Get all Chapter’s Instant download by email at etutorsource@gmail.com
SOLUTIONS MANUAL FOR
APPLIED PROBABILITY
AND STOCHASTIC
PROCESSES
SECOND EDITION
by
Frank Beichelt
University of the
Witwatersrand
Johannesburg, South Africa
Get all Chapter’s Instant download by email at etutorsource@gmail.com
Get all Chapter’s Instant download by email at etutorsource@gmail.com
SOLUTIONS MANUAL FOR
APPLIED PROBABILITY
AND STOCHASTIC
PROCESSES
SECOND EDITION
by
Frank Beichelt
University of the Witwatersrand
Johannesburg, South Africa
Get all Chapter’s Instant download by email at etutorsource@gmail.com
Get all Chapter’s Instant download by email at etutorsource@gmail.com
CHAPTER 1
Random Events and their Probabilities
1.1) A random experiment consists of simultaneously flipping three coins.
(1) What is the corresponding sample space?
(2) Give the following events in terms of elementary events:
A = 'head appears at least two times,'
B = 'head appears not more than once,'
C = 'no head appears.'
(3) Characterize verbally the complementary events of A, B, and C.
Solution
(1) 1 head, 0 tail (head not). Ω = {(i, j, k); i, j, k = 0, 1}. Ω has 8 elements.
(2) A = {(1, 1, 0), (1, 0, 1), (0, 1, 1), (0, 1, 1)},
B = {(1, 0, 0), (0, 1, 0), (0, 0, 1), (0, 0, 0)},
C = {(0, 0, 0)}.
(3) A = 'head appears not more than once' (= B ).
B = 'head appears at least two times' (= A ).
C = 'at least one head appears'.
1.2) A random experiment consists of flipping a die to the first appearance of a '6.'
What is the corresponding sample space?
Solution
The (countably infinite) sample space consists of all vectors (z 1 , z 2 , ..., z k−1 , z k ) with property that
z k = 6 and all z 1 , z 2 , ..., z k−1 are integers between 1 and 5; k = 1, 2, ... .
1.3) Castings are produced weighing either 1, 5, 10, or 20 kg. Let A, B, and C be the events that a
casting weighs 1 or 5kg, exactly 10kg, and at least 10kg, respectively.
Characterize verbally the events A ∩ B, A ∪ B, A ∩ C, and (A ∪ B) ∩ C.
Solution
A∩B
A∪B
A∩C
(A ∪ B) ∩ C
Impossible event.
A casting weighs 1, 5, or 10kg.
A casting weighs 1 or 5kg.
A casting weighs at least 10kg.
1.4) Three randomly chosen persons are to be tested for the presence of gene g. Three random
events are introduced:
A = 'none of them has gene g,'
B = 'at least one of them has gene g,'
C = 'not more than one of them has gene g.'
Determine the corresponding sample space and characterize the events A ∩ B, B
by elementary events.
C, and B ∩ C
Get all Chapter’s Instant download by email at etutorsource@gmail.com
Get all Chapter’s Instant download by email at etutorsource@gmail.com
2
SOLUTIONS MANUAL
Solution
Let 1 and 0 indicate whether a person has gene g or not, respectively. Then the sample space Ω
consists of all the 2 3 = 8 vectors (z 1 , z 2 , z 3 ) with
zi =
1 if a person has gene g,
0
otherwise.
Ω is the same sample space as in exercise 1.1.
A = {(0, 0, 0)}, B = A = Ω \A, C = {(0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1)}.
A ∩ B = ∅ (impossible event),
B
C=B
(since C ⊂ B ),
B ∩ C = B ∪ C = A ∪ C = C (de Morgan rule, A ⊂ C).
1.5) Under which conditions are the following relations between events A and B true:
(1) A ∩ B = Ω , (2) A ∪ B = Ω , (3) A ∪ B = A ∩ B ?
Solution
(1) A = B = Ω .
(2) A = B or B = A . More generally if A ⊇ B or A ⊇ B.
(3) A = B .
1.6) Visualize by a Venn diagram that the following relations between random events A, B, and C
are true:
(1) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) , (2) (A ∩ B) ∪ (A ∩ B) = A , (3) A ∪ B = B ∪ (A ∩ B) .
B
A∩B∩C
A
(A\B) ∩ C
C
1.7) (1) Verify by a Venn diagram that for three random events A, B, and C the following relation
is true: (A\B) ∩ C = (A ∩ C)\(B ∩ C) .
(2) Verify by the same Venn diagram that the relation (A ∩ B)\C = (A\C) ∩ (B\C) is true as well.
1.8) The random events A and B belong to a σ−algebra E. What events, generated by A and B,
must belong to E (see definition 1.2)?
Solution
Ω, ∅, A, A, B, B, A ∪ B, A ∩ B, A ∪ B = A ∩ B, A ∩ B = A ∪ B (de Morgan rules (1.1)).
