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**Probability and Stochastic Processes
**

A Friendly Introduction for Electrical and Computer Engineers

SECOND EDITION

**Problem Solutions
September 28, 2005 Draft
**

Roy D. Yates, David J. Goodman, David Famolari

September 28, 2005

*• *This solution manual remains under construction. The current count is that 678 (out of 687)
problems have solutions. The unsolved problems are

12.1.7, 12.1.8, 12.5.8, 12.5.9, 12.11.5 – 12.11.9.

If you volunteer a solution for one of those problems, we’ll be happy to include it . . . and, of course, “your wildest dreams will come true.”

*• *Of course, the correctness of every single solution reamins unconfirmed. If you find errors or
have suggestions or comments, please send email: ** ryates@winlab.rutgers.edu**.

*• *If you need to make solution sets for your class, you might like the *Solution Set Constructor
*at the instructors site *www.winlab.rutgers.edu/probsolns*. If you need access, send email:
** ryates@winlab.rutgers.edu**.

*• *Matlab functions written as solutions to homework problems can be found in the archive
matsoln.zip (available to instructors) or in the directory matsoln. Other Matlab functions
used in the text or in these homework solutions can be found in the archive matcode.zip
or directory matcode. The .m files in matcode are available for download from the Wiley
website. Two other documents of interest are also available for download:

**– **A manual probmatlab.pdf describing the matcode .m functions is also available.

**– **The quiz solutions manual quizsol.pdf.

*• *A web-based solution set constructor for the second edition is available to instructors at
*http://www.winlab.rutgers.edu/probsolns
*

*• ***The next update of this solution manual is likely to occur in January, 2006.
**

1

**Problem Solutions – Chapter 1
**

**Problem 1.1.1 Solution
**Based on the Venn diagram

*M O
*

*T
*

the answers are fairly straightforward:

(a) Since *T ∩M *= *φ*, *T *and *M *are not mutually exclusive.
(b) Every pizza is either Regular (*R*), or Tuscan (*T *). Hence *R ∪ T *= *S *so that *R *and *T *are

collectively exhaustive. Thus its also (trivially) true that *R ∪ T ∪M *= *S*. That is, *R*, *T *and
*M *are also collectively exhaustive.

(c) From the Venn diagram, *T *and *O *are mutually exclusive. In words, this means that Tuscan
pizzas never have onions or pizzas with onions are never Tuscan. As an aside, “Tuscan” is
a fake pizza designation; one shouldn’t conclude that people from Tuscany actually dislike
onions.

(d) From the Venn diagram, *M ∩T *and *O *are mutually exclusive. Thus Gerlanda’s doesn’t make
Tuscan pizza with mushrooms and onions.

(e) Yes. In terms of the Venn diagram, these pizzas are in the set (*T ∪M ∪O*)*c*.

**Problem 1.1.2 Solution
**Based on the Venn diagram,

*M O
*

*T
*

the complete Gerlandas pizza menu is
*• *Regular without toppings
*• *Regular with mushrooms
*• *Regular with onions
*• *Regular with mushrooms and onions
*• *Tuscan without toppings
*• *Tuscan with mushrooms

**Problem 1.2.1 Solution
**

(a) An outcome specifies whether the fax is high (*h*), medium (*m*), or low (*l*) speed, and whether
the fax has two (*t*) pages or four (*f*) pages. The sample space is

*S *= *{ht, hf,mt,mf, lt, lf} . *(1)

2

(b) The event that the fax is medium speed is *A*1 = *{mt,mf}*.
(c) The event that a fax has two pages is *A*2 = *{ht,mt, lt}*.
(d) The event that a fax is either high speed or low speed is *A*3 = *{ht, hf, lt, lf}*.
(e) Since *A*1 *∩A*2 = *{mt} *and is not empty, *A*1, *A*2, and *A*3 are not mutually exclusive.
(f) Since

*A*1 *∪A*2 *∪A*3 = *{ht, hf,mt,mf, lt, lf} *= *S, *(2)
the collection *A*1, *A*2, *A*3 is collectively exhaustive.

**Problem 1.2.2 Solution
**

(a) The sample space of the experiment is

*S *= *{aaa, aaf, afa, faa, ffa, faf, aff, fff} . *(1)

(b) The event that the circuit from *Z *fails is

*ZF *= *{aaf, aff, faf, fff} . *(2)

The event that the circuit from *X *is acceptable is

*XA *= *{aaa, aaf, afa, aff} . *(3)

(c) Since *ZF ∩XA *= *{aaf, aff} *= *φ*, *ZF *and *XA *are not mutually exclusive.
(d) Since *ZF ∪XA *= *{aaa, aaf, afa, aff, faf, fff} *= *S*, *ZF *and *XA *are not collectively exhaus-

tive.

(e) The event that more than one circuit is acceptable is

*C *= *{aaa, aaf, afa, faa} . *(4)

The event that at least two circuits fail is

*D *= *{ffa, faf, aff, fff} . *(5)

(f) Inspection shows that *C ∩D *= *φ *so *C *and *D *are mutually exclusive.
(g) Since *C ∪D *= *S*, *C *and *D *are collectively exhaustive.

**Problem 1.2.3 Solution
**The sample space is

*S *= *{A♣, . . . ,K♣, A♦, . . . ,K♦, A♥, . . . ,K♥, A♠, . . . ,K♠} . *(1)

The event *H *is the set
*H *= *{A♥, . . . ,K♥} . *(2)

3

**Problem 1.2.4 Solution
**The sample space is

*S *=

⎧⎨ ⎩

1*/*1 *. . . *1*/*31*, *2*/*1 *. . . *2*/*29*, *3*/*1 *. . . *3*/*31*, *4*/*1 *. . . *4*/*30*,
*5*/*1 *. . . *5*/*31*, *6*/*1 *. . . *6*/*30*, *7*/*1 *. . . *7*/*31*, *8*/*1 *. . . *8*/*31*,
*9*/*1 *. . . *9*/*31*, *10*/*1 *. . . *10*/*31*, *11*/*1 *. . . *11*/*30*, *12*/*1 *. . . *12*/*31

⎫⎬
⎭ *. *(1)

The event *H *defined by the event of a July birthday is described by following 31 sample points.

*H *= *{*7*/*1*, *7*/*2*, . . . , *7*/*31*} . *(2)

**Problem 1.2.5 Solution
**Of course, there are many answers to this problem. Here are four event spaces.

1. We can divide students into engineers or non-engineers. Let *A*1 equal the set of engineering
students and *A*2 the non-engineers. The pair *{A*1*, A*2*} *is an event space.

2. We can also separate students by GPA. Let *Bi *denote the subset of students with GPAs *G
*satisfying *i − *1 *≤ G < i*. At Rutgers, *{B*1*, B*2*, . . . , B*5*} *is an event space. Note that *B*5 is
the set of all students with perfect 4.0 GPAs. Of course, other schools use different scales for
GPA.

3. We can also divide the students by age. Let *Ci *denote the subset of students of age *i *in years.
At most universities, *{C*10*, C*11*, . . . , C*100*} *would be an event space. Since a university may
have prodigies either under 10 or over 100, we note that *{C*0*, C*1*, . . .} *is always an event space

4. Lastly, we can categorize students by attendance. Let *D*0 denote the number of students who
have missed zero lectures and let *D*1 denote all other students. Although it is likely that *D*0
is an empty set, *{D*0*, D*1*} *is a well defined event space.

