Preview Extract

Problem Solutions โ Chapter 2
Problem 2.1.1 Solution
A sequential sample space for this experiment is
1/4 H1
1/4 H2 โขH1 H2
T2 โขH1 T2
1/16
3/16
3/4
X
XXX
1/4
X
H2 โขT1 H2 3/16
3/4 X T1 XX
XXX
3/4 X T2 โขT1 T2 9/16
(a) From the tree, we observe
P [H2 ] = P [H1 H2 ] + P [T1 H2 ] = 1/4.
(1)
1/16
P [H1 H2 ]
=
= 1/4.
P [H2 ]
1/4
(2)
This implies
P [H1 |H2 ] =
(b) The probability that the first flip is heads and the second flip is tails is
P[H1 T2 ] = 3/16.
Problem 2.1.2 Solution
The tree with adjusted probabilities is
3/4 G2 โขG1 G2
XXX
1/4 X R2 โขG1 R2
3/8
1/2 G1 XX
HH
H
H
1/4 G2 โขR1 G2
XXX
X
3/4 X R2 โขR1 R2
1/8
1/8
1/2HH R1 X
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3/8
From the tree, the probability the second light is green is
P [G2 ] = P [G1 G2 ] + P [R1 G2 ] = 3/8 + 1/8 = 1/2.
(1)
The conditional probability that the first light was green given the second light was
green is
P [G1 |G2 ] =
P [G1 G2 ]
P [G2 |G1 ] P [G1 ]
=
= 3/4.
P [G2 ]
P [G2 ]
(2)
Finally, from the tree diagram, we can directly read that P[G2 |G1 ] = 3/4.
Problem 2.1.3 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
3/4 G2 โขG1 G2
1/2 G1 XXX
XX
1/4 X B2 โขG1 B2
HH
H
H
1/4 G2 โขB1 G2
XXX
X
3/4 X B2 โขB1 B2
3/8
1/8
1/8
1/2HH B1 X
3/8
The game goes into overtime if exactly one free throw is made. This event has
probability
P [O] = P [G1 B2 ] + P [B1 G2 ] = 1/8 + 1/8 = 1/4.
Problem 2.1.4 Solution
The tree for this experiment is
1/2 A
1/4 H โขAH
1/8
T โขAT
3/8
X
XXX
3/4
X
โขBH
1/2 X B XX
XXX H
1/4 X T โขBT
1/8
3/4
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3/8
(1)
The probability that you guess correctly is
P [C] = P [AT ] + P [BH] = 3/8 + 3/8 = 3/4.
(1)
Problem 2.1.5 Solution
The P[โ |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[+|H c ]. Since the test is correct 99% of the time,
P [โ|H] = P [+|H c ] = 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 [H|+] =
=
.
(2)
P [+]
P [+, H] + P [+, H c ]
We can use Bayesโ formula to evaluate these joint probabilities.
P [+|H] P [H]
P [+|H] P [H] + P [+|H c ] P [H c ]
(0.99)(0.0002)
=
(0.99)(0.0002) + (0.01)(0.9998)
= 0.0194.
P [H|+] =
(3)
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.
Problem 2.1.6 Solution
Let Ai and Di indicate whether the ith photodetector is acceptable or defective.
3/5 A1
4/5 A2 โขA1 A2
12/25
D2 โขA1 D2
3/25
X
X
XXX
2/5
A2 โขD1 A2
2/5 X D1 XX
XXX
3/5 X D2 โขD1 D2
4/25
1/5
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6/25
(a) We wish to find the probability P[E1 ] that exactly one photodetector is acceptable. From the tree, we have
P [E1 ] = P [A1 D2 ] + P [D1 A2 ] = 3/25 + 4/25 = 7/25.
(1)
(b) The probability that both photodetectors are defective is P[D1 D2 ] = 6/25.
Problem 2.1.7 Solution
The tree for this experiment is
3/4 H2 โขA1 H1 H2
1/4 H1
T2 โขA1 H1 T2
1/4
3/4
T1 XX
H2 โขA1 T1 H2
1/2 A1
3/4
XX
X
X
T2 โขA1 T1 T2
1/4
H
HH
H
3/32
1/32
9/32
3/32
1/4 H2 โขB1 H1 H2
3/32
T2 โขB1 H1 T2
XXX
3/4
1/4
X
X
T1 XX
H2 โขB1 T1 H2
1/4
XXX
3/4 X T2 โขB1 T1 T2
1/32
1/2HH B1 X
3/4
H1
9/32
3/32
The event H1 H2 that heads occurs on both flips has probability
P [H1 H2 ] = P [A1 H1 H2 ] + P [B1 H1 H2 ] = 6/32.
(1)
The probability of H1 is
P [H1 ] = P [A1 H1 H2 ] + P [A1 H1 T2 ] + P [B1 H1 H2 ] + P [B1 H1 T2 ]
= 1/2.
(2)
Similarly,
P [H2 ] = P [A1 H1 H2 ] + P [A1 T1 H2 ] + P [B1 H1 H2 ] + P [B1 T1 H2 ]
= 1/2.
(3)
Thus P[H1 H2 ] 6= P[H1 ] P[H2 ], implying H1 and H2 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.