Other events arise if in these events A and/or B are replaced with A and B :
A ∩ B = B\A, A ∩ B = A\B, A ∪ B = A\B, A ∪ B = B\A.
Any unions of two or more of these events do not give rise to an event, which is different from the
listed ones.
Get all Chapter’s Instant download by email at etutorsource@gmail.com
Get all Chapter’s Instant download by email at etutorsource@gmail.com
1 RANDOM EVENTS AND THEIR PROBABILITIES
3
1.9) Two dice D 1 and D 2 are simultaneously thrown. The respective outcomes of D 1 and D 2 are
ω 1 and ω 2 . Thus, the sample space is Ω = {(ω 1 , ω 2 ); ω 1 , ω 2 = 1, 2, ..., 6}. Let events A and B be
defined as follows:
A = 'ω 1 is even and ω 2 is odd,'
B = 'ω 1 and ω 2 satisfy ω 1 + ω 2 = 9. '
What is the σ−algebra E generated by the events A and B ?
Solution
A = {(2, 1), (2, 3), (2, 5), (4, 1), (4, 3), (4, 5), (6, 1), (6, 3), (6, 5)},
B = {(3, 6), (4, 5), (5, 4), (6, 3)}.
With these events A and B, the σ− algebra consists of all the events listed under exercise 1.8.
1.10) Let A and B be two disjoint random events, A ⊂ Ω , B ⊂ Ω .
Check whether the set of events {A, B, A ∩ B, and A ∩ B } is (1) an exhaustive and (2) a disjoint
set of events (Venn diagram).
Solution
This set is neither exhaustive nor disjoint.
1.11) A coin is flipped 5 times in a row. What is the probability of the event A that 'head' appears
at least 3 times one after the other?
Solution
The underlying random experiment is a Laplace experiment the state space Ω of which has
2 5 = 32 elementary events.
'head' five times in a row: 1 elementary event
'head' four times in a row: 2 elementary events
'head' three times in a row: 3 elementary events
Thus, 6 elementary events are favorable for the occurrence of A. Hence, P(A) = 6/32.
1.12) A die is thrown. Let A = {1, 2, 3} and B = {3, 4, 6} be two random events. Determine the
probabilities P(A ∪ B), P(A ∩ B), and P(B\A).
Solution
P(A) = P(B) = 0.5. P(A ∩ B) = P({3}) = 1/6.
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.5 + 0.5 − 1/6 = 5/6.
P(B\A) = P(B) − P(A ∩ B) = 0.5 − 1/6 = 1/3.
1.13) A die is thrown 3 times. Determine the probability of the event A that the resulting sequence
of three integers is strictly increasing.
Solution
The state space Ω of this random experiment comprises 6 3 = 216 elementary events. There are the
following favorable elementary events:
(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 3, 4), (1, 3, 5), (1, 3, 6), (1, 4, 5), (1, 4, 6), (2, 3, 4), (2, 3, 5),
(2, 3, 6), (2, 4.5), (2, 4, 6), (2, 5, 6), (3, 4, 5), (3, 4, 6), (3, 5, 6), (4, 5, 6).
Hence, P(A) = 19/216.
Get all Chapter’s Instant download by email at etutorsource@gmail.com
We Don’t reply in this website, you need to contact by email for all chapters
Instant download. Just send email and get all chapters download.
Get all Chapters Solutions Manual/Test Bank Instant Download by email at
etutorsource@gmail.com
You can also order by WhatsApp
https://api.whatsapp.com/send/?phone=%2B447507735190&text&type=ph
one_number&app_absent=0
Send email or WhatsApp with complete Book title, Edition Number and
Author Name.
Get all Chapter’s Instant download by email at etutorsource@gmail.com
4
SOLUTIONS MANUAL
1.14) Two dice are thrown simultaneously. Let (ω 1 , ω 2 ) be an outcome of this random experiment
' A = ω 1 + ω 2 ≤ 10 ' and B = 'ω 1 ⋅ ω 2 ≥ 19 .' Determine the probability P(A ∩ B).
Solution
A = {(5, 6), (6, 5), (6, 6)}, B = {(4, 5), (5, 4), (5, 5), (5, 6), (6, 5), (6, 6)}.
P(A) = 3/36, P(B) = 6/36.
B = (A ∩ B) ∪ (A ∩ B) = (A ∩ B) ∪ A since A ⊂ B .
Hence, P(B) = P(A ∩ B) + P(A) so that P(A ∩ B) = 1/12.
1.15) What is the probability p 3 to get 3 numbers right with one ticket in the '6 out of 49' number
lottery?
Solution
Hypergeometric distribution with N = 49, M = n = 6, m = 3 :
⎛ 6 ⎞ ⎛ 43 ⎞
⎝3⎠ ⎝ 3 ⎠
p3 =
= 0.01765.