**Problem 1.2.6 Solution
**Let *R*1 and *R*2 denote the measured resistances. The pair (*R*1*, R*2) is an outcome of the experiment.
Some event spaces include

1. If we need to check that neither resistance is too high, an event space is

*A*1 = *{R*1 *< *100*, R*2 *< *100*} , A*2 = *{*either *R*1 *≥ *100 or *R*2 *≥ *100*} . *(1)
2. If we need to check whether the first resistance exceeds the second resistance, an event space

is
*B*1 = *{R*1 *> R*2*} B*2 = *{R*1 *≤ R*2*} . *(2)

3. If we need to check whether each resistance doesn’t fall below a minimum value (in this case
50 ohms for *R*1 and 100 ohms for *R*2), an event space is

*C*1 = *{R*1 *< *50*, R*2 *< *100*} , C*2 = *{R*1 *< *50*, R*2 *≥ *100*} , *(3)
*C*3 = *{R*1 *≥ *50*, R*2 *< *100*} , C*4 = *{R*1 *≥ *50*, R*2 *≥ *100*} . *(4)

4. If we want to check whether the resistors in parallel are within an acceptable range of 90 to 110 ohms, an event space is

*D*1 =
{
(1*/R*1 + 1*/R*2)*−*1 *< *90

}
*, *(5)

*D*2 =
{
90 *≤ *(1*/R*1 + 1*/R*2)*−*1 *≤ *110

}
*, *(6)

*D*2 =
{
110 *< *(1*/R*1 + 1*/R*2)*−*1

}
*. *(7)

4

**Problem 1.3.1 Solution
**The sample space of the experiment is

*S *= *{LF,BF,LW,BW} . *(1)

From the problem statement, we know that *P *[*LF *] = 0*.*5, *P *[*BF *] = 0*.*2 and *P *[*BW *] = 0*.*2. This
implies *P *[*LW *] = 1*− *0*.*5*− *0*.*2*− *0*.*2 = 0*.*1. The questions can be answered using Theorem 1.5.
(a) The probability that a program is slow is

*P *[*W *] = *P *[*LW *] + *P *[*BW *] = 0*.*1 + 0*.*2 = 0*.*3*. *(2)

(b) The probability that a program is big is

*P *[*B*] = *P *[*BF *] + *P *[*BW *] = 0*.*2 + 0*.*2 = 0*.*4*. *(3)

(c) The probability that a program is slow or big is

*P *[*W ∪B*] = *P *[*W *] + *P *[*B*]*− P *[*BW *] = 0*.*3 + 0*.*4*− *0*.*2 = 0*.*5*. *(4)

**Problem 1.3.2 Solution
**A sample outcome indicates whether the cell phone is handheld (*H*) or mobile (*M*) and whether
the speed is fast (*F *) or slow (*W *). The sample space is

*S *= *{HF,HW,MF,MW} . *(1)

The problem statement tells us that *P *[*HF *] = 0*.*2, *P *[*MW *] = 0*.*1 and *P *[*F *] = 0*.*5. We can use
these facts to find the probabilities of the other outcomes. In particular,

*P *[*F *] = *P *[*HF *] + *P *[*MF *] *. *(2)

This implies
*P *[*MF *] = *P *[*F *]*− P *[*HF *] = 0*.*5*− *0*.*2 = 0*.*3*. *(3)

Also, since the probabilities must sum to 1,

*P *[*HW *] = 1*− P *[*HF *]*− P *[*MF *]*− P *[*MW *] = 1*− *0*.*2*− *0*.*3*− *0*.*1 = 0*.*4*. *(4)

Now that we have found the probabilities of the outcomes, finding any other probability is easy.

(a) The probability a cell phone is slow is

*P *[*W *] = *P *[*HW *] + *P *[*MW *] = 0*.*4 + 0*.*1 = 0*.*5*. *(5)

(b) The probability that a cell hpone is mobile and fast is *P *[*MF *] = 0*.*3.

(c) The probability that a cell phone is handheld is

*P *[*H*] = *P *[*HF *] + *P *[*HW *] = 0*.*2 + 0*.*4 = 0*.*6*. *(6)

5

**Problem 1.3.3 Solution
**A reasonable probability model that is consistent with the notion of a shuffled deck is that each
card in the deck is equally likely to be the first card. Let *Hi *denote the event that the first card
drawn is the *i*th heart where the first heart is the ace, the second heart is the deuce and so on. In
that case, *P *[*Hi*] = 1*/*52 for 1 *≤ i ≤ *13. The event *H *that the first card is a heart can be written
as the disjoint union

*H *= *H*1 *∪H*2 *∪ · · · ∪H*13*. *(1)
Using Theorem 1.1, we have

*P *[*H*] =
13∑
*i*=1

*P *[*Hi*] = 13*/*52*. *(2)

This is the answer you would expect since 13 out of 52 cards are hearts. The point to keep in mind is that this is not just the common sense answer but is the result of a probability model for a shuffled deck and the axioms of probability.

**Problem 1.3.4 Solution
**Let *si *denote the outcome that the down face has *i *dots. The sample space is *S *= *{s*1*, . . . , s*6*}*.
The probability of each sample outcome is *P *[*si*] = 1*/*6. From Theorem 1.1, the probability of the
event *E *that the roll is even is

*P *[*E*] = *P *[*s*2] + *P *[*s*4] + *P *[*s*6] = 3*/*6*. *(1)

**Problem 1.3.5 Solution
**Let *si *equal the outcome of the student’s quiz. The sample space is then composed of all the
possible grades that she can receive.

*S *= *{*0*, *1*, *2*, *3*, *4*, *5*, *6*, *7*, *8*, *9*, *10*} . *(1)
Since each of the 11 possible outcomes is equally likely, the probability of receiving a grade of *i*, for
each *i *= 0*, *1*, . . . , *10 is *P *[*si*] = 1*/*11. The probability that the student gets an A is the probability
that she gets a score of 9 or higher. That is

*P *[Grade of A] = *P *[9] + *P *[10] = 1*/*11 + 1*/*11 = 2*/*11*. *(2)

The probability of failing requires the student to get a grade less than 4.

*P *[Failing] = *P *[3] + *P *[2] + *P *[1] + *P *[0] = 1*/*11 + 1*/*11 + 1*/*11 + 1*/*11 = 4*/*11*. *(3)

**Problem 1.4.1 Solution
**From the table we look to add all the disjoint events that contain *H*0 to express the probability
that a caller makes no hand-offs as

*P *[*H*0] = *P *[*LH*0] + *P *[*BH*0] = 0*.*1 + 0*.*4 = 0*.*5*. *(1)

In a similar fashion we can express the probability that a call is brief by

*P *[*B*] = *P *[*BH*0] + *P *[*BH*1] + *P *[*BH*2] = 0*.*4 + 0*.*1 + 0*.*1 = 0*.*6*. *(2)

The probability that a call is long or makes at least two hand-offs is

*P *[*L ∪H*2] = *P *[*LH*0] + *P *[*LH*1] + *P *[*LH*2] + *P *[*BH*2] (3)
= 0*.*1 + 0*.*1 + 0*.*2 + 0*.*1 = 0*.*5*. *(4)

6

**Problem 1.4.2 Solution
**

(a) From the given probability distribution of billed minutes, *M *, the probability that a call is
billed for more than 3 minutes is

*P *[*L*] = 1*− P *[3 or fewer billed minutes] (1)
= 1*− P *[*B*1]*− P *[*B*2]*− P *[*B*3] (2)
= 1*− α− α*(1*− α*)*− α*(1*− α*)2 (3)
= (1*− α*)3 = 0*.*57*. *(4)

(b) The probability that a call will billed for 9 minutes or less is

*P *[9 minutes or less] =
9∑

*i*=1

*α*(1*− α*)*i−*1 = 1*− *(0*.*57)3*. *(5)

**Problem 1.4.3 Solution
**The first generation consists of two plants each with genotype *yg *or *gy*. They are crossed to produce
the following second generation genotypes, *S *= *{yy, yg, gy, gg}*. Each genotype is just as likely as
any other so the probability of each genotype is consequently 1/4. A pea plant has yellow seeds if
it possesses at least one dominant *y *gene. The set of pea plants with yellow seeds is

*Y *= *{yy, yg, gy} . *(1)

So the probability of a pea plant with yellow seeds is

*P *[*Y *] = *P *[*yy*] + *P *[*yg*] + *P *[*gy*] = 3*/*4*. *(2)

**Problem 1.4.4 Solution
**Each statement is a consequence of part 4 of Theorem 1.4.

(a) Since *A ⊂ A ∪B*, *P *[*A*] *≤ P *[*A ∪B*].
(b) Since *B ⊂ A ∪B*, *P *[*B*] *≤ P *[*A ∪B*].
(c) Since *A ∩B ⊂ A*, *P *[*A ∩B*] *≤ P *[*A*].
(d) Since *A ∩B ⊂ B*, *P *[*A ∩B*] *≤ P *[*B*].