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Problem 2.1.8 Solution
We start with a tree diagram:
0.999 T + โขH
0.99 A
10โ4 D XXXX
X
0.01 X Ac
HH
H
H
0.001
Tโ
+
0.1 A XX 0.001
XXX T โขH
X
0.999
Tโ
XX
XXX c
0.9
A
1โ10โ4HH Dc X
(a) Here we are asked to calculate the conditional probability P[D|A]. In this
part, its simpler to ignore the last branches of the tree that indicate the lab
test result. This yields
P [D|A] =
P [DA]
P [AD]
=
P [A]
P [DA] + P [Dc A]
(10โ4 )(0.99)
=
(10โ4 )(0.99) + (0.1)(1 โ 10โ4 )
= 9.89 ร 10โ4 .
(1)
The probability of the defect D given the arrhythmia A is still quite low
because the probability of the defect is so small.
(b) Since the heart surgery occurs if and only if the event T + occurs, H and T +
are the same event and (from the previous part)
P [DT + ]
P [H|D] = P T + |D =
P [D]
10โ4 (0.99)(0.999)
=
= (0.99)(0.999).
10โ4
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(2)
(c) Since the heart surgery occurs if and only if the event T + occurs, H and T +
are the same event and (from the previous part)
P [Dc T + ]
P [H|Dc ] = P T + |Dc =
P [Dc ]
(1 โ 10โ4 )(0.1)(0.001)
=
= 10โ4 .
1 โ 10โ4
(3)
(d) Heart surgery occurs with probability
P [H] = P [H|D] P [D] + P [H|Dc ] P [Dc ]
= (0.99)(0.999)(10โ4 ) + (10โ4 )(1 โ 10โ4 )
= 1.99 ร 10โ4 .
(4)
(e) Given that heart surgery was performed, the probability the child had no
defect is
P [Dc |H] =
P [Dc H]
P [H]
(1 โ 10โ4 )(0.1)(0.001)
(0.99)(0.999)(10โ4 ) + (10โ4 )(1 โ 10โ4 )
1 โ 10โ4
=
= 0.5027.
2 โ 10โ2 โ 10โ3 + 10โ4
=
(5)
Because the arrythmia is fairly common and the lab test is not fully reliable,
roughly half of all the heart surgeries are performed on healthy infants.
Problem 2.1.9 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.
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H1 โขH1
1/2 H3 โขT1 H2 H3
1/2
1/8
1/2
1/2
H4 โขT1 H2 T3 H4 1/16
T3
T4 โขT1 H2 T3 T4 1/16
1/2 H2
1/2
1/2
XX
XXX
1/2
1/2
1/16
T2
H 3 XX
H โขT T H H
T1
1/2
1/2
Z
XXX 4 1 2 3 4
โขT
T
H
T
1/16
T
1
2
3
4
4
Z
1/2
Z
Z
1/2 Z
T3 โขT1 T2 T3 1/8
(b) From the tree,
P [H3 ] = P [T1 H2 H3 ] + P [T1 T2 H3 H4 ] + P [T1 T2 H3 H4 ]
= 1/8 + 1/16 + 1/16 = 1/4.
(1)
Similarly,
P [T3 ] = P [T1 H2 T3 H4 ] + P [T1 H2 T3 T4 ] + P [T1 T2 T3 ]
= 1/8 + 1/16 + 1/16 = 1/4.
(2)
(c) The event that Dagwood must diet is
D = (T1 H2 T3 T4 ) โช (T1 T2 H3 T4 ) โช (T1 T2 T3 ).
(3)
The probability that Dagwood must diet is
P [D] = P [T1 H2 T3 T4 ] + P [T1 T2 H3 T4 ] + P [T1 T2 T3 ]
= 1/16 + 1/16 + 1/8 = 1/4.
(4)
The conditional probability of heads on flip 1 given that Dagwood must diet
is
P [H1 |D] =
P [H1 D]
= 0.
P [D]
(5)
Remember, if there was heads on flip 1, then Dagwood always postpones his
diet.
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(d) From part (b), we found that P[H3 ] = 1/4. To check independence, we
calculate
P [H2 ] = P [T1 H2 H3 ] + P [T1 H2 T3 ] + P [T1 H2 T4 T4 ] = 1/4
P [H2 H3 ] = P [T1 H2 H3 ] = 1/8.
(6)
Now we find that
P [H2 H3 ] = 1/8 6= P [H2 ] P [H3 ] .
(7)
Hence, H2 and H3 are dependent events. In fact, P[H3 |H2 ] = 1/2 while
P[H3 ] = 1/4. The reason for the dependence is that given H2 occurred, then
we know there will be a third flip which may result in H3 . That is, knowledge
of H2 tells us that the experiment didnโt end after the first flip.
Problem 2.1.10 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 2.1.6,
we can easily calculate p.
p = P [both diodes are defective] = P [D1 D2 ] = 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.
n
n
Y
6
(2)
P [Zn ] =
p = pn =
25
i=1
(b) Another way to phrase this question is to ask how many pairs must we test
until P[Zn ] โค 0.01. Since P[Zn ] = (6/25)n , we require
n
6
ln 0.01
โค 0.01 โ n โฅ
= 3.23.