⎛ 49 ⎞
⎝6⎠
1.16) A sample of 300 students showed the following results with regard to physical fitness and
body weight:
weight [kg]
60 <
[60-80]
80 >
good
48
64
11
fitness satisfactory
22
42
29
bad
19
17
48
One student is randomly chosen. It happens to be Paul.
(1) What is the probability that the fitness of Paul is satisfactory?
(2) What is the probability that the weight of Paul is greater than 80 kg?
(3) What is the probability that the fitness of Paul is bad and that his weight is less than 60 kg?
Solution
(1) p = (22 + 42 + 29) /300 = 0.3100.
(2) p = (11 + 29 + 48) /300 = 0.2933.
(3) p = 19/300 = 0.0633.
1.17) Paul writes four letters and addresses the four accompanying envelopes. After having had a
bottle of whisky, he puts the letters randomly into the envelopes. Determine the probabilities p k
that k letters are in the 'correct' envelopes, k = 0, 1, 2, 3.
Solution
There are 4! = 24 possibilities (elementary events) to put the letters into the envelopes
k = 0 : There are 9 favorable elementary events: p 0 = 9/24 ≈ 0.3750.
k = 1 : There are 8 favorable elementary events: p 1 = 8/24 = 0.3333.
k = 2 : There are 6 favorable elementary events: p 2 = 6/24 = 0.2500.
k = 3 : p 3 = 1 − p 0 − p 1 − p 2 = 1/24 = 0.0416.
Get all Chapter’s Instant download by email at etutorsource@gmail.com
Get all Chapter’s Instant download by email at etutorsource@gmail.com
1 RANDOM EVENTS AND THEIR PROBABILITIES
5
1.18) A straight stick is broken at two randomly chosen positions. What is the probability that the
resulting three parts of the stick allow the construction of a triangle?
Solution
Without loss of generality, let us assume that the stick has length 1. The breaks have occurred at x
and y. A triangle can be constructed if max(x, y) > 1/2, i.e., if the point (x, y) is in the hatched part
of the Figure. By the formula of the geometric probability (1.8), the desired probability is 0.75.
1
y
0.5
0
0.5
x
1
1.19) (Encounter problem) Two hikers climb to the top of a mountain from different directions.
Their arrival time points are between 0:00 and 1:00 a.m., and they stay on the top for 30 minutes.
For each hiker, every time point between 0 and 1 has the same chance to be the arrival time. What
is the probability p that the hikers meet on the top?
Solution
Let x and y be the arrival time points of the hikers on the top. They will meet there if and only if
y − x ≤ 30,
i.e., they will meet if (x, y) belongs to the hatched part of the Figure. Hence, by formula (1.8), the
desired probability is p = 0.75.
1 y
0.5
10
0.5
0
0.5
x
0.5
Exercise 1.19
1
5
0
5
10
Exercise 1.20
1.20) A fence consists of horizontal and vertical wooden rods with a distance of 10 cm between
them (measured from the center of the rods). The rods have a circular sectional view with a
diameter of 2 cm. Thus, the arising 'empty' squares have an edge length of 8 cm. Children throw
balls with a diameter of 5 cm horizontally at the fence. What is the probability p that a ball passes
the fence without touching the rods?
Solution
The center of the ball must hit a point in the shaded square of the Figure. Otherwise the ball would
hit a rod. The shaded square has edge length 3 cm. Hence, p = 9/100 = 0.09.
1.21) Determine the probability p that the quadratic equation x 2 + 2 a x = b − 1 has real solutions
if the pair (a,b) is randomly chosen from the quarter circle {(a, b) ; a, b > 0, a 2 + b 2 < 1}.
Get all Chapter’s Instant download by email at etutorsource@gmail.com
Get all Chapter’s Instant download by email at etutorsource@gmail.com
6
SOLUTIONS MANUAL
Solution
The solutions of the quadratic equation x 2 + 2 a x = b − 1 are
x 1/2 = ± a + b − 1 − a .
These solutions are real iff a + b − 1 ≥ 0 or, equivalently, b ≥ 1 − a. The corresponding area with
the points (a, b) in the quarter circle, which satisfy this inequality, is hatched in the Figure. Hence,
the desired probability is p = π/4 − 0.5 ≈ 0.2854. (Compare to example 1.9, page 16).
1
b
1 − a2
b=
b=1−a
a
0
Exercise 1.21
1
1.22) Let A and B be disjoint events with P(A) = 0.3 and P(B) = 0.45. Determine the probabilities
P(A ∪ B), P(A ∪ B), P(A ∪ B), and P(A ∩ B).