**Problem 1.4.5 Solution
**Specifically, we will use Theorem 1.7(c) which states that for any events *A *and *B*,

*P *[*A ∪B*] = *P *[*A*] + *P *[*B*]*− P *[*A ∩B*] *. *(1)

To prove the union bound by induction, we first prove the theorem for the case of *n *= 2 events. In
this case, by Theorem 1.7(c),

*P *[*A*1 *∪A*2] = *P *[*A*1] + *P *[*A*2]*− P *[*A*1 *∩A*2] *. *(2)

7

By the first axiom of probability, *P *[*A*1 *∩A*2] *≥ *0. Thus,

*P *[*A*1 *∪A*2] *≤ P *[*A*1] + *P *[*A*2] *. *(3)

which proves the union bound for the case *n *= 2. Now we make our induction hypothesis that the
union-bound holds for any collection of *n − *1 subsets. In this case, given subsets *A*1*, . . . , An*, we
define

*A *= *A*1 *∪A*2 *∪ · · · ∪An−*1*, B *= *An. *(4)
By our induction hypothesis,

*P *[*A*] = *P *[*A*1 *∪A*2 *∪ · · · ∪An−*1] *≤ P *[*A*1] + *· · ·*+ *P *[*An−*1] *. *(5)

This permits us to write

*P *[*A*1 *∪ · · · ∪An*] = *P *[*A ∪B*] (6)
*≤ P *[*A*] + *P *[*B*] (by the union bound for *n *= 2) (7)
= *P *[*A*1 *∪ · · · ∪An−*1] + *P *[*An*] (8)
*≤ P *[*A*1] + *· · ·P *[*An−*1] + *P *[*An*] (9)

which completes the inductive proof.

**Problem 1.4.6 Solution
**

(a) For convenience, let *pi *= *P *[*FHi*] and *qi *= *P *[*V Hi*]. Using this shorthand, the six unknowns
*p*0*, p*1*, p*2*, q*0*, q*1*, q*2 fill the table as

*H*0 *H*1 *H*2
*F p*0 *p*1 *p*2
*V q*0 *q*1 *q*2

*. *(1)

However, we are given a number of facts:

*p*0 + *q*0 = 1*/*3*, p*1 + *q*1 = 1*/*3*, *(2)
*p*2 + *q*2 = 1*/*3*, p*0 + *p*1 + *p*2 = 5*/*12*. *(3)

Other facts, such as *q*0 + *q*1 + *q*2 = 7*/*12, can be derived from these facts. Thus, we have
four equations and six unknowns, choosing *p*0 and *p*1 will specify the other unknowns. Un-
fortunately, arbitrary choices for either *p*0 or *p*1 will lead to negative values for the other
probabilities. In terms of *p*0 and *p*1, the other unknowns are

*q*0 = 1*/*3*− p*0*, p*2 = 5*/*12*− *(*p*0 + *p*1)*, *(4)
*q*1 = 1*/*3*− p*1*, q*2 = *p*0 + *p*1 *− *1*/*12*. *(5)

Because the probabilities must be nonnegative, we see that

0 *≤ p*0 *≤ *1*/*3*, *(6)
0 *≤ p*1 *≤ *1*/*3*, *(7)

1*/*12 *≤ p*0 + *p*1 *≤ *5*/*12*. *(8)

8

Although there are an infinite number of solutions, three possible solutions are:

*p*0 = 1*/*3*, p*1 = 1*/*12*, p*2 = 0*, *(9)
*q*0 = 0*, q*1 = 1*/*4*, q*2 = 1*/*3*. *(10)

and

*p*0 = 1*/*4*, p*1 = 1*/*12*, p*2 = 1*/*12*, *(11)
*q*0 = 1*/*12*, q*1 = 3*/*12*, q*2 = 3*/*12*. *(12)

and

*p*0 = 0*, p*1 = 1*/*12*, p*2 = 1*/*3*, *(13)
*q*0 = 1*/*3*, q*1 = 3*/*12*, q*2 = 0*. *(14)

(b) In terms of the *pi, qi *notation, the new facts are *p*0 = 1*/*4 and *q*1 = 1*/*6. These extra facts
uniquely specify the probabilities. In this case,

*p*0 = 1*/*4*, p*1 = 1*/*6*, p*2 = 0*, *(15)
*q*0 = 1*/*12*, q*1 = 1*/*6*, q*2 = 1*/*3*. *(16)

**Problem 1.4.7 Solution
**It is tempting to use the following proof:

Since *S *and *φ *are mutually exclusive, and since *S *= *S ∪ φ*,

1 = *P *[*S ∪ φ*] = *P *[*S*] + *P *[*φ*] *. *(1)

Since *P *[*S*] = 1, we must have *P *[*φ*] = 0.

The above “proof” used the property that for mutually exclusive sets *A*1 and *A*2,

*P *[*A*1 *∪A*2] = *P *[*A*1] + *P *[*A*2] *. *(2)

The problem is that this property is a consequence of the three axioms, and thus must be proven.
For a proof that uses just the three axioms, let *A*1 be an arbitrary set and for *n *= 2*, *3*, . . .*, let
*An *= *φ*. Since *A*1 = *∪∞i*=1*Ai*, we can use Axiom 3 to write

*P *[*A*1] = *P *[*∪∞i*=1*Ai*] = *P *[*A*1] + *P *[*A*2] +
*∞*∑
*i*=3

*P *[*Ai*] *. *(3)

By subtracting *P *[*A*1] from both sides, the fact that *A*2 = *φ *permits us to write

*P *[*φ*] +
*∞*∑
*n*=3

*P *[*Ai*] = 0*. *(4)

By Axiom 1, *P *[*Ai*] *≥ *0 for all *i*. Thus,
∑*∞
*

*n*=3 *P *[*Ai*] *≥ *0. This implies *P *[*φ*] *≤ *0. Since Axiom 1
requires *P *[*φ*] *≥ *0, we must have *P *[*φ*] = 0.

9

**Problem 1.4.8 Solution
**Following the hint, we define the set of events *{Ai|i *= 1*, *2*, . . .} *such that *i *= 1*, . . . ,m*, *Ai *= *Bi *and
for *i > m*, *Ai *= *φ*. By construction, *∪mi*=1*Bi *= *∪∞i*=1*Ai*. Axiom 3 then implies

*P *[*∪mi*=1*Bi*] = *P *[*∪∞i*=1*Ai*] =
*∞*∑
*i*=1

*P *[*Ai*] *. *(1)

For *i > m*, *P *[*Ai*] = *P *[*φ*] = 0, yielding the claim *P *[*∪mi*=1*Bi*] =
∑*m
*

*i*=1 *P *[*Ai*] =
∑*m
*

*i*=1 *P *[*Bi*].
Note that the fact that *P *[*φ*] = 0 follows from Axioms 1 and 2. This problem is more challenging

if you just use Axiom 3. We start by observing

*P *[*∪mi*=1*Bi*] =
*m−*1∑
*i*=1

*P *[*Bi*] +
*∞*∑

*i*=*m
*

*P *[*Ai*] *. *(2)

Now, we use Axiom 3 again on the countably infinite sequence *Am, Am*+1*, . . . *to write

*∞*∑
*i*=*m
*

*P *[*Ai*] = *P *[*Am ∪Am*+1 *∪ · · ·*] = *P *[*Bm*] *. *(3)

Thus, we have used just Axiom 3 to prove Theorem 1.4: *P *[*∪mi*=1*Bi*] =
∑*m
*

*i*=1 *P *[*Bi*]*.
*

**Problem 1.4.9 Solution
**Each claim in Theorem 1.7 requires a proof from which we can check which axioms are used.
However, the problem is somewhat hard because there may still be a simpler proof that uses fewer
axioms. Still, the proof of each part will need Theorem 1.4 which we now prove.