(3)
25
ln 6/25
Since n must be an integer, n = 4 pairs must be tested.
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Problem 2.1.11 Solution
The starting point is to draw a tree of the experiment. We define the events W
that the plant is watered, L that the plant lives, and D that the plant dies. The
tree diagram is
0.7 W
0.8 L โขW L
0.56
D โขW D
0.14
0.2
X
XXX
X
c
0.3 X W c XX 0.1
XXX L โขW L 0.03
X
0.9
D โขW c D 0.27
It follows that
(a) P[L] = P[W L] + P[W c L] = 0.56 + 0.03 = 0.59.
(b)
P [W c |D] =
P [W c D]
0.27
27
=
= .
P [D]
0.14 + 0.27
41
(1)
(c) P[D|W c ] = 0.9.
In informal conversation, it can be confusing to distinguish between P[D|W c ] and
P[W c |D]; however, they are simple once you draw the tree.
Problem 2.1.12 Solution
The experiment ends as soon as a fish is caught. The tree resembles
p C1
1โp
C1c
p C2
1โp
C2c
p C3
1โp
C3c …
From the tree, P[C1 ] = p and P[C2 ] = (1 โ p)p. Finally, a fish is caught on the nth
cast if no fish were caught on the previous n โ 1 casts. Thus,
P [Cn ] = (1 โ p)nโ1 p.
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(1)
Problem 2.2.1 Solution
Technically, a gumball machine has a finite number of gumballs, but the problem
description models the drawing of gumballs as sampling from the machine without
replacement. This is a reasonable model when the machine has a very large gumball
capacity and we have no knowledge beforehand of how many gumballs of each color
are in the machine. Under this model, the requested probability is given by the
multinomial probability
2 2 2 2
8!
1
1
1
1
P [R2 Y2 G2 B2 ] =
2! 2! 2! 2! 4
4
4
4
8!
= 10 โ 0.0385.
(1)
4
Problem 2.2.2 Solution
In this model of a starburst package, the pieces in a package are collected by
sampling without replacement from a giant collection of starburst pieces.
(a) Each piece is โberry or cherryโ with probability p = 1/2. The probability of
only berry or cherry pieces is p12 = 1/4096.
(b) Each piece is โnot cherryโ with probability 3/4. The probability all 12 pieces
are โnot pinkโ is (3/4)12 = 0.0317.
(c) For i = 1, 2, . . . , 6, let Ci denote the event that all 12 pieces are flavor i. Since
each piece is flavor i with probability 1/4, P[Ci ] = (1/4)12 . Since Ci and Cj
are mutually exclusive,
P[F1 ] = P[C1 โช C2 โช ยท ยท ยท โช C4 ] =
4
X
P[Ci ] = 4 P[C1 ] = (1/4)11 .
i=1
Problem 2.2.3 Solution
(a) Let Bi , Li , Oi and Ci denote the events that the ith piece is Berry, Lemon,
Orange, and Cherry respectively. Let F1 denote the event that all three pieces
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draw are the same flavor. Thus,
F1 = {S1 S2 S3 , L1 L2 L3 , O1 O2 O3 , C1 C2 C3 }
(1)
P [F1 ] = P [S1 S2 S3 ] + P [L1 L2 L3 ] + P [O1 O2 O3 ] + P [C1 C2 C3 ]
Note that
P[L1 L2 L3 ] =
1
3 2 1
ยท
ยท
=
12 11 10
220
and by symmetry,
P[F1 ] = 4 P[L1 L2 L3 ] =
(2)
(3)
1
.
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(4)
(b) Let Di denote the event that the ith piece is a different flavor from all the
prior pieces. Let Si denote the event that piece i is the same flavor as a
previous piece. A tree for this experiment is
2/11
1
D1
9/11
S2
D2
4/10
S3
6/10
D3
Note that:
โข P[D1 ] = 1 because the first piece is โdifferentโ since there havenโt been
any prior pieces.
โข The second piece is the same as the first piece with probability 2/11
because in the remaining 11 pieces there are 2 pieces that are the same
as the first piece. Alternatively, out of 11 pieces left, there are 3 colors
each with 3 pieces (that is, 9 pieces out of 11) that are different from
the first piece.
โข Given the first two pieces are different, there are 2 colors, each with 3
pieces (6 pieces) out of 10 remaining pieces that are a different flavor
from the first two pieces. Thus P[D3 |D2 D1 ] = 6/10.
It follows that the three pieces are different with probability
6
27
9
= .
P [D1 D2 D3 ] = 1
11
10
55
40
(5)
Problem 2.2.4 Solution
(a) Since there are only three pieces of each flavor, we cannot draw four pieces
of all the same flavor. Hence P[F1 ] = 0.
(b) Let Di denote the event that the ith piece is a different flavor from all the
prior pieces. Let Si denote the event that piece i is the same flavor as a
previous piece. A tree for this experiment is relatively simple because we
stop the experiment as soon as we draw a piece that is the same as a previous
piece. The tree is
2/11
1
D1
9/11
S2
D2
4/10
S3
6/9
6/10
D3
S4
3/9
D4
Note that:
โข P[D1 ] = 1 because the first piece is โdifferentโ since there havenโt been
any prior pieces.