Solution
By the rules provided in section 1.3.3 and taking into account P(A ∩ B) = 0 :
P(A ∪ B) = 0.75, P(A ∪ B) = 0.25, P(A ∪ B) = 1 − P(A ∪ B) = 1 − P(A ∩ B) = 1,
P(A ∩ B) = P(B\A) = P(B) − P(A ∩ B) = P(B) = 0.45.
1.23) Let P(A ∩ B) = 0.3 and P(B) = 0.6. Determine P(A ∪ B) .
Solution
P(A ∩ B) = P(A\B) = P(A) − P(A ∩ B) = 0.3.
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.4 − 0.3 = 0.1.
1.24) Is it possible that for two random events A and B with P(A) = 0.4 and P(B) = 0.2 the relation
P(A ∩ B) = 0.3 is true?
Solution
No, since A ∩ B ⊆ B so that 0.3 = P(A ∩ B) ≤ P(B). But, by assumption, P(B) = 0.2.
1.25) Check whether for 3 arbitrary random events A, B, and C the following constellations of
probabilities can be true:
(1) P(A) = 0.6, P(A ∩ B) = 0.2, and P(A ∩ B) = 0.5,
(2) P(A) = 0.6, P(B) = 0.4, P(A ∩ B) = 0, and P(A ∩ B ∩ C) = 0.1,
(3) P(A ∪ B ∪ C) = 0.68 and P(A ∩ B) = P(A ∩ C) = 1.
Solution
(1) No, since 0.5 = P(A ∩ B) = P(A) − P(A ∩ B) ≠ 0.6 − 0.2 = 0.4.
(2) No, since A ∩ B ∩ C ⊆ A ∩ B so that P(A ∩ B ∩ C) ≤ P(A ∩ B).
(3) No, since the assumption P(A ∩ B) = P(A ∩ C) = 1 implies that A = B = C = Ω.
Get all Chapter’s Instant download by email at etutorsource@gmail.com
Get all Chapter’s Instant download by email at etutorsource@gmail.com
1 RANDOM EVENTS AND THEIR PROBABILITIES
7
1.26) Show that for two arbitrary random events A and B the following inequalities are true:
P(A ∩ B) ≤ P(A) ≤ P(A ∪ B) ≤ P(A) + P(B).
Solution
This chain of inequalities follows from A ∩ B ⊆ A and P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
1.27) Let A, B, and C be 3 arbitrary random events.
(1) Express the event 'A occurs and both B and C do not occur' in terms of suitable relations
between these events and their complements.
(2) Prove: the probability p of the event 'exactly one of the events A, B, or C occurs' is
P(A) + P(B) + P(C) − 2P(A ∩ B) − 2P(A ∩ C) − 2P(B ∩ C) + 3P(A ∩ B ∩ C).
Solution
(1) A ∩ B ∩ C
(2) p = P( A ∩ B ∩ C) + P( B ∩ A ∩ C) + P( C ∩ A ∩ B)
By applying the de Morgan rule (1.2), formula (1.17), and
P( A ∪ B ∪ C) + P(A ∪ B ∪ C) = P( B ∪ C) ,
the following sum representation of the first term is obtained:
P( A ∩ B ∩ C) = P(A) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C).
Analogously one gets (simply replace A, B, C twice with B, C, A , respectively)
P( B ∩ A ∩ C) = P(B) − P(B ∩ C) − P(A ∩ B) − P(A ∩ C) + P(A ∩ B ∩ C),
P( C ∩ A ∩ B) = P(C) − P(A ∩ C) − P(B ∩ C) − P(A ∩ B) + P(A ∩ B ∩ C).
Adding up (a), (b), and (c) gives the desired result.
(a)
(b)
(c)
Section 1.4
1.28) Two dice are simultaneously thrown. The result is (ω 1 , ω 2 ). What is the probability p of the
event 'ω 2 = 6 ' on condition that 'ω 1 + ω 2 = 8 ?'
Solution
The condition reduces the space of the elementary events to Ω r = {(2, 6), (3, 5), (4, 4), (6, 2)}.
Only one elementary event from Ω r is favorable for the occurrence of 'ω 2 = 6 '. Hence, p = 0.2.
1.29) Two dice are simultaneously thrown. By means of formula (1.24) determine the probability
p of the event A that the dice show the same number.
Solution
Let B i denote the probability that die 1 shows 'i'. Then P(B i ) = P(A B i ) = 1/6. Hence,
6
6
p = Σ i=1 P(A B i )P(B i ) = Σ 16 ⋅ 16 = 16 .
i=1
1.30) A publishing house offers a new book as a standard or a luxury edition and with or without a
CD. The publisher analyzes the first 1000 orders:
luxury edition
yes
with CD
no
yes
324
82
no
48
546
Get all Chapter’s Instant download by email at etutorsource@gmail.com
We Don’t reply in this website, you need to contact by email for all chapters
Instant download. Just send email and get all chapters download.