For the mutually exclusive events *B*1*, . . . , Bm*, let *Ai *= *Bi *for *i *= 1*, . . . ,m *and let *Ai *= *φ *for
*i > m*. In that case, by Axiom 3,

*P *[*B*1 *∪B*2 *∪ · · · ∪Bm*] = *P *[*A*1 *∪A*2 *∪ · · ·*] (1)

=
*m−*1∑
*i*=1

*P *[*Ai*] +
*∞*∑

*i*=*m
*

*P *[*Ai*] (2)

=
*m−*1∑
*i*=1

*P *[*Bi*] +
*∞*∑

*i*=*m
*

*P *[*Ai*] *. *(3)

Now, we use Axiom 3 again on *Am, Am*+1*, . . . *to write

*∞*∑
*i*=*m
*

*P *[*Ai*] = *P *[*Am ∪Am*+1 *∪ · · ·*] = *P *[*Bm*] *. *(4)

Thus, we have used just Axiom 3 to prove Theorem 1.4:

*P *[*B*1 *∪B*2 *∪ · · · ∪Bm*] =
*m*∑
*i*=1

*P *[*Bi*] *. *(5)

(a) To show *P *[*φ*] = 0, let *B*1 = *S *and let *B*2 = *φ*. Thus by Theorem 1.4,

*P *[*S*] = *P *[*B*1 *∪B*2] = *P *[*B*1] + *P *[*B*2] = *P *[*S*] + *P *[*φ*] *. *(6)
Thus, *P *[*φ*] = 0. Note that this proof uses only Theorem 1.4 which uses only Axiom 3.

10

(b) Using Theorem 1.4 with *B*1 = *A *and *B*2 = *Ac*, we have

*P *[*S*] = *P *[*A ∪Ac*] = *P *[*A*] + *P *[*Ac*] *. *(7)
Since, Axiom 2 says *P *[*S*] = 1, *P *[*Ac*] = 1*− P *[*A*]. This proof uses Axioms 2 and 3.

(c) By Theorem 1.2, we can write both *A *and *B *as unions of disjoint events:

*A *= (*AB*) *∪ *(*ABc*) *B *= (*AB*) *∪ *(*AcB*)*. *(8)
Now we apply Theorem 1.4 to write

*P *[*A*] = *P *[*AB*] + *P *[*ABc*] *, P *[*B*] = *P *[*AB*] + *P *[*AcB*] *. *(9)

We can rewrite these facts as

*P *[*ABc*] = *P *[*A*]*− P *[*AB*]*, P *[*AcB*] = *P *[*B*]*− P *[*AB*]*. *(10)
Note that so far we have used only Axiom 3. Finally, we observe that *A ∪B *can be written
as the union of mutually exclusive events

*A ∪B *= (*AB*) *∪ *(*ABc*) *∪ *(*AcB*)*. *(11)
Once again, using Theorem 1.4, we have

*P *[*A ∪B*] = *P *[*AB*] + *P *[*ABc*] + *P *[*AcB*] (12)
Substituting the results of Equation (10) into Equation (12) yields

*P *[*A ∪B*] = *P *[*AB*] + *P *[*A*]*− P *[*AB*] + *P *[*B*]*− P *[*AB*] *, *(13)
which completes the proof. Note that this claim required only Axiom 3.

(d) Observe that since *A ⊂ B*, we can write *B *as the disjoint union *B *= *A ∪ *(*AcB*). By
Theorem 1.4 (which uses Axiom 3),

*P *[*B*] = *P *[*A*] + *P *[*AcB*] *. *(14)

By Axiom 1, *P *[*AcB*] *≥ *0, hich implies *P *[*A*] *≤ P *[*B*]. This proof uses Axioms 1 and 3.

**Problem 1.5.1 Solution
**Each question requests a conditional probability.

(a) Note that the probability a call is brief is

*P *[*B*] = *P *[*H*0*B*] + *P *[*H*1*B*] + *P *[*H*2*B*] = 0*.*6*. *(1)

The probability a brief call will have no handoffs is

*P *[*H*0*|B*] = *P *[*H*0*B*]
*P *[*B*]

=
0*.*4
0*.*6

=
2
3
*. *(2)

(b) The probability of one handoff is *P *[*H*1] = *P *[*H*1*B*] + *P *[*H*1*L*] = 0*.*2. The probability that a
call with one handoff will be long is

*P *[*L|H*1] = *P *[*H*1*L*]
*P *[*H*1]

=
0*.*1
0*.*2

=
1
2
*. *(3)

(c) The probability a call is long is *P *[*L*] = 1*− P *[*B*] = 0*.*4. The probability that a long call will
have one or more handoffs is

*P *[*H*1 *∪H*2*|L*] = *P *[*H*1*L ∪H*2*L*]
*P *[*L*]

=
*P *[*H*1*L*] + *P *[*H*2*L*]

*P *[*L*]
=

0*.*1 + 0*.*2
0*.*4

=
3
4
*. *(4)

11

**Problem 1.5.2 Solution
**Let *si *denote the outcome that the roll is *i*. So, for 1 *≤ i ≤ *6, *Ri *= *{si}*. Similarly, *Gj *=
*{sj*+1*, . . . , s*6*}*.
(a) Since *G*1 = *{s*2*, s*3*, s*4*, s*5*, s*6*} *and all outcomes have probability 1*/*6, *P *[*G*1] = 5*/*6. The event

*R*3*G*1 = *{s*3*} *and *P *[*R*3*G*1] = 1*/*6 so that

*P *[*R*3*|G*1] = *P *[*R*3*G*1]
*P *[*G*1]

=
1
5
*. *(1)

(b) The conditional probability that 6 is rolled given that the roll is greater than 3 is

*P *[*R*6*|G*3] = *P *[*R*6*G*3]
*P *[*G*3]

=
*P *[*s*6]

*P *[*s*4*, s*5*, s*6]
=

1*/*6
3*/*6

*. *(2)

(c) The event *E *that the roll is even is *E *= *{s*2*, s*4*, s*6*} *and has probability 3*/*6. The joint
probability of *G*3 and *E *is

*P *[*G*3*E*] = *P *[*s*4*, s*6] = 1*/*3*. *(3)

The conditional probabilities of *G*3 given *E *is

*P *[*G*3*|E*] = *P *[*G*3*E*]
*P *[*E*]

=
1*/*3
1*/*2

=
2
3
*. *(4)

(d) The conditional probability that the roll is even given that it’s greater than 3 is

*P *[*E|G*3] = *P *[*EG*3]
*P *[*G*3]

=
1*/*3
1*/*2

=
2
3
*. *(5)

**Problem 1.5.3 Solution
**Since the 2 of clubs is an even numbered card, *C*2 *⊂ E *so that *P *[*C*2*E*] = *P *[*C*2] = 1*/*3. Since
*P *[*E*] = 2*/*3,

*P *[*C*2*|E*] = *P *[*C*2*E*]
*P *[*E*]

=
1*/*3
2*/*3

= 1*/*2*. *(1)

The probability that an even numbered card is picked given that the 2 is picked is

*P *[*E|C*2] = *P *[*C*2*E*]
*P *[*C*2]

=
1*/*3
1*/*3

= 1*. *(2)

**Problem 1.5.4 Solution
**DefineD as the event that a pea plant has two dominant *y *genes. To find the conditional probability
of *D *given the event *Y *, corresponding to a plant having yellow seeds, we look to evaluate

*P *[*D|Y *] = *P *[*DY *]
*P *[*Y *]

*. *(1)

Note that *P *[*DY *] is just the probability of the genotype *yy*. From Problem 1.4.3, we found that
with respect to the color of the peas, the genotypes *yy*, *yg*, *gy*, and *gg *were all equally likely. This
implies

*P *[*DY *] = *P *[*yy*] = 1*/*4 *P *[*Y *] = *P *[*yy, gy, yg*] = 3*/*4*. *(2)

Thus, the conditional probability can be expressed as

*P *[*D|Y *] = *P *[*DY *]
*P *[*Y *]

=
1*/*4
3*/*4

= 1*/*3*. *(3)

12

**Problem 1.5.5 Solution
**The sample outcomes can be written *ijk *where the first card drawn is *i*, the second is *j *and the
third is *k*. The sample space is

*S *= *{*234*, *243*, *324*, *342*, *423*, *432*} . *(1)
and each of the six outcomes has probability 1*/*6. The events *E*1*, E*2*, E*3*, O*1*, O*2*, O*3 are

*E*1 = *{*234*, *243*, *423*, *432*} , O*1 = *{*324*, *342*} , *(2)
*E*2 = *{*243*, *324*, *342*, *423*} , O*2 = *{*234*, *432*} , *(3)
*E*3 = *{*234*, *324*, *342*, *432*} , O*3 = *{*243*, *423*} . *(4)