โข For the second piece, there are 11 pieces left and 9 of those pieces are
different from the first piece drawn.
โข Given the first two pieces are different, there are 2 colors, each with 3
pieces (6 pieces) out of 10 remaining pieces that are a different flavor
from the first two pieces. Thus P[D3 |D2 D1 ] = 6/10.
โข Finally, given the first three pieces are different flavors, there are 3 pieces
remaining that are a different flavor from the pieces previously picked.
Thus P[D4 |D3 D2 D1 ] = 3/9. It follows that the three pieces are different with
probability
6 3
9
9
= .
(1)
P [D1 D2 D3 D4 ] = 1
11
10 9
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An alternate approach to this problem is to assume that each piece is distinguishable, say by numbering the pieces 1, 2, 3 in each flavor. In addition,
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we define the outcome of the experiment to be a 4-permutation of the 12
distinguishable pieces. Under this model, there are (12)4 = 12!
8! equally likely
outcomes in the sample space. The number of outcomes in which all four
pieces are different is n4 = 12 ยท 9 ยท 6 ยท 3 since there are 12 choices for the first
piece drawn, 9 choices for the second piece from the three remaining flavors,
6 choices for the third piece and three choices for the last piece. Since all
outcomes are equally likely,
P [F4 ] =
n4
12 ยท 9 ยท 6 ยท 3
9
=
=
(12)4
12 ยท 11 ยท 10 ยท 9
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(2)
(c) The second model of distinguishable starburst pieces makes it easier to solve
this last question. In this case, let the outcome of the experiment be the
12
4 = 495 combinations or pieces. In this case, we are ignoring the order in
which the pieces were selected. Now we count the number of combinations
in which we have two pieces of each of two flavors. We can do this with the
following steps:
1. Choose two of the four flavors.
2. Choose 2 out of 3 pieces of one of the two chosen flavors.
3. Choose 2 out of 3 pieces of the other of the two chosen flavors.
Let ni equal the number of ways to execute step i. We see that
4
3
3
n1 =
= 6,
n2 =
= 3,
n3 =
= 3.
2
2
2
(3)
There are n1 n2 n3 = 54 possible ways to execute this sequence of steps. Since
all combinations are equally likely,
P [F2 ] =
n1 n2 n3
54
6
=
= .
12
495
55
4
(4)
Problem 2.2.5 Solution
Since there are H = 52
7 equiprobable seven-card hands, each probability is just
the number of hands of each type divided by H.
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(a) Since there are 26 red cards, there are 26
seven-card hands with all red
7
cards. The probability of a seven-card hand of all red cards is
26
26! 45!
7
P [R7 ] = 52 =
= 0.0049.
(1)
52! 19!
7
(b) There are 12 face cards in a 52 card deck and there are 12
7 seven card hands
with all face cards. The probability of drawing only face cards is
12
12! 45!
7
P [F ] = 52 =
= 5.92 ร 10โ6 .
(2)
5!
52!
7
(c) There are 6 red face cards (J, Q, K of diamonds and hearts) in a 52 card deck.
Thus it is impossible to get a seven-card hand of red face cards: P[R7 F ] = 0.
Problem 2.2.6 Solution
There are H5 = 52
5 equally likely five-card hands. Dividing the number of hands
of a particular type by H will yield the probability of a hand of that type.
(a) There are 26
5 five-card hands of all red cards. Thus the probability getting
a five-card hand of all red cards is
26
26! 47!
5
= 0.0253.
(1)
P [R5 ] = 52 =
21! 52!
5
Note that this can be rewritten as
P[R5 ] =
26 25 24 23 22
,
52 51 50 49 48
which shows the successive probabilities of receiving a red card.
(b) The following sequence of subexperiments will generate all possible โfull
houseโ
1. Choose a kind for three-of-a-kind.
2. Choose a kind for two-of-a-kind.
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3. Choose three of the four cards of the three-of-a-kind kind.
4. Choose two of the four cards of the two-of-a-kind kind.
The number of ways of performing subexperiment i is
13
12
4
4
n1 =
,
n2 =
,
n3 =
,
n4 =
. (2)
1
1
3
2
Note that n2 = 12
1 because after choosing a three-of-a-kind, there are twelove
kinds left from which to choose two-of-a-kind. is
The probability of a full house is
P [full house] =
n1 n2 n3 n4
=
52
5
3, 744
= 0.0014.
2, 598, 960
(3)
Problem 2.2.7 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 53 = 10 codes with exactly 3 zeros.
Problem 2.2.8 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 ยท 3 ยท 2 = 24 possible codes.
Problem 2.2.9 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 N1 =
10
1
= 10 ways to do this.
2. Choose 1 of the 15 field players to be the designated hitter (DH). There are
N2 = 15
1 = 15 ways to do this.
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3. Of the remaining 14 field players, choose 8 for the remaining field positions.
There are N3 = 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 N4 = 9! ways to do this.
So the total number of different starting lineups when the DH is selected among
the field players is
14
N = N1 N2 N3 N4 = (10)(15)
9! = 163,459,296,000.
