs are given by

p

I(i) =mX

j=1p

I;J(i;j) =miP

n k=1mk; i= 1;:::;n; p

J(j) =nX

i=1p

I;J(i;j) =ljP

n k=1mk; j= 1;:::;m; whereljis the number of students that have answered questionj, i.e., studentsiwith jmi. (b) The expected value of the score of studentiis the sum of the expected values p ija+ (1pij)bof the scores on questionsjwithj= 1;:::;mi, i.e., m iX j=1 p ija+ (1pij)b
: Solution to Problem 2.26.(a) The possible values of the random variableXare the ten numbers 101;:::;110, and the PMF is given by p

X(k) =P(X > k1)P(X > k);ifk= 101;:::110,

0;otherwise.

We haveP(X >100) = 1 and fork= 101;:::110,

P(X > k) =P(X1> k;X2> k;X3> k)

=P(X1> k)P(X2> k)P(X3> k) = (110k)310 3: 30

It follows that

p

X(k) =((111k)3(110k)310

3;ifk= 101;:::110,

0;otherwise.

(An alternative solution is based on the notion of a CDF, which will be introduced in

Chapter 3.)

(b) SinceXiis uniformly distributed over the integers in the range [101;110], we have E[Xi] = (101 + 110)=2 = 105:5. The expected value ofXis

E[X] =1X

k=1k
pX(k) =110X k=101k
px(k) =110X k=101k
(111k)3(110k)310 3: The above expression can be evaluated to be equal to 103.025. The expected improve- ment is therefore 105.5 - 103.025 = 2.475. Solution to Problem 2.31.The marginal PMFpYis given by the binomial formula p

Y(y) =4

y 16  y56  4y ; y= 0;1;:::;4: To compute the conditional PMFpXjY, note that given thatY=y,Xis the number of 1's in the remaining 4yrolls, each of which can take the 5 values 1;3;4;5;6 with equal probability 1/5. Thus, the conditional PMFpXjYis binomial with parameters

4yandp= 1=5:

p

XjY(xjy) =4y

x 15  x45  4yx ; for all nonnegative integersxandysuch that 0x+y4. The joint PMF is now given by p

X;Y(x;y) =pY(y)pXjY(xjy)

=4 y 16  y56 

4y4y

x 15  x45  4yx ; for all nonnegative integersxandysuch that 0x+y4. For other values ofxand y, we havepX;Y(x;y) = 0. Solution to Problem 2.32.LetXibe the random variable taking the value 1 or 0 depending on whether the rst partner of theith couple has survived or not. LetYi be the corresponding random variable for the second partner of theith couple. Then, we haveS=Pm i=1XiYi, and by using the total expectation theorem,

E[SjA=a] =mX

i=1E[XiYijA=a] =mE[X1Y1jA=a] =mE[Y1jX1= 1; A=a]P(X1= 1jA=a) =mP(Y1= 1jX1= 1; A=a)P(X1= 1jA=a): 31

We have

P(Y1= 1jX1= 1; A=a) =a12m1;P(X1= 1jA=a) =a2m:

Thus

E[SjA=a] =ma12m1
a2m=a(a1)2(2m1):

Note thatE[SjA=a] does not depend onp.

Solution to Problem 2.38.(a) LetXbe the number of red lights that Alice encounters. The PMF ofXis binomial withn= 4 andp= 1=2. The mean and the variance ofXareE[X] =np= 2 and var(X) =np(1p) = 4
(1=2)
(1=2) = 1. (b) The variance of Alice's commuting time is the same as the variance of the time by which Alice is delayed by the red lights. This is equal to the variance of 2X, which is

4var(X) = 4.

Solution to Problem 2.39.LetXibe the number of eggs Harry eats on dayi. Then, theXiare independent random variables, uniformly distributed over the set f1;:::;6g. We haveX=P10 i=1Xi, and

E[X] =E

10X i=1X i! =10X i=1E[Xi] = 35:

Similarly, we have

var(X) = var 10X i=1X i! =10X i=1var(Xi); since theXiare independent. Using the formula of Example 2.6, we have var(Xi) =(61)(61 + 2)12 2:9167; so that var(X)29:167. Solution to Problem 2.40.Associate a success with a paper that receives a grade that has not been received before. LetXibe the number of papers between theith success and the (i+ 1)st success. Then we haveX= 1 +P5 i=1Xiand hence

E[X] = 1 +5X

i=1E[Xi]: After receivingi1 dierent grades so far (i1 successes), each subsequent paper has probability (6i)=6 of receiving a grade that has not been received before. Therefore, the random variableXiis geometric with parameterpi= (6i)=6, soE[Xi] = 6=(6i).