(a) The conditional probability the second card is even given that the first card is even is

*P *[*E*2*|E*1] = *P *[*E*2*E*1]
*P *[*E*1]

=
*P *[243*, *423]

*P *[234*, *243*, *423*, *432]
=

2*/*6
4*/*6

= 1*/*2*. *(5)

(b) The conditional probability the first card is even given that the second card is even is

*P *[*E*1*|E*2] = *P *[*E*1*E*2]
*P *[*E*2]

=
*P *[243*, *423]

*P *[243*, *324*, *342*, *423]
=

2*/*6
4*/*6

= 1*/*2*. *(6)

(c) The probability the first two cards are even given the third card is even is

*P *[*E*1*E*2*|E*3] = *P *[*E*1*E*2*E*3]
*P *[*E*3]

= 0*. *(7)

(d) The conditional probabilities the second card is even given that the first card is odd is

*P *[*E*2*|O*1] = *P *[*O*1*E*2]
*P *[*O*1]

=
*P *[*O*1]
*P *[*O*1]

= 1*. *(8)

(e) The conditional probability the second card is odd given that the first card is odd is

*P *[*O*2*|O*1] = *P *[*O*1*O*2]
*P *[*O*1]

= 0*. *(9)

**Problem 1.5.6 Solution
**The problem statement yields the obvious facts that *P *[*L*] = 0*.*16 and *P *[*H*] = 0*.*10. The words
“10% of the ticks that had either Lyme disease or HGE carried both diseases” can be written as

*P *[*LH|L ∪H*] = 0*.*10*. *(1)
(a) Since *LH ⊂ L ∪H*,

*P *[*LH|L ∪H*] = *P *[*LH ∩ *(*L ∪H*)]
*P *[*L ∪H*] =

*P *[*LH*]
*P *[*L ∪H*] = 0*.*10*. *(2)

Thus,
*P *[*LH*] = 0*.*10*P *[*L ∪H*] = 0*.*10 (*P *[*L*] + *P *[*H*]*− P *[*LH*]) *. *(3)

Since *P *[*L*] = 0*.*16 and *P *[*H*] = 0*.*10,

*P *[*LH*] =
0*.*10 (0*.*16 + 0*.*10)

1*.*1
= 0*.*0236*. *(4)

13

(b) The conditional probability that a tick has HGE given that it has Lyme disease is

*P *[*H|L*] = *P *[*LH*]
*P *[*L*]

=
0*.*0236
0*.*16

= 0*.*1475*. *(5)

**Problem 1.6.1 Solution
**This problems asks whether *A *and *B *can be independent events yet satisfy *A *= *B*? By definition,
events *A *and *B *are independent if and only if *P *[*AB*] = *P *[*A*]*P *[*B*]. We can see that if *A *= *B*, that
is they are the same set, then

*P *[*AB*] = *P *[*AA*] = *P *[*A*] = *P *[*B*] *. *(1)

Thus, for *A *and *B *to be the same set and also independent,

*P *[*A*] = *P *[*AB*] = *P *[*A*]*P *[*B*] = (*P *[*A*])2 *. *(2)

There are two ways that this requirement can be satisfied:

*• P *[*A*] = 1 implying *A *= *B *= *S*.
*• P *[*A*] = 0 implying *A *= *B *= *φ*.

**Problem 1.6.2 Solution
**

*A
*

*B
*

In the Venn diagram, assume the sample space has area 1 correspond-
ing to probability 1. As drawn, both *A *and *B *have area 1*/*4 so that
*P *[*A*] = *P *[*B*] = 1*/*4. Moreover, the intersection *AB *has area 1*/*16
and covers 1*/*4 of *A *and 1*/*4 of *B*. That is, *A *and *B *are independent
since

*P *[*AB*] = *P *[*A*]*P *[*B*] *. *(1)

**Problem 1.6.3 Solution
**

(a) Since *A *and *B *are disjoint, *P *[*A ∩B*] = 0. Since *P *[*A ∩B*] = 0,

*P *[*A ∪B*] = *P *[*A*] + *P *[*B*]*− P *[*A ∩B*] = 3*/*8*. *(1)

A Venn diagram should convince you that *A ⊂ Bc *so that *A ∩Bc *= *A*. This implies

*P *[*A ∩Bc*] = *P *[*A*] = 1*/*4*. *(2)

It also follows that *P *[*A ∪Bc*] = *P *[*Bc*] = 1*− *1*/*8 = 7*/*8.
(b) Events *A *and *B *are dependent since *P *[*AB*] = *P *[*A*]*P *[*B*].

14

(c) Since *C *and *D *are independent,

*P *[*C ∩D*] = *P *[*C*]*P *[*D*] = 15*/*64*. *(3)
The next few items are a little trickier. From Venn diagrams, we see

*P *[*C ∩Dc*] = *P *[*C*]*− P *[*C ∩D*] = 5*/*8*− *15*/*64 = 25*/*64*. *(4)
It follows that

*P *[*C ∪Dc*] = *P *[*C*] + *P *[*Dc*]*− P *[*C ∩Dc*] (5)
= 5*/*8 + (1*− *3*/*8)*− *25*/*64 = 55*/*64*. *(6)

Using DeMorgan’s law, we have

*P *[*Cc ∩Dc*] = *P *[(*C ∪D*)*c*] = 1*− P *[*C ∪D*] = 15*/*64*. *(7)
(d) Since *P *[*CcDc*] = *P *[*Cc*]*P *[*Dc*], *Cc *and *Dc *are independent.

**Problem 1.6.4 Solution
**

(a) Since *A ∩B *= *∅*, *P *[*A ∩B*] = 0. To find *P *[*B*], we can write
*P *[*A ∪B*] = *P *[*A*] + *P *[*B*]*− P *[*A ∩B*] (1)

5*/*8 = 3*/*8 + *P *[*B*]*− *0*. *(2)
Thus, *P *[*B*] = 1*/*4. Since *A *is a subset of *Bc*, *P *[*A ∩Bc*] = *P *[*A*] = 3*/*8. Furthermore, since
*A *is a subset of *Bc*, *P *[*A ∪Bc*] = *P *[*Bc*] = 3*/*4.

(b) The events *A *and *B *are dependent because

*P *[*AB*] = 0 = 3*/*32 = *P *[*A*]*P *[*B*] *. *(3)
(c) Since *C *and *D *are independent *P *[*CD*] = *P *[*C*]*P *[*D*]. So

*P *[*D*] =
*P *[*CD*]
*P *[*C*]

=
1*/*3
1*/*2

= 2*/*3*. *(4)

In addition, *P *[*C ∩Dc*] = *P *[*C*] *− P *[*C ∩D*] = 1*/*2 *− *1*/*3 = 1*/*6. To find *P *[*Cc ∩Dc*], we first
observe that

*P *[*C ∪D*] = *P *[*C*] + *P *[*D*]*− P *[*C ∩D*] = 1*/*2 + 2*/*3*− *1*/*3 = 5*/*6*. *(5)
By De Morgan’s Law, *Cc ∩Dc *= (*C ∪D*)*c*. This implies

*P *[*Cc ∩Dc*] = *P *[(*C ∪D*)*c*] = 1*− P *[*C ∪D*] = 1*/*6*. *(6)
Note that a second way to find *P *[*Cc ∩Dc*] is to use the fact that if *C *and *D *are independent,
then *Cc *and *Dc *are independent. Thus

*P *[*Cc ∩Dc*] = *P *[*Cc*]*P *[*Dc*] = (1*− P *[*C*])(1*− P *[*D*]) = 1*/*6*. *(7)
Finally, since *C *and *D *are independent events, *P *[*C|D*] = *P *[*C*] = 1*/*2.