(1)
8
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 managerโs 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 excerpt from Major League
Baseball Rule 6.10:
The Designated Hitter may be used defensively, continuing to bat
in the same position 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 youโre curious, you can find the complete rule on the web.
Problem 2.2.10 Solution
When the DH can be chosen among all the players, including the pitchers, there
are two cases:
โข The DH is a field player. In this case, the number of possible lineups, NF ,
is given in Problem 2.2.9. In this case, the designated hitter must be chosen
45
from the 15 field players. We repeat the solution of Problem 2.2.9 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 N1 = 10
1 = 10 ways to do this.
2. Choose 1 of the
15 field players to be the designated hitter (DH). There
are N2 = 15
1 = 15 ways to do this.
3. Of the remaining 14 field players, choose 8 for the remaining field positions. There are N3 = 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 N4 = 9! ways to do this.
So the total number of different starting lineups when the DH is selected
among the field players is
14
N = N1 N2 N3 N4 = (10)(15)
9! = 163,459,296,000.
(1)
8
โข 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
15
0
N = (10)(10)
9! = 233, 513, 280, 000.
(2)
8
The total number of ways of choosing a lineup is N + N 0 = 396,972,576,000.
Problem 2.2.11 Solution
(a) This is just the multinomial probability
19 19 2
40
19
19
2
P [A] =
19, 19, 2
40
40
40
19 19 2
40!
19
19
2
=
.
19!19!2! 40
40
40
46
(1)
(b) Each spin is either green (with probability 19/40) or not (with probability
21/40). If we call landing on greeen a success, then G19 is the probability of
19 successes in 40 trials. Thus
19 21
21
40
19
.
(2)
P [G19 ] =
40
40
19
(c) If you bet on red, the probability you win is 19/40. If you bet green, the
probability that you win is 19/40. If you first make a random choice to
bet red or green, (say by flipping a coin), the probability you win is still
p = 19/40.
Problem 2.2.12 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 doesnโt 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 N1 + N2 + N3 . 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
3 4 4
N1 =
= 72.
(1)
1 2 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
3 4 4
N2 =
= 72.
(2)
1 1 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
3 4 4
N3 =
= 108,
(3)
1 2 2
and the total number of lineups is N1 + N2 + N3 = 252.
47
Problem 2.2.13 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 โ 1 remaining boxes. Continuing
this argument, the probability that a ticket is a winner is
p=
5+k4+k 3+k 2+k 1+k
(k + 5)!(n โ 5)!
=
.
n nโ1nโ2nโ3nโ4
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 2.3.1 Solution
(a) Since the probability of a zero is 0.8, we can express the probability of the code
word 00111 as 2 occurrences of a 0 and three occurrences of a 1. Therefore
P [00111] = (0.8)2 (0.2)3 = 0.00512.
(b) The probability that a code word has exactly three 1โs is
5
P [three 1โs] =
(0.8)2 (0.2)3 = 0.0512.
3
48
(1)
(2)
Problem 2.3.2 Solution
Given that the probability that the Celtics win a single championship in any given
year is 0.32, we can find the probability that they win 8 straight NBA championships.
P [8 straight championships] = (0.32)8 = 0.00011.
The probability that they win 10 titles in 11 years is
11
P [10 titles in 11 years] =
(.32)10 (.68) = 0.00084.
10
(1)
(2)
The probability of each of these events is less than 1 in 1000! Given that these
events took place in the relatively short fifty year history of the NBA, it should
seem that these probabilities should be much higher. What the model overlooks
is that the sequence of 10 titles in 11 years started when Bill Russell joined the
Celtics. In the years with Russell (and a strong supporting cast) the probability of
a championship was much higher.
Problem 2.3.3 Solution
We know that the probability of a green and red light is 7/16, and that of a yellow
light is 1/8. Since there are always 5 lights, G, Y , and R obey the multinomial
probability law:
2 2
5!
7
1
7
P [G = 2, Y = 1, R = 2] =
.
(1)
2!1!2! 16
8
16
The probability that the number of green lights equals the number of red lights
P [G = R] = P [G = 1, R = 1, Y = 3] + P [G = 2, R = 2, Y = 1]
+ P [G = 0, R = 0, Y = 5]
3
2 2
5!
7
7
1
7
7
1
5!
=
+
1!1!3! 16
16
8
2!1!2! 16
16
8
5
5!
1
+
0!0!5! 8
โ 0.1449.
49
(2)
Problem 2.3.4 Solution
For the team with the homecourt advantage, let Wi and Li denote whether game
i was a win or a loss. Because games 1 and 3 are home games and game 2 is an
away game, the tree is
1โp W2 โขW1 W2
p W1
X
XXX
X
1โp X L1
p
1โp
HH
H
p(1โp)
p W3 โขW1 L2 W3
L2
1โp
W2 XX
p
p3
L3 โขW1 L2 L3
p2 (1โp)
W โขL W W
p(1โp)2
3
1 2 3
XX
X
1โp X L3 โขL1 W2 L3 (1โp)3
HH
p H
L2 โขL1 L2
p(1โp)
The probability that the team with the home court advantage wins is
P [H] = P [W1 W2 ] + P [W1 L2 W3 ] + P [L1 W2 W3 ]
= p(1 โ p) + p3 + p(1 โ p)2 .