It follows that

E[X] = 1 +5X

i=166i= 1 + 65X i=11i = 14:7: 32
Solution to Problem 2.41.(a) The PMF ofXis the binomial PMF with parameters p= 0:02 andn= 250. The mean isE[X] =np= 250
0:02 = 5. The desired probability is

P(X= 5) =250

5 (0:02)5(0:98)245= 0:1773: (b) The Poisson approximation has parameter=np= 5, so the probability in (a) is approximated by e 55! = 0:1755: (c) LetYbe the amount of money you pay in trac tickets during the year. Then

E[Y] =5X

i=150
E[Yi]; whereYiis the amount of money you pay on theith day. The PMF ofYiis

P(Yi=y) =8

>< > :0:98;ify= 0,

0:01;ify= 10,

0:006;ify= 20,

0:004;ify= 50.

The mean is

E[Yi] = 0:01
10 + 0:006
20 + 0:004
50 = 0:42:

The variance is

var(Yi) =E[Y2i]E[Yi]

2= 0:01
(10)2+0:006
(20)2+0:004
(50)2(0:42)2= 13:22:

The mean ofYis

E[Y] = 250
E[Yi] = 105;

and using the independence of the random variablesYi, the variance ofYis var(Y) = 250
var(Yi) = 3;305: (d) The variance of the sample mean is p(1p)250 so assuming thatjp^pjis within 5 times the standard deviation, the possible values ofpare those that satisfyp2[0;1] and (p0:02)225p(1p)250 : 33
This is a quadratic inequality that can be solved for the interval of values ofp. After some calculation, the inequality can be written as 275p235p+ 0:10, which holds if and only ifp2[0:0025;0:1245].

Solution to Problem 2.42.(a) Noting that

P(Xi= 1) =Area(S)Area

[0;1][0;1]
= Area(S); we obtain

E[Sn] =E"

1n n X i=1X i# = 1n n X i=1E[Xi] =E[Xi] = Area(S); and var(Sn) = var 1n n X i=1X i! = 1n 2n X i=1var(Xi) =1n var(Xi) =1n 1Area(S)
Area(S); which tends to zero asntends to innity. (b) We have S n=n1n

Sn1+1n

Xn: (c) We can generateS10000(up to a certain precision) as follows :

1:InitializeSto zero.

2:Fori= 1 to 10000

3:Randomly select two real numbersaandb(up to a certain precision)

independently and uniformly from the interval [0;1].

4:If (a0:5)2+ (b0:5)2<0:25, setxto 1 else setxto 0.

5:SetS:= (i1)S=i+x=i :

6:ReturnS.

By running the above algorithm, a value ofS10000equal to 0.7783 was obtained (the exact number depends on the random number generator). We know from part (a) that the variance ofSntends to zero asntends to innity, so the obtained value ofS10000 is an approximation ofE[S10000]. ButE[S10000] = Area(S) ==4, this leads us to the following approximation of:

4
0:7783 = 3:1132:

(d) We only need to modify the test done at step 4. We have to test whether or not

0cosa+ sinb1. The obtained approximation of the area was 0.3755.

34

C H A P T E R 3

Solution to Problem 3.1.The random variableY=g(X) is discrete and its PMF is given by p

Y(1) =P(X1=3) = 1=3; pY(2) = 1pY(1) = 2=3:

Thus,

E[Y] =13

1 +23
2 =53 : The same result is obtained using the expected value rule:

E[Y] =Z

1 0 g(x)fX(x)dx=Z 1=3 0 dx+Z 1

1=32dx=53

:

Solution to Problem 3.2.We have

Z 1 1 f

X(x)dx=Z

1 12 ejxjdx= 2
12 Z 1 0 exdx= 2
12 = 1; where we have used the fact R1

0exdx= 1;i.e., the normalization property of the

exponential PDF. By symmetry of the PDF, we haveE[X] = 0. We also have

E[X2] =Z

1 1 x22 ejxjdx=Z 1 0 x2exdx=2 2; where we have used the fact that the second moment of the exponential PDF is 2=2. Thus var(X) =E[X2]E[X]

2= 2=2:

Solution to Problem 3.5.LetA=bh=2 be the area of the given triangle, where bis the length of the base, andhis the height of the triangle. From the randomly chosen point, draw a line parallel to the base, and letAxbe the area of the triangle thus formed. The height of this triangle ishxand its base has lengthb(hx)=h.