(d) Note that we found *P *[*C ∪D*] = 5*/*6. We can also use the earlier results to show
*P *[*C ∪Dc*] = *P *[*C*] + *P *[*D*]*− P *[*C ∩Dc*] = 1*/*2 + (1*− *2*/*3)*− *1*/*6 = 2*/*3*. *(8)

(e) By Definition 1.7, events *C *and *Dc *are independent because

*P *[*C ∩Dc*] = 1*/*6 = (1*/*2)(1*/*3) = *P *[*C*]*P *[*Dc*] *. *(9)

15

**Problem 1.6.5 Solution
**For a sample space *S *= *{*1*, *2*, *3*, *4*} *with equiprobable outcomes, consider the events

*A*1 = *{*1*, *2*} A*2 = *{*2*, *3*} A*3 = *{*3*, *1*} . *(1)

Each event *Ai *has probability 1*/*2. Moreover, each pair of events is independent since

*P *[*A*1*A*2] = *P *[*A*2*A*3] = *P *[*A*3*A*1] = 1*/*4*. *(2)

However, the three events *A*1*, A*2*, A*3 are not independent since

*P *[*A*1*A*2*A*3] = 0 = *P *[*A*1]*P *[*A*2]*P *[*A*3] *. *(3)

**Problem 1.6.6 Solution
**There are 16 distinct equally likely outcomes for the second generation of pea plants based on a
first generation of *{rwyg, rwgy, wryg, wrgy}*. They are listed below

*rryy rryg rrgy rrgg
rwyy rwyg rwgy rwgg
wryy wryg wrgy wrgg
wwyy wwyg wwgy wwgg
*

(1)

A plant has yellow seeds, that is event *Y *occurs, if a plant has at least one dominant *y *gene. Except
for the four outcomes with a pair of recessive *g *genes, the remaining 12 outcomes have yellow seeds.
From the above, we see that

*P *[*Y *] = 12*/*16 = 3*/*4 (2)

and
*P *[*R*] = 12*/*16 = 3*/*4*. *(3)

To find the conditional probabilities *P *[*R|Y *] and *P *[*Y |R*], we first must find *P *[*RY *]. Note that
*RY *, the event that a plant has rounded yellow seeds, is the set of outcomes

*RY *= *{rryy, rryg, rrgy, rwyy, rwyg, rwgy, wryy, wryg, wrgy} . *(4)

Since *P *[*RY *] = 9*/*16,

*P *[*Y |R *] = *P *[*RY *]
*P *[*R*]

=
9*/*16
3*/*4

= 3*/*4 (5)

and
*P *[*R |Y *] = *P *[*RY *]

*P *[*Y *]
=

9*/*16
3*/*4

= 3*/*4*. *(6)

Thus *P *[*R|Y *] = *P *[*R*] and *P *[*Y |R*] = *P *[*Y *] and *R *and *Y *are independent events. There are four
visibly different pea plants, corresponding to whether the peas are round (*R*) or not (*Rc*), or yellow
(*Y *) or not (*Y c*). These four visible events have probabilities

*P *[*RY *] = 9*/*16 *P *[*RY c*] = 3*/*16*, *(7)
*P *[*RcY *] = 3*/*16 *P *[*RcY c*] = 1*/*16*. *(8)

16

**Problem 1.6.7 Solution
**

(a) For any events *A *and *B*, we can write the law of total probability in the form of

*P *[*A*] = *P *[*AB*] + *P *[*ABc*] *. *(1)

Since *A *and *B *are independent, *P *[*AB*] = *P *[*A*]*P *[*B*]. This implies

*P *[*ABc*] = *P *[*A*]*− P *[*A*]*P *[*B*] = *P *[*A*] (1*− P *[*B*]) = *P *[*A*]*P *[*Bc*] *. *(2)

Thus *A *and *Bc *are independent.

(b) Proving that *Ac *and *B *are independent is not really necessary. Since *A *and *B *are arbitrary
labels, it is really the same claim as in part (a). That is, simply reversing the labels of *A *and
*B *proves the claim. Alternatively, one can construct exactly the same proof as in part (a)
with the labels *A *and *B *reversed.

(c) To prove that *Ac *and *Bc *are independent, we apply the result of part (a) to the sets *A *and
*Bc*. Since we know from part (a) that *A *and *Bc *are independent, part (b) says that *Ac *and
*Bc *are independent.

**Problem 1.6.8 Solution
**

*A AC
*

*AB ABC C
*

*BCB
*

In the Venn diagram at right, assume the sample space has area 1 cor-
responding to probability 1. As drawn, *A*, *B*, and *C *each have area 1*/*2
and thus probability 1*/*2. Moreover, the three way intersection *ABC *has
probability 1*/*8. Thus *A*, *B*, and *C *are mutually independent since

*P *[*ABC*] = *P *[*A*]*P *[*B*]*P *[*C*] *. *(1)

**Problem 1.6.9 Solution
**

*A AB B
*

*AC C BC
*

In the Venn diagram at right, assume the sample space has area 1 cor-
responding to probability 1. As drawn, *A*, *B*, and *C *each have area
1*/*3 and thus probability 1*/*3. The three way intersection *ABC *has zero
probability, implying *A*, *B*, and *C *are not mutually independent since

*P *[*ABC*] = 0 = *P *[*A*]*P *[*B*]*P *[*C*] *. *(1)

However, *AB*, *BC*, and *AC *each has area 1*/*9. As a result, each pair of events is independent
since

*P *[*AB*] = *P *[*A*]*P *[*B*] *, P *[*BC*] = *P *[*B*]*P *[*C*] *, P *[*AC*] = *P *[*A*]*P *[*C*] *. *(2)

17

**Problem 1.7.1 Solution
**A sequential sample space for this experiment is

*H*1
1*/*4

*T*13*/*4

*H*2
1*/*4

*T*23*/*4

*H*2
1*/*4 *T*23*/*4

*•H*1*H*2 1*/*16
*•H*1*T*2 3*/*16

*•T*1*H*2 3*/*16
*•T*1*T*2 9*/*16

(a) From the tree, we observe

*P *[*H*2] = *P *[*H*1*H*2] + *P *[*T*1*H*2] = 1*/*4*. *(1)

This implies

*P *[*H*1*|H*2] = *P *[*H*1*H*2]
*P *[*H*2]

=
1*/*16
1*/*4

= 1*/*4*. *(2)

(b) The probability that the first flip is heads and the second flip is tails is *P *[*H*1*T*2] = 3*/*16.

**Problem 1.7.2 Solution
**The tree with adjusted probabilities is

*G*11*/*2

*R*11*/*2

*G*2
3*/*4

*R*21*/*4

*G*2
1*/*4

*R*23*/*4

*•G*1*G*2 3*/*8

*•G*1*R*2 1*/*8

*•R*1*G*2 1*/*8

*•R*1*R*2 3*/*8

From the tree, the probability the second light is green is

*P *[*G*2] = *P *[*G*1*G*2] + *P *[*R*1*G*2] = 3*/*8 + 1*/*8 = 1*/*2*. *(1)

The conditional probability that the first light was green given the second light was green is

*P *[*G*1*|G*2] = *P *[*G*1*G*2]
*P *[*G*2]

=
*P *[*G*2*|G*1]*P *[*G*1]

*P *[*G*2]
= 3*/*4*. *(2)

Finally, from the tree diagram, we can directly read that *P *[*G*2*|G*1] = 3*/*4.

**Problem 1.7.3 Solution
**Let *Gi *and *Bi *denote events indicating whether free throw *i *was good (*Gi*) or bad (*Bi*). The tree
for the free throw experiment is

18

*G*11*/*2

*B*11*/*2

*G*2
3*/*4

*B*21*/*4

*G*2
1*/*4

*B*23*/*4

*•G*1*G*2 3*/*8

*•G*1*B*2 1*/*8

*•B*1*G*2 1*/*8

*•B*1*B*2 3*/*8

The game goes into overtime if exactly one free throw is made. This event has probability

*P *[*O*] = *P *[*G*1*B*2] + *P *[*B*1*G*2] = 1*/*8 + 1*/*8 = 1*/*4*. *(1)

**Problem 1.7.4 Solution
**The tree for this experiment is

*A
*1*/*2

*B*1*/*2

*H
*1*/*4

*T*3*/*4

*H
*3*/*4 *T*1*/*4

*•AH *1*/*8
*•AT *3*/*8

*•BH *3*/*8
*•BT *1*/*8

The probability that you guess correctly is

*P *[*C*] = *P *[*AT *] + *P *[*BH*] = 3*/*8 + 3*/*8 = 3*/*4*. *(1)

**Problem 1.7.5 Solution
**The *P *[*− |H *] is the probability that a person who has HIV tests negative for the disease. This is
referred to as a false-negative result. The case where a person who does not have HIV but tests
positive for the disease, is called a false-positive result and has probability *P *[+*|Hc*]. Since the test
is correct 99% of the time,

*P *[*−|H*] = *P *[+*|Hc*] = 0*.*01*. *(1)
Now the probability that a person who has tested positive for HIV actually has the disease is

*P *[*H|*+] = *P *[+*, H*]
*P *[+]

=
*P *[+*, H*]

*P *[+*, H*] + *P *[+*, Hc*]
*. *(2)

We can use Bayes’ formula to evaluate these joint probabilities.