(1)
Note that P[H] โค p for 1/2 โค p โค 1. Since the team with the home court advantage
would win a 1 game playoff with probability p, the home court team is less likely
to win a three game series than a 1 game playoff!
Problem 2.3.5 Solution
(a) There are 3 group 1 kickers and 6 group 2 kickers. Using Gi to denote that
a group i kicker was chosen, we have
P [G1 ] = 1/3,
P [G2 ] = 2/3.
(1)
In addition, the problem statement tells us that
P [K|G1 ] = 1/2,
P [K|G2 ] = 1/3.
(2)
Combining these facts using the Law of Total Probability yields
P [K] = P [K|G1 ] P [G1 ] + P [K|G2 ] P [G2 ]
= (1/2)(1/3) + (1/3)(2/3) = 7/18.
50
(3)
(b) To solve this part, we need to identify the groups from which the first and
second kicker were chosen. Let ci indicate whether a kicker was chosen from
group i and let Cij indicate that the first kicker was chosen from group i
and the second kicker from group j. The experiment to choose the kickers is
described by the sample tree:
2/8 c1 โขC11
3/9 c1
c2 โขC12
6/8
XXX
XX
3/8
c1 โขC21
6/9 X c2 XX
XXX
X
c
2 โขC22
5/8
1/12
1/4
1/4
5/12
Since a kicker from group 1 makes a kick with probability 1/2 while a kicker
from group 2 makes a kick with probability 1/3,
P [K1 K2 |C11 ] = (1/2)2 ,
P [K1 K2 |C12 ] = (1/2)(1/3),
2
P [K1 K2 |C21 ] = (1/3)(1/2),
P [K1 K2 |C22 ] = (1/3) .
(4)
(5)
By the law of total probability,
P [K1 K2 ] = P [K1 K2 |C11 ] P [C11 ] + P [K1 K2 |C12 ] P [C12 ]
+ P [K1 K2 |C21 ] P [C21 ] + P [K1 K2 |C22 ] P [C22 ]
1 1
11 11 1 5
=
+
+
+
= 15/96.
4 12 6 4 6 4 9 12
(6)
It should be apparent that P[K1 ] = P[K] from part (a). Symmetry should
also make it clear that P[K1 ] = P[K2 ] since for any ordering of two kickers,
the reverse ordering is equally likely. If this is not clear, we derive this result
by calculating P[K2 |Cij ] and using the law of total probability to calculate
P[K2 ].
P [K2 |C11 ] = 1/2,
P [K2 |C12 ] = 1/3,
P [K2 |C21 ] = 1/2,
P [K2 |C22 ] = 1/3.
51
(7)
(8)
By the law of total probability,
P [K2 ] = P [K2 |C11 ] P [C11 ] + P [K2 |C12 ] P [C12 ]
+ P [K2 |C21 ] P [C21 ] + P [K2 |C22 ] P [C22 ]
1 1
11 11 1 5
7
=
+
+
+
= .
2 12 3 4 2 4 3 12
18
(9)
We observe that K1 and K2 are not independent since
15
P [K1 K2 ] =
6=
96
7
18
2
= P [K1 ] P [K2 ] .
(10)
Note that 15/96 and (7/18)2 are close but not exactly the same. The reason
K1 and K2 are dependent is that if the first kicker is successful, then it is
more likely that kicker is from group 1. This makes it more likely that the
second kicker is from group 2 and is thus more likely to miss.
(c) Once a kicker is chosen, each of the 10 field goals is an independent trial. If
the kicker is from group 1, then the success probability is 1/2. If the kicker
is from group 2, the success probability is 1/3. Out of 10 kicks, there are 5
misses iff there are 5 successful kicks. Given the type of kicker chosen, the
probability of 5 misses is
10
10
5
5
P [M |G1 ] =
(1/2) (1/2) ,
P [M |G2 ] =
(1/3)5 (2/3)5 .
(11)
5
5
We use the Law of Total Probability to find
P [M ] = P [M |G1 ] P [G1 ] + P [M |G2 ] P [G2 ]
10
=
(1/3)(1/2)10 + (2/3)(1/3)5 (2/3)5 .
5
(12)
Problem 2.4.1 Solution
From the problem statement, we can conclude that the device components are
configured in the following way.
52
W1
W3
W2
W5
W4
W6
To find the probability that the device works, we replace series devices 1, 2, and 3,
and parallel devices 5 and 6 each with a single device labeled with the probability
that it works. In particular,
P [W1 W2 W3 ] = (1 โ q)3 ,
(1)
P [W5 โช W6 ] = 1 โ P [W5c W6c ] = 1 โ q 2 .
(2)
This yields a composite device of the form
(1-q)
3
2
1-q
1-q
The probability P[W 0 ] that the two devices in parallel work is 1 minus the probability that neither works:
P W 0 = 1 โ q(1 โ (1 โ q)3 ).
(3)
Finally, for the device to work, both composite device in series must work. Thus,
the probability the device works is
P [W ] = [1 โ q(1 โ (1 โ q)3 )][1 โ q 2 ].