Get all Chapters Solutions Manual/Test Bank Instant Download by email at
etutorsource@gmail.com
You can also order by WhatsApp
https://api.whatsapp.com/send/?phone=%2B447507735190&text&type=ph
one_number&app_absent=0
Send email or WhatsApp with complete Book title, Edition Number and
Author Name.
Get all Chapter’s Instant download by email at etutorsource@gmail.com
8
SOLUTIONS MANUAL
Let A (B) the random event that a book, randomly choosen from these 1000, is a luxury one (comes
with a CD). (1) Determine the probabilities
P(A), P(B), P(A ∪ B), P(A ∩ B), P(A B), P(B A), P(A ∪ B B), and P(A B).
(2) Are the events A and B independent?
Solution
(1) P(A) = 0.372, P(B) = 0.406
P(A ∪ B) = (324 + 48 + 82)/1000 = 0.454, P(A ∩ B) = 0.324, P(A B) = 324/406 ≈ 0.7980,
P(B A) = 324/372 ≈ 0.8710, P(A ∪ B B) = 48/594 = 0.8080, P(A |B ) = 546/(48 + 546) = 0.9191.
(2) No since P(A) ⋅ P(B) = 0.372 ⋅ 0.406 = 0.1510 ≠ P(A ∩ B) = 0.324.
1.31) A manufacturer equips its newly developed car of type Treekill optionally with or without a
tracking device and with or without speed limitation technology. He analyzes the first 1200 orders:
speed limitation
yes
tracking device
no
yes
74
642
no
48
436
Let A (B) be the random event that a car, randomly chosen from these 1200, has speed limitation
(comes with a tracking device).
(1) Calculate the probabilities P(A), P(B), and P(A ∩ B) from the figures in the table.
(2) Based on the probabilities determined under a) and only by using the rules developed in section
1.3.3, determine the probabilities
P(A ∪ B), P(A B), P(B A), P(A ∪ B B), and P(A B)
Solution
(1) P(A) = 132/1200 = 0.1016, P(B) = 716/1200 = 0.5966, P(A ∩ B) = 74/1200 = 0.0616.
(2) P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = (132 + 716 − 74)/1200 = 0.645,
P(A B) = 74/(642 + 74) ≈ 0.10335, P(B A) = 74/(48 + 74) ≈ 0.6066,
P((A∪B)∩B)
P(A∩B)
P(A)−P(A∩B)
−74
=
=
= 132
≈ 0.1198,
484
P(B)
P(B)
P(B)
P(A∩B)
1−P(A∪B)
1−0.645
=
=
= 484/1200
≈ 0.8802.
P(B)
P(B)
P(A ∪ B |B) =
P(A B)
1.32) A bowl contains m white marbles and n red marbles. A marble is taken randomly from the
bowl and returned to the bowl together with r marbles of the same color. This procedure continues to infinity.
(1) What is the probability that the second marble taken is red?
(2) What is the probability that the first marble taken is red on condition that the second marble
taken is red as well? (This is a variant of Pólya 's urn problem.)
Solution
(1) Let A 1 (A 2 ) be the event that the first (second) drawn marble is red. On condition A 1 there are
n + r red marbles in the bowl and on condition A 1 there are n read marbles in the bowl. Hence,
n+r
n
P(A 2 A 1 ) = m+n+r
and P(A 2 A 1 ) = m+n+r
.
By the total probability rule (1.24),
P(A 2 ) = P(A 2 A 1 ) P(A 1 ) + P(A 2 A 1 ) P(A 1 )
Get all Chapter’s Instant download by email at etutorsource@gmail.com
Get all Chapter’s Instant download by email at etutorsource@gmail.com
1 RANDOM EVENTS AND THEIR PROBABILITIES
=
9
n+r ⋅ n +
n
⋅ n + r = n = P(A 1 ).
m+n+r m+n m+n+r m+n+r m+n
Thus, P(A 1 ) = P(A 2 ).
(2) By formula (1.22), the desired probability P(A 1 A 2 ) is given by the ratio
P(A 1 ∩ A 2 ) P(A 2 A 1 ) ⋅ P(A 1 )
=
.
P(A 2 )
P(A 2 )
The results obtained under (1) are applicable and yield
P(A 1 A 2 ) = n + r .
m+n+r
P(A 1 A 2 ) =
1.33) A test procedure for diagnosing faults in circuits indicates no fault with probablility 0.99 if
the circuit is faultless. It indicates a fault with probability 0.90 if the circuit is faulty. Let the probability of a circuit to be faulty be 0.02.
(1) What is the probability that a circuit is faulty if the test procedure indicates a fault?
(2) What is the probability that a circuit is faultless if the test procedure indicates that it is faultless?
Solution
Let A be the random event that a circuit (selected at random from the population) is faulty, and B
be the random event that the test indicates a fault. From the probabilities given,
P(A) = 0.02, P(B A) = 0.90, P(B A) = 0.10, P(B A) = 0.01, P(B |A) = 0.99.