*P *[*H|*+] = *P *[+*|H*]*P *[*H*]
*P *[+*|H*]*P *[*H*] + *P *[+*|Hc*]*P *[*Hc*] (3)

=
(0*.*99)(0*.*0002)

(0*.*99)(0*.*0002) + (0*.*01)(0*.*9998)
(4)

= 0*.*0194*. *(5)

Thus, even though the test is correct 99% of the time, the probability that a random person who tests positive actually has HIV is less than 0.02. The reason this probability is so low is that the a priori probability that a person has HIV is very small.

19

**Problem 1.7.6 Solution
**Let *Ai *and *Di *indicate whether the *i*th photodetector is acceptable or defective.

*A*1
3*/*5

*D*12*/*5

*A*2
4*/*5

*D*21*/*5

*A*2
2*/*5*D*23*/*5

*•A*1*A*2 12*/*25
*•A*1*D*2 3*/*25

*•D*1*A*2 4*/*25
*•D*1*D*2 6*/*25

(a) We wish to find the probability *P *[*E*1] that exactly one photodetector is acceptable. From
the tree, we have

*P *[*E*1] = *P *[*A*1*D*2] + *P *[*D*1*A*2] = 3*/*25 + 4*/*25 = 7*/*25*. *(1)

(b) The probability that both photodetectors are defective is *P *[*D*1*D*2] = 6*/*25.

**Problem 1.7.7 Solution
**The tree for this experiment is

*A*11*/*2

*B*11*/*2

*H*1
1*/*4

*T*13*/*4

*H*1
3*/*4 *T*11*/*4

*H*2
3*/*4

*T*21*/*4
*H*2

3*/*4 *T*21*/*4

*H*2
1*/*4

*T*23*/*4
*H*2

1*/*4 *T*23*/*4

*•A*1*H*1*H*2 3*/*32
*•A*1*H*1*T*2 1*/*32
*•A*1*T*1*H*2 9*/*32
*•A*1*T*1*T*2 3*/*32

*•B*1*H*1*H*2 3*/*32
*•B*1*H*1*T*2 9*/*32
*•B*1*T*1*H*2 1*/*32
*•B*1*T*1*T*2 3*/*32

The event *H*1*H*2 that heads occurs on both flips has probability

*P *[*H*1*H*2] = *P *[*A*1*H*1*H*2] + *P *[*B*1*H*1*H*2] = 6*/*32*. *(1)

The probability of *H*1 is

*P *[*H*1] = *P *[*A*1*H*1*H*2] + *P *[*A*1*H*1*T*2] + *P *[*B*1*H*1*H*2] + *P *[*B*1*H*1*T*2] = 1*/*2*. *(2)

Similarly,

*P *[*H*2] = *P *[*A*1*H*1*H*2] + *P *[*A*1*T*1*H*2] + *P *[*B*1*H*1*H*2] + *P *[*B*1*T*1*H*2] = 1*/*2*. *(3)

Thus *P *[*H*1*H*2] = *P *[*H*1]*P *[*H*2], implying *H*1 and *H*2 are not independent. This result should not
be surprising since if the first flip is heads, it is likely that coin *B *was picked first. In this case, the
second flip is less likely to be heads since it becomes more likely that the second coin flipped was
coin *A*.

20

**Problem 1.7.8 Solution
**

(a) The primary difficulty in this problem is translating the words into the correct tree diagram. The tree for this problem is shown below.

*H*11*/*2

*T*11*/*2

*H*21*/*2

*T*21*/*2
*H*3

1*/*2

*T*31*/*2

*H*4
1*/*2*T*41*/*2

*H*31*/*2

*T*31*/*2

*H*41*/*2
*T*41*/*2

*•H*1 1*/*2
*•T*1*H*2*H*3 1*/*8

*•T*1*H*2*T*3*H*4 1*/*16
*•T*1*H*2*T*3*T*4 1*/*16

*•T*1*T*2*H*3*H*4 1*/*16
*•T*1*T*2*H*3*T*4 1*/*16

*•T*1*T*2*T*3 1*/*8

(b) From the tree,

*P *[*H*3] = *P *[*T*1*H*2*H*3] + *P *[*T*1*T*2*H*3*H*4] + *P *[*T*1*T*2*H*3*H*4] (1)
= 1*/*8 + 1*/*16 + 1*/*16 = 1*/*4*. *(2)

Similarly,

*P *[*T*3] = *P *[*T*1*H*2*T*3*H*4] + *P *[*T*1*H*2*T*3*T*4] + *P *[*T*1*T*2*T*3] (3)
= 1*/*8 + 1*/*16 + 1*/*16 = 1*/*4*. *(4)

(c) The event that Dagwood must diet is

*D *= (*T*1*H*2*T*3*T*4) *∪ *(*T*1*T*2*H*3*T*4) *∪ *(*T*1*T*2*T*3)*. *(5)
The probability that Dagwood must diet is

*P *[*D*] = *P *[*T*1*H*2*T*3*T*4] + *P *[*T*1*T*2*H*3*T*4] + *P *[*T*1*T*2*T*3] (6)
= 1*/*16 + 1*/*16 + 1*/*8 = 1*/*4*. *(7)

The conditional probability of heads on flip 1 given that Dagwood must diet is

*P *[*H*1*|D*] = *P *[*H*1*D*]
*P *[*D*]

= 0*. *(8)

Remember, if there was heads on flip 1, then Dagwood always postpones his diet.

(d) From part (b), we found that *P *[*H*3] = 1*/*4. To check independence, we calculate

*P *[*H*2] = *P *[*T*1*H*2*H*3] + *P *[*T*1*H*2*T*3] + *P *[*T*1*H*2*T*4*T*4] = 1*/*4 (9)
*P *[*H*2*H*3] = *P *[*T*1*H*2*H*3] = 1*/*8*. *(10)

Now we find that
*P *[*H*2*H*3] = 1*/*8 = *P *[*H*2]*P *[*H*3] *. *(11)

Hence, *H*2 and *H*3 are dependent events. In fact, *P *[*H*3*|H*2] = 1*/*2 while *P *[*H*3] = 1*/*4. The
reason for the dependence is that given *H*2 occurred, then we know there will be a third flip
which may result in *H*3. That is, knowledge of *H*2 tells us that the experiment didn’t end
after the first flip.

21

**Problem 1.7.9 Solution
**

(a) We wish to know what the probability that we find no good photodiodes in *n *pairs of diodes.
Testing each pair of diodes is an independent trial such that with probability *p*, both diodes
of a pair are bad. From Problem 1.7.6, we can easily calculate *p*.

*p *= *P *[both diodes are defective] = *P *[*D*1*D*2] = 6*/*25*. *(1)

The probability of *Zn*, the probability of zero acceptable diodes out of *n *pairs of diodes is *pn
*

because on each test of a pair of diodes, both must be defective.

*P *[*Zn*] =
*n*∏

*i*=1

*p *= *pn *=
(

6 25

)*n
*(2)

(b) Another way to phrase this question is to ask how many pairs must we test until *P *[*Zn*] *≤ *0*.*01.
Since *P *[*Zn*] = (6*/*25)*n*, we require(

6 25

)*n
≤ *0*.*01 *⇒ n ≥ *ln 0*.*01

ln 6*/*25
= 3*.*23*. *(3)

Since *n *must be an integer, *n *= 4 pairs must be tested.