ThusAx=b(hx)2=(2h). Forx2[0;h], we have

F

X(x) = 1P(X > x) = 1AxA

= 1b(hx)2=(2h)bh=2= 1hxh  2 ; whileFX(x) = 0 forx <0 andFX(x) = 1 forx > h. The PDF is obtained by dierentiating the CDF. We have f

X(x) =dFXdx

(x) =(

2(hx)h

2;if 0xh,

0;otherwise.

35
Solution to Problem 3.6.LetXbe the waiting time andYbe the number of customers found. Forx <0, we haveFX(x) = 0, while forx0, F

X(x) =P(Xx) =12

P(XxjY= 0) +12

P(XxjY= 1):

Since

P(XxjY= 0) = 1;

P(XxjY= 1) = 1ex;

we obtain F

X(x) =(

12 (2ex);ifx0,

0;otherwise.

Note that the CDF has a discontinuity atx= 0. The random variableXis neither discrete nor continuous. Solution to Problem 3.7.(a) We rst calculate the CDF ofX. Forx2[0;r], we have F

X(x) =P(Xx) =x2r

2=xr  2: Forx <0, we haveFX(x) = 0, and forx > r, we haveFX(x) = 1. By dierentiating, we obtain the PDF f

X(x) =(

2xr

2;if 0xr,

0;otherwise.

We have

E[X] =Z

r 02x2r

2dx=2r3

: Also

E[X2] =Z

r 02x3r

2dx=r22

; so var(X) =E[X2]E[X]
2=r22 4r29 =r218 : (b) Alvin gets a positive score in the range [1=t;1) if and only ifXt, and otherwise he gets a score of 0. Thus, fors <0, the CDF ofSisFS(s) = 0. For 0s <1=t, we have F S(s) =P(Ss) =P(Alvin's hit is outside the inner circle) = 1P(Xt) = 1t2r 2:

For 1=t < s, the CDF ofSis given by

F S(s) =P(Ss) =P(Xt)P(SsjXt) +P(X > t)P(SsjX > t): 36

We have

P(Xt) =t2r

2;P(X > t) = 1t2r

2; and sinceS= 0 whenX > t,

P(SsjX > t) = 1:

Furthermore,

P(SsjXt) =P(1=XsjXt) =P(1=sXt)P(Xt)=t

2(1=s)2r

2t 2r

2= 11s

2t2:

Combining the above equations, we obtain

P(Ss) =t2r

2 11s 2t2 + 1t2r

2= 11s

2r2: Collecting the results of the preceding calculations, the CDF ofSis F

S(s) =8

>>< > >:0;ifs <0, 1t2r

2;if 0s <1=t,

11s

2r2;if 1=ts.

BecauseFShas a discontinuity ats= 0, the random variableSis not continuous. Solution to Problem 3.8.(a) By the total probability theorem, we have F X(x) =P(Xx) =pP(Yx) + (1p)P(Zx) =pFY(x) + (1p)FZ(x):

By dierentiating, we obtain

f

X(x) =pfY(x) + (1p)fZ(x):

(b) Consider the random variableYthat has PDF f

Y(y) =

ey;ify <0

0;otherwise,

and the random variableZthat has PDF f

Z(z) =

ez;ify0

0;otherwise.

We note that the random variablesYandZare exponential. Using the CDF of the exponential random variable, we see that the CDFs ofYandZare given by F

Y(y) =

ey;ify <0,

1;ify0,

37
F

Z(z) =n0;ifz <0,

1ez;ifz0.

We havefX(x) =pfY(x) + (1p)fZ(x), and consequentlyFX(x) =pFY(x) + (1 p)FZ(x). It follows that F

X(x) =pex;ifx <0,

p+ (1p)(1ex);ifx0, =pex;ifx <0,

1(1p)ex;ifx0.

Solution to Problem 3.11.(a)Xis a standard normal, so by using the normal table, we haveP(X1:5) = (1:5) = 0:9332. AlsoP(X 1) = 1(1) =

10:8413 = 0:1587:

(b) The random variable (Y1)=2 is obtained by subtracting fromYits mean (which is 1) and dividing by the standard deviation (which is 2), so the PDF of (Y1)=2 is the standard normal. (c) We have, using the normal table,

P(1Y1) =P1(Y1)=20

=P(1Z0) =P(0Z1) = (1)(0) = 0:84130:5 = 0:3413; whereZis a standard normal random variable. Solution to Problem 3.12.The random variableZ=X=is a standard normal, so

P(Xk) =P(Zk) = 1(k):

From the normal tables we have

(1) = 0:8413;(2) = 0:9772;(3) = 0:9986: ThusP(X) = 0:1587,P(X2) = 0:0228,P(X3) = 0:0014.