(1) By the total probability rule (1.24), the probability that the test indicates a fault is
P(B) = P(B A) P(A) + P(B A) P(A) = 0.90 ⋅ 0.02 + 0.01 ⋅ 0.98 = 0.0278.
By Bayes' formula (1.25), the desired probability is
P(B A) P(A) 0.90 ⋅ 0.02
P(A B) =
=
= 0.6475.
P(B)
0.0278
(2) Again by Bayes' formula (1.25), the desired probability is
P(B A) P(A) 0.99 ⋅ 0.98
=
= 0.9970.
1 − 0.0278
P(B)
Contrary to the result obtained under (1), this probability is a strong argument in favor of the test.
P(A B) =
1.34) Suppose 2% of cotton fabric rolls and 3% of nylon fabric rolls contain flaws. Of the rolls
used by a manufacturer, 70% are cotton and 30% are nylon.
(1) What is the probability that a randomly selected roll used by the manufacturer contains flaws?
(2) Given that a randomly selected roll used by the manufacturer does not contain flaws, what is
the probability that it is a nylon fabric roll?
Solution
A roll is selected at random from the ones used by the manufacturer. Let A be the random event
that this roll contains flaws, and B be the random event that this roll is cotton. Then,
P(B) = 0.7, P(B) = 0.3, P(A B) = 0.02, P(A B) = 0.98, P(A B) = 0.03, P(A B) = 0.97.
(1) By the total probability rule,
P(A) = P(A B)P(B) + P(A B) P(B) = 0.02 ⋅ 0.70 + 0.03 ⋅ 0.3 = 0.0230.
(2) By Bayes' formula:
P(A B) P(B) 0.97 ⋅ 0.30
P ⎛⎝ B A) =
=
= 0.2979.
1 − 0.0230
P(A)
Get all Chapter’s Instant download by email at etutorsource@gmail.com
Get all Chapter’s Instant download by email at etutorsource@gmail.com
10
SOLUTIONS MANUAL
1.35) A group of 8 students arrives at an examination. Of these students 1 is very well prepared, 2
are well prepared, 3 are satisfactorily prepared, and 2 are insufficiently prepared. There is a total
of 16 questions. A very well prepared student can answer all of them, a well prepared 12, a satisfactorily prepared 8, and an insufficiently prepared 4. Each student has to draw randomly 4 questions. Frank could answer all the 4 questions. What is the probability that Frank
(1) was very well prepared, (2) was insufficiently prepared?
Solution
Let A 1 , A 2 , A 3 , A 4 be the events that in this order a randomly chosen student from the group is
very well, well, satisfactorily, and insufficiently prepared. Then, by the figures given,
P(A 1 ) = 1/8, P(A 2 ) = 1/4, P(A 3 ) = 3/8, P(A 4 ) = 1/4.
Let further B be the event that Frank got all 4 answers right. Then,
P(B A 1 ) = 1, P(B A 2 ) = 12
⋅ 11 ⋅ 10 ⋅ 9 ≈ 0.271978,
16 15 14 13
8
7
6
5
4
3
2
1
P(B A 3 ) = 16
⋅ 15
⋅ 14
⋅ 13
≈ 0.038462, P(B A 4 ) = 16
⋅ 15
⋅ 14
⋅ 13
≈ 0.000549.
Hence, by the total probability rule (1.24),
P(B) ≈ 0.207555..
The desired probabilities are obtained from the Formula of Bayes (1.25):
P(B A 1 ) ⋅ P(A 1 ) 1 ⋅ 0.125
P(A 1 B) =
≈
≈ 0.60225.
(1)
P(B)
0.207555
(2)
P(A 4 B) =
P(B A 4 ) ⋅ P(A 4 ) 0.000549 ⋅ 0.25
≈
≈ 0.00066.
P(B)
0.207555
1.36) Symbols 0 and 1 are transmitted independently from each other in proportion 1 : 4 . Random
noise may cause transmission failures: If a 0 was sent, then a 1 will arrive at the sink with probability 0.1. If a 1 was sent, then a 0 will arrive at the sink with probability 0.05 (Figure).
(1) What is the probability that a received symbol is '1'?
(2) '1' has been received. What is the probability that '1' had been sent?
(3) '0' has been received. What is the probability that '1' had been sent?
transmitter
0
receiver
0
0.9
0.05
0.1
1
0.95
1
Solution
Let A be the event that a 1 has arrived at the sink and B be the event that a 1 had been sent. Then
A ( B ) is the event that a 0 has arrived at the sink (had been sent). The proportion P(B)/P( B) = 4
implies P(B) = 0.80. From the probabilities given,
P(A B) = 0.95, P( A B) = 0.05,
P(A B) = 0.10, P( A B) = 0.90,
P(A) = P(A B) P(B) + P(A B) P(B) = 0.95 ⋅ 0.80 + 0.10 ⋅ 0.20 = 0.7800.