(4)
Problem 2.4.2 Solution
Suppose that the transmitted bit was a 1. We can view each repeated transmission
as an independent trial. We call each repeated bit the receiver decodes as 1 a
success. Using Sk,5 to denote the event of k successes in the five trials, then the
probability k 1โs are decoded at the receiver is
5 k
P [Sk,5 ] =
p (1 โ p)5โk , k = 0, 1, . . . , 5.
(1)
k
53
The probability a bit is decoded correctly is
P [C] = P [S5,5 ] + P [S4,5 ] = p5 + 5p4 (1 โ p) = 0.91854.
(2)
The probability a deletion occurs is
P [D] = P [S3,5 ] + P [S2,5 ] = 10p3 (1 โ p)2 + 10p2 (1 โ p)3 = 0.081.
(3)
The probability of an error is
P [E] = P [S1,5 ] + P [S0,5 ] = 5p(1 โ p)4 + (1 โ p)5 = 0.00046.
(4)
Note that if a 0 is transmitted, then 0 is sent five times and we call decoding a 0 a
success. You should convince yourself that this a symmetric situation with the same
deletion and error probabilities. Introducing deletions reduces the probability of
an error by roughly a factor of 20. However, the probability of successfull decoding
is also reduced.
Problem 2.4.3 Solution
Note that each digit 0 through 9 is mapped to the 4 bit binary representation of
the digit. That is, 0 corresponds to 0000, 1 to 0001, up to 9 which corresponds to
1001. Of course, the 4 bit binary numbers corresponding to numbers 10 through
15 go unused, however this is unimportant to our problem. the 10 digit number
results in the transmission of 40 bits. For each bit, an independent trial determines
whether the bit was correct, a deletion, or an error. In Problem 2.4.2, we found
the probabilities of these events to be
P [C] = ฮณ = 0.91854,
P [D] = ฮด = 0.081,
P [E] = = 0.00046.
(1)
Since each of the 40 bit transmissions is an independent trial, the joint probability
of c correct bits, d deletions, and e erasures has the multinomial probability
(
40! c d e
ฮณ ฮด c + d + e = 40; c, d, e โฅ 0,
P [C = c, D = d, E = d] = c!d!e!
(2)
0
otherwise.
54
Problem 2.4.4 Solution
From the statement of Problem 2.4.1, the configuration of device components is
W1
W2
W3
W5
W4
W6
By symmetry, note that the reliability of the system is the same whether we replace
component 1, component 2, or component 3. Similarly, the reliability is the same
whether we replace component 5 or component 6. Thus we consider the following
cases:
I Replace component 1 In this case
q
P [W1 W2 W3 ] = (1 โ )(1 โ q)2 ,
2
P [W4 ] = 1 โ q,
2
P [W5 โช W6 ] = 1 โ q .
(1)
(2)
(3)
This implies
P [W1 W2 W3 โช W4 ] = 1 โ (1 โ P [W1 W2 W3 ])(1 โ P [W4 ])
q2
(5 โ 4q + q 2 ).
2
In this case, the probability the system works is
=1โ
P [WI ] = P [W1 W2 W3 โช W4 ] P [W5 โช W6 ]
q2
2
= 1 โ (5 โ 4q + q ) (1 โ q 2 ).
2
(4)
(5)
II Replace component 4 In this case,
P [W1 W2 W3 ] = (1 โ q)3 ,
q
P [W4 ] = 1 โ ,
2
P [W5 โช W6 ] = 1 โ q 2 .
55
(6)
(7)
(8)
This implies
P [W1 W2 W3 โช W4 ] = 1 โ (1 โ P [W1 W2 W3 ])(1 โ P [W4 ])
q q
= 1 โ + (1 โ q)3 .
2 2
(9)
In this case, the probability the system works is
P [WII ] = P [W1 W2 W3 โช W4 ] P [W5 โช W6 ]
i
h
q q
= 1 โ + (1 โ q)3 (1 โ q 2 ).
2 2
(10)
III Replace component 5 In this case,
P [W1 W2 W3 ] = (1 โ q)3 ,
P [W4 ] = 1 โ q,
P [W5 โช W6 ] = 1 โ
q2
2
.
(11)
(12)
(13)
This implies
P [W1 W2 W3 โช W4 ] = 1 โ (1 โ P [W1 W2 W3 ])(1 โ P [W4 ])
= (1 โ q) 1 + q(1 โ q)2 .
(14)
In this case, the probability the system works is
P [WIII ] = P [W1 W2 W3 โช W4 ] P [W5 โช W6 ]
q2
1 + q(1 โ q)2 .
= (1 โ q) 1 โ
2
(15)
From these expressions, its hard to tell which substitution creates the most reliable
circuit. First, we observe that P[WII ] > P[WI ] if and only if
1โ
q2
q q
+ (1 โ q)3 > 1 โ (5 โ 4q + q 2 ).
2 2
2
(16)
Some algebra will show that P[WII ] > P[WI ] if and only if q 2
P[WIII ] for all values of 0 โค q โค 1. Thus the best policy is to replace component 4.