**Problem 1.7.10 Solution
**The experiment ends as soon as a fish is caught. The tree resembles

*C*1*p
*

*Cc*11*−p *

*C*2*p
*

*Cc*21*−p *

*C*3*p
*

*Cc*31*−p ...
*

From the tree, *P *[*C*1] = *p *and *P *[*C*2] = (1*− p*)*p*. Finally, a fish is caught on the *n*th cast if no fish
were caught on the previous *n− *1 casts. Thus,

*P *[*Cn*] = (1*− p*)*n−*1*p. *(1)

**Problem 1.8.1 Solution
**There are 25 = 32 different binary codes with 5 bits. The number of codes with exactly 3 zeros
equals the number of ways of choosing the bits in which those zeros occur. Therefore there are(
5
3

) = 10 codes with exactly 3 zeros.

**Problem 1.8.2 Solution
**Since each letter can take on any one of the 4 possible letters in the alphabet, the number of 3
letter words that can be formed is 43 = 64. If we allow each letter to appear only once then we
have 4 choices for the first letter and 3 choices for the second and two choices for the third letter.
Therefore, there are a total of 4 *· *3 *· *2 = 24 possible codes.

22

**Problem 1.8.3 Solution
**

(a) The experiment of picking two cards and recording them in the order in which they were selected can be modeled by two sub-experiments. The first is to pick the first card and record it, the second sub-experiment is to pick the second card without replacing the first and recording it. For the first sub-experiment we can have any one of the possible 52 cards for a total of 52 possibilities. The second experiment consists of all the cards minus the one that was picked first(because we are sampling without replacement) for a total of 51 possible outcomes. So the total number of outcomes is the product of the number of outcomes for each sub-experiment.

52 *· *51 = 2652 outcomes. (1)

(b) To have the same card but different suit we can make the following sub-experiments. First we need to pick one of the 52 cards. Then we need to pick one of the 3 remaining cards that are of the same type but different suit out of the remaining 51 cards. So the total number outcomes is

52 *· *3 = 156 outcomes. (2)

(c) The probability that the two cards are of the same type but different suit is the number of outcomes that are of the same type but different suit divided by the total number of outcomes involved in picking two cards at random from a deck of 52 cards.

*P *[same type, different suit] =
156
2652

= 1 17

*. *(3)

(d) Now we are not concerned with the ordering of the cards. So before, the outcomes (*K♥, *8*♦*)
and (8*♦,K♥*) were distinct. Now, those two outcomes are not distinct and are only considered
to be the single outcome that a King of hearts and 8 of diamonds were selected. So every
pair of outcomes before collapses to a single outcome when we disregard ordering. So we can
redo parts (a) and (b) above by halving the corresponding values found in parts (a) and (b).
The probability however, does not change because both the numerator and the denominator
have been reduced by an equal factor of 2, which does not change their ratio.

**Problem 1.8.4 Solution
**We can break down the experiment of choosing a starting lineup into a sequence of subexperiments:

1. Choose 1 of the 10 pitchers. There are *N*1 =
(
10
1

) = 10 ways to do this.

2. Choose 1 of the 15 field players to be the designated hitter (DH). There are *N*2 =
(
15
1

) = 15

ways to do this.

3. Of the remaining 14 field players, choose 8 for the remaining field positions. There are
*N*3 =

( 14 8

) to do this.

4. For the 9 batters (consisting of the 8 field players and the designated hitter), choose a batting
lineup. There are *N*4 = 9! ways to do this.

23

So the total number of different starting lineups when the DH is selected among the field players is

*N *= *N*1*N*2*N*3*N*4 = (10)(15)
(
14
8

)
9! = 163*,*459*,*296*,*000*. *(1)

Note that this overestimates the number of combinations the manager must really consider because most field players can play only one or two positions. Although these constraints on the manager reduce the number of possible lineups, it typically makes the manager’s job more difficult. As for the counting, we note that our count did not need to specify the positions played by the field players. Although this is an important consideration for the manager, it is not part of our counting of different lineups. In fact, the 8 nonpitching field players are allowed to switch positions at any time in the field. For example, the shortstop and second baseman could trade positions in the middle of an inning. Although the DH can go play the field, there are some coomplicated rules about this. Here is an an excerpt from Major league Baseball Rule 6.10:

The Designated Hitter may be used defensively, continuing to bat in the same posi- tion in the batting order, but the pitcher must then bat in the place of the substituted defensive player, unless more than one substitution is made, and the manager then must designate their spots in the batting order.

If you’re curious, you can find the complete rule on the web.

**Problem 1.8.5 Solution
**When the DH can be chosen among all the players, including the pitchers, there are two cases:

*• *The DH is a field player. In this case, the number of possible lineups, *NF *, is given in
Problem 1.8.4. In this case, the designated hitter must be chosen from the 15 field players.
We repeat the solution of Problem 1.8.4 here: We can break down the experiment of choosing
a starting lineup into a sequence of subexperiments:

1. Choose 1 of the 10 pitchers. There are *N*1 =
(
10
1

) = 10 ways to do this.

2. Choose 1 of the 15 field players to be the designated hitter (DH). There are *N*2 =
(
15
1

) =

15 ways to do this.

3. Of the remaining 14 field players, choose 8 for the remaining field positions. There are
*N*3 =

( 14 8

) to do this.

4. For the 9 batters (consisting of the 8 field players and the designated hitter), choose a
batting lineup. There are *N*4 = 9! ways to do this.

So the total number of different starting lineups when the DH is selected among the field players is

*N *= *N*1*N*2*N*3*N*4 = (10)(15)
(
14
8

)
9! = 163*,*459*,*296*,*000*. *(1)

*• *The DH is a pitcher. In this case, there are 10 choices for the pitcher, 10 choices for the
DH among the pitchers (including the pitcher batting for himself),

( 15 8

) choices for the field

players, and 9! ways of ordering the batters into a lineup. The number of possible lineups is

*N ′ *= (10)(10)
(
15
8

)
9! = 233*, *513*, *280*, *000*. *(2)

The total number of ways of choosing a lineup is *N *+*N ′ *= 396*,*972*,*576*,*000*.
*

24

**Problem 1.8.6 Solution
**

(a) We can find the number of valid starting lineups by noticing that the swingman presents
three situations: (1) the swingman plays guard, (2) the swingman plays forward, and (3) the
swingman doesn’t play. The first situation is when the swingman can be chosen to play the
guard position, and the second where the swingman can only be chosen to play the forward
position. Let *Ni *denote the number of lineups corresponding to case *i*. Then we can write
the total number of lineups as *N*1 +*N*2 +*N*3. In the first situation, we have to choose 1 out
of 3 centers, 2 out of 4 forwards, and 1 out of 4 guards so that

*N*1 =
(
3
1

)( 4 2

)( 4 1

)
= 72*. *(1)

In the second case, we need to choose 1 out of 3 centers, 1 out of 4 forwards and 2 out of 4 guards, yielding

*N*2 =
(
3
1

)( 4 1

)( 4 2

)
= 72*. *(2)

Finally, with the swingman on the bench, we choose 1 out of 3 centers, 2 out of 4 forward, and 2 out of four guards. This implies

*N*3 =
(
3
1

)( 4 2

)( 4 2

)
= 108*, *(3)

and the total number of lineups is *N*1 +*N*2 +*N*3 = 252.

**Problem 1.8.7 Solution
**What our design must specify is the number of boxes on the ticket, and the number of specially
marked boxes. Suppose each ticket has *n *boxes and 5+ *k *specially marked boxes. Note that when
*k > *0, a winning ticket will still have *k *unscratched boxes with the special mark. A ticket is a
winner if each time a box is scratched off, the box has the special mark. Assuming the boxes are
scratched off randomly, the first box scratched off has the mark with probability (5 + *k*)*/n *since
there are 5 + *k *marked boxes out of *n *boxes. Moreover, if the first scratched box has the mark,
then there are 4 + *k *marked boxes out of *n − *1 remaining boxes. Continuing this argument, the
probability that a ticket is a winner is

*p *=
5 + *k
n
*

4 + *k
n− *1

3 + *k
n− *2

2 + *k
n− *3

1 + *k
n− *4 =

(*k *+ 5)!(*n− *5)!
*k*!*n*!

*. *(1)

By careful choice of *n *and *k*, we can choose *p *close to 0*.*01. For example,

*n *9 11 14 17
*k *0 1 2 3
*p *0*.*0079 0*.*012 0*.*0105 0*.*0090

(2)

A gamecard with *N *= 14 boxes and 5 + *k *= 7 shaded boxes would be quite reasonable.

**Problem 1.9.1 Solution
**

25