Get all Chapter’s Instant download by email at etutorsource@gmail.com
Get all Chapter’s Instant download by email at etutorsource@gmail.com
1 RANDOM EVENTS AND THEIR PROBABILITIES
11
(1) By Bayes' formula (1.25), the desired probability is
P(B A) =
P(A B) P(B)
⋅0.80
= 0.95
= 0.9744.
0.78
P(A)
P(B A) =
P(A B) P(B)
⋅0.80
= 0.05
= 0.1818.
0.22
P(A)
(2) Again by Bayes' formula,
1.37) The companies 1, 2, and 3 have 60, 80, and 100 employees with 45, 40, and 25 women,
respectively. In every company, employees have the same chance to be retrenched. It is known
that a woman had been retrenched (event B). What is the probability that she had worked in
company 1, 2, and 3, respectively?
Solution
Let A 1 , A 2 , A 3 be the events that a randomly selected employee from the 240 total ones is with
company 1, 2, 3, respectively, and B be the event the chosen employee is a woman. Then,
P(A 1 ) = 0.25, P(A 2 ) = 0.30, P(A 3 ) = 5/12.
Given A i , the conditional probabilities P(B A i ) are
P(B A 1 ) = 0.75, P(B A 2 ) = 0.5, P(B A 3 ) = 0.25.
Hence,
P(B) = 0.75 ⋅ 0.25 + 0.5 ⋅ 0.3 + 0.25 ⋅ (5/12) = 0.44167.
From Bayes' formula:
P(A 1 B) =
P(B A 1 )P(A 1 ) 0.75 ⋅ 0.25
=
≈ 0.42452.
P(B)
0.44167
Analogously, P(A 2 B) = 0.33962, P(A 3 B) = 0.23585.
1.38) John needs to take an examination, which is organized as follows: To each question 5
answers are given. But John knows the correct answer only with probability 0.6. Thus, with
probability 0.4 he has to guess the right answer. In this case, John guesses the correct answer with
probability 1/5 (that means, he chooses an answer by chance). What is the probability that John
knew the answer to a question given that he did answer the question correctly?
Solution
event A: John knows the answer.
P(A) = 0.6.
event A : John does not know the answer. P(A) = 0.4.
even B : The answer of John was correct: P(B A) = 1, P(B A) = 0.2.
P(B) = P(B A) ⋅ P(A) + P(B A) ⋅ P(A) = 1 ⋅ 0.6 + 15 ⋅ 0.4 = 0.68
Hence, P(A B) =
P(B A) ⋅ P(A) 0.6
=
≈ 0.88235.
P(B)
0.68
1.39) A delivery of 25 parts is subject to a quality control according to the following scheme:
A sample of size 5 is drawn. If at least one part is faulty, then the delivery is rejected. If all 5 parts
are o.k., then they are returned to the lot, and a sample of size 10 is randomly taken from the again
25 parts. The delivery is rejected if at least 1 part from the 10 is faulty.
Determine the probabilities that a delivery is rejected on condition that
(1) it contains 2 defective parts,
(2) it contains 4 defective parts.
Get all Chapter’s Instant download by email at etutorsource@gmail.com
We Don’t reply in this website, you need to contact by email for all chapters
Instant download. Just send email and get all chapters download.
Get all Chapters Solutions Manual/Test Bank Instant Download by email at
etutorsource@gmail.com
You can also order by WhatsApp
https://api.whatsapp.com/send/?phone=%2B447507735190&text&type=ph
one_number&app_absent=0
Send email or WhatsApp with complete Book title, Edition Number and
Author Name.
Get all Chapter’s Instant download by email at etutorsource@gmail.com
12
SOLUTIONS MANUAL
Solution
Let B be the event that the delivery is rejected, and let A be the event that the delivery is rejected
after having taken a sample of 5. Then
P(B) = P(B A) ⋅ P(A) + P(B A) ⋅ P(A).
(1) Since
P(B A) = 1, P(A) = 23 ⋅ 22 ⋅ 21 ⋅ 20 ⋅ 19 ≈ 0.636, P(A) = 0.366,
25 ⋅ 24 ⋅ 23 ⋅ 22 ⋅ 21
and P(B A) is the probability of rejection of the delivery based on a sample of size 10, i.e,
P(B A) = 1 − 23 ⋅ 22 ⋅ 21 ⋅ 20 ⋅ 19 ⋅ 18 ⋅ 17 ⋅ 16 ⋅ 15 ⋅ 14 ≈ 0.6482,
25 ⋅ 24 ⋅ 23 ⋅ 22 ⋅ 21 ⋅ 20 ⋅ 19 ⋅ 18 ⋅ 17 ⋅ 16
the probability of rejection of the delivery is
P(B) = 1 ⋅ 0.3667 + 0.6482 ⋅ 0.6367 ≈ 0.7794..