We also have

P jXj k
=PjZj k
= (k)P(Z k) = (k)1(k)
= 2(k)1:

Using the normal table values above, we obtain

P(jXj ) = 0:6826;P(jXj 2) = 0:9544;P(jXj 3) = 0:9972; wheretis a standard normal random variable. 38
Solution to Problem 3.13.LetXandYbe the temperature in Celsius and Fahrenheit, respectively, which are related byX= 5(Y32)=9. Therefore, 59 degrees Fahrenheit correspond to 15 degrees Celsius. So, ifZis a standard normal random variable, we have usingE[X] =X= 10,

P(Y59) =P(X15) =P

Z15E[X]

X =P(Z0:5) = (0:5): From the normal tables we have (0:5) = 0:6915, soP(Y59) = 0:6915: Solution to Problem 3.15.(a) Since the area of the semicircle isr2=2, the joint PDF ofXandYisfX;Y(x;y) = 2=r2, for (x;y) in the semicircle, andfX;Y(x;y) = 0, otherwise. (b) To nd the marginal PDF ofY, we integrate the joint PDF over the range of X. For any possible valueyofY, the range of possible values ofXis the interval [pr

2y2;pr

2y2], and we have

f

Y(y) =Z

pr 2y2 pr

2y22r

2dx=8 < :4pr

2y2r

2;if 0yr,

0;otherwise.

Thus,

E[Y] =4r

2Z r 0 ypr

2y2dy=4r3;

where the integration is performed using the substitutionz=r2y2. (c) There is no need to nd the marginal PDFfYin order to ndE[Y]. LetDdenote the semicircle. We have, using polar coordinates

E[Y] =Z Z

(x;y)2Dyf

X;Y(x;y)dxdy=Z

 0Z r 02r

2s(sin)sdsd=4r3:

Solution to Problem 3.16.LetAbe the event that the needle will cross a horizontal line, and letBbe the probability that it will cross a vertical line. From the analysis of

Example 3.11, we have that

P(A) =2la

;P(B) =2lb : Since at most one horizontal (or vertical) line can be crossed, the expected number of horizontal lines crossed isP(A) [orP(B), respectively]. Thus the expected number of crossed lines is

P(A) +P(B) =2la

+2lb =2l(a+b)ab : The probability that at least one line will be crossed is

P(A[B) =P(A) +P(B)P(A\B):

39
LetX(orY) be the distance from the needle's center to the nearest horizontal (or vertical) line. Let  be the angle formed by the needle's axis and the horizontal lines as in Example 3.11. We have

P(A\B) =P

Xlsin2

; Ylcos2  : We model the triple (X;Y;) as uniformly distributed over the set of all (x;y;) that satisfy 0xa=2, 0yb=2, and 0=2. Hence, within this set, we have f

X;Y;(x;y;) =8ab

:

The probabilityP(A\B) is

P X(l=2)sin; Y(l=2)cos
=Z Z x(l=2) sin y(l=2) cosf

X;Y;(x;y;)dxdy d

= 8ab Z =2 0Z (l=2) cos 0Z (l=2) sin 0 dxdy d =

2l2ab

Z =2 0 cossind = l2ab :

Thus we have

P(A[B) =P(A) +P(B)P(A\B) =2la

+2lb l2ab =lab 2(a+b)l
:

Solution to Problem 3.18.(a) We have

E[X] =Z

3 1x 24
dx=x312  3 1 =2712 112 =2612 =136 ;

P(A) =Z

3 2x4 dx=x28  3 2 =98 48 =58 :

We also have

f

XjA(x) =(

fX(x)P(A);ifx2A,

0;otherwise,

= ( 2x5 ;if 2x3,

0;otherwise,

40
from which we obtain

E[XjA] =Z

3 2 x
2x5 dx=2x315  3 2 =5415 1615 =3815 : (b) We have

E[Y] =E[X2] =Z

3 1x 34
dx= 5; and

E[Y2] =E[X4] =Z

3 1x 54
dx=913 : Thus, var(Y) =E[Y2]E[Y]
2=913 52=163 : Solution to Problem 3.19.(a) We have, using the normalization property, Z 2 1 cx2dx= 1; or c=1Z 2 1 x2dx= 2: (b) We have

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