56
Problem 2.5.1 Solution
Rather than just solve the problem for 50 trials, we can write a function that
generates vectors C and H for an arbitrary number of trials n. The code for this
task is
function [C,H]=twocoin(n);
C=ceil(2*rand(n,1));
P=1-(C/4);
H=(rand(n,1)< P);
The first line produces the n ร 1 vector C such that C(i) indicates whether coin 1
or coin 2 is chosen for trial i. Next, we generate the vector P such that P(i)=0.75
if C(i)=1; otherwise, if C(i)=2, then P(i)=0.5. As a result, H(i) is the simulated
result of a coin flip with heads, corresponding to H(i)=1, occurring with probability
P(i).
Problem 2.5.2 Solution
Rather than just solve the problem for 100 trials, we can write a function that
generates n packets for an arbitrary number of trials n. The code for this task is
function C=bit100(n);
% n is the number of 100 bit packets sent
B=floor(2*rand(n,100));
P=0.03-0.02*B;
E=(rand(n,100)< P);
C=sum((sum(E,2)q;
D=(W(:,1)&W(:,2)&W(:,3))|W(:,4);
D=D&(W(:,5)|W(:,6));
N=sum(D);
The n ร 6 matrix W is a logical matrix such that W(i,j)=1 if component j of device
i works properly. Because W is a logical matrix, we can use the Matlab logical
operators | and & to implement the logic requirements for a working device. By
applying these logical operators to the n ร 1 columns of W, we simulate the test of n
circuits. Note that D(i)=1 if device i works. Otherwise, D(i)=0. Lastly, we count
the number N of working devices. The following code snippet produces ten sample
runs, where each sample run tests n=100 devices for q = 0.2.
>> for n=1:10, w(n)=reliable6(100,0.2); end
>> w
w =
82
87
87
92
91
85
85
83
90
>>
89
As we see, the number of working devices is typically around 85 out of 100. Solving
Problem 2.4.1, will show that the probability the device works is actually 0.8663.
Problem 2.5.4 Solution
The code
58
function n=countequal(x,y)
%Usage: n=countequal(x,y)
%n(j)= # elements of x = y(j)
[MX,MY]=ndgrid(x,y);
%each column of MX = x
%each row of MY = y
n=(sum((MX==MY),1))โ;
for countequal is quite short (just two lines excluding comments) but needs some
explanation. The key is in the operation
[MX,MY]=ndgrid(x,y).
The Matlab built-in function ndgrid facilitates plotting a function g(x, y) as a
surface over the x, y plane.The x, y plane is represented by a grid of all pairs of
points x(i), y(j). When x has n elements, and y has m elements, ndgrid(x,y)
creates a grid (an n ร m array) of all possible pairs [x(i) y(j)]. This grid is
represented by two separate n ร m matrices: MX and MY which indicate the x and
y values at each grid point. Mathematically,
MX(i,j) = x(i),
MY(i,j)=y(j).
Next, C=(MX==MY) is an n ร m array such that C(i,j)=1 if x(i)=y(j); otherwise
C(i,j)=0. That is, the jth column of C indicates indicates which elements of x
equal y(j). Lastly, we sum along each column j to count number of elements of x
equal to y(j). That is, we sum along column j to count the number of occurrences
(in x) of y(j).
Problem 2.5.5 Solution
For arbitrary number of trials n and failure probability q, the following functions
evaluates replacing each of the six components by an ultrareliable device.
59
function N=ultrareliable6(n,q);
% n is the number of 6 component devices
%N is the number of working devices
for r=1:6,
W=rand(n,6)>q;
R=rand(n,1)>(q/2);
W(:,r)=R;
D=(W(:,1)&W(:,2)&W(:,3))|W(:,4);
D=D&(W(:,5)|W(:,6));
N(r)=sum(D);
end
This code is based on the code for the solution of Problem 2.5.3. The n ร 6
matrix W is a logical matrix such that W(i,j)=1 if component j of device i works
properly. Because W is a logical matrix, we can use the Matlab logical operators
| and & to implement the logic requirements for a working device. By applying
these logical opeators to the n ร 1 columns of W, we simulate the test of n circuits.
Note that D(i)=1 if device i works. Otherwise, D(i)=0. Note that in the code,
we first generate the matrix W such that each component has failure probability
q. To simulate the replacement of the jth device by the ultrareliable version by
replacing the jth column of W by the column vector R in which a device has failure
probability q/2. Lastly, for each column replacement, we count the number N of
working devices. A sample run for n = 100 trials and q = 0.2 yielded these results:
>> ultrareliable6(100,0.2)
ans =
93
89
91
92
90
93
From the above, we see, for example, that replacing the third component with
an ultrareliable component resulted in 91 working devices. The results are fairly
inconclusive in that replacing devices 1, 2, or 3 should yield the same probability
of device failure. If we experiment with n = 10, 000 runs, the results are more
definitive:
>> ultrareliable6(10000,0.2)
ans =
8738
8762
8806
>> >> ultrareliable6(10000,0.2)
ans =
8771
8795
8806
>>
60
9135
8800
8796
9178
8886
8875
In both cases, it is clear that replacing component 4 maximizes the device reliability.
The somewhat complicated solution of Problem 2.4.4 will confirm this observation.
61

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