(2)
P(B A) = 1, P(A) = 21 ⋅ 20 ⋅ 19 ⋅ 18 ⋅ 17 ≈ 0.3830, P(A) = 0.6170,
25 ⋅ 24 ⋅ 23 ⋅ 22 ⋅ 21
⋅ 19 ⋅ 18 ⋅ 17 ⋅ 16 ⋅ 15 ⋅ 14 ⋅ 13 ⋅ 12 ≈ 0.8921.
21
⋅
20
P(B A) = 1 −
25 ⋅ 24 ⋅ 23 ⋅ 22 ⋅ 21 ⋅ 20 ⋅ 19 ⋅ 18 ⋅ 17 ⋅ 16
P(B) = 1 ⋅ 0.61670 + 0.8921 ⋅ 0.3830 = 0.9583.
1.40) The random events A 1 , A 2 , ..., A n are assumed to be independent. Show that
P(A 1 ∪ A 2 ∪ . . . ∪ A n ) = 1 − (1 − P(A 1 ))(1 − P(A 2 )) . . . (1 − P(A n )).
Solution
Applying the de Morgan rule gives
P(A 1 ∪ A 2 ∪ . . . ∪ A n ) = 1 − P(A 1 ∩ A 2 ∩ . . . ∩ A n ).
Since the A i are independent, P(A 1 ∩ A 2 ∩ . . . ∩ A n ) = P(A 1 ) ⋅ P(A 1 ) ⋅ . . . ⋅ P(A n ), which implies
the desired result.
1.41) n hunters shoot at a target independently of each other, and each of them hits it with probability 0.8. Determine the smallest n with property that the target is hit with probability 0.99 by at
least one hunter.
Solution
Let A i be the event that hunter i hits the target. Then P(A i ) = 0.8 , i = 1, 2, ..., n, and the probability p n that at least one of n hunters hits the target is given by P(A 1 ∪ A 2 ∪ . . . ∪ A n ). Hence, by
exercise 1.40,
p n = 1 − (0.2) n .
The smallest integer n with p n ≥ 0.99 is n min = 3.
1.42) Starting a car of type Treekill is successful with probability 0.6. What is the probability p 4
that the driver needs no more than 4 start trials to be able to leave?
Solution
Let A i be the event that i trials are necessary. Then, since the A i are disjoint,
p 4 = P(A 1 ∪ A 2 ∪ A 3 ∪ A 4 ) = P(A 1 ) + P(A 2 ) + P(A 3 ) + P(A 4 )
= 0.6 + 0.4 ⋅ 0.6 + 0.4 2 ⋅ 0.6 + 0.4 3 ⋅ 0.6 = 0.9744.
Get all Chapter’s Instant download by email at etutorsource@gmail.com
Get all Chapter’s Instant download by email at etutorsource@gmail.com
1 RANDOM EVENTS AND THEIR PROBABILITIES
13
1.43) Let A and B be two subintervals of [0, 1]. A point x is randomly chosen from [0, 1]. Now A
and B can be interpreted as random events, which occur if x ∈ A or x ∈ B, respectively. Under
which condition are A and B independent?
Solution
P(A ∩ B) = P(x ∈ A and x ∈ B) = P(A) ⋅ P(B) iff the intervals A and B are disjoint.
1.44) A tank is shot at by 3 independently acting antitank helicopter 1, 2, and 3 with one antitank
missile each. Each missile hits the tank with probability 0.6. If the tank is hit by 1 missile, it is put
out of action with probability 0.8. If the tank is hit by at least 2 missiles, it is put out of action with
probability 1. What is the probability that the tank is put out of action by this attack?
Solution
The underlying sample space regarding the random number of missiles hitting the tank is
{(i 1 , i 2 , i 3 ), i k = 0, 1},
where i k = 0 (1) indicates that helicopter k has missed (hit) the target, k = 0, 1. Let A n be the event
that the tank is hit by n missiles. Hence,
A 0 = {(0, 0, 0)}, A 1 = {(1, 0, 0), (0, 1, 0), (0, 0, 1)},
A 2 = {(1, 1, 0), (1, 0, 1), (0, 1, 1)}, A 3 = {(1, 1, 1)}
so that
P(A 0 ) = 0.4 3 , P(A 1 ) = 3 ⋅ 0.4 2 ⋅ 0.6, P(A 2 ) = 3 ⋅ 0.4 1 ⋅ 0.6 2 , P(A 3 ) = 0.6 3 .
Let B be the event that the tank is destroyed by the attack. By the total probability rule,
P(B) = 0 ⋅ P(A 0 ) + 0.8 ⋅ P(A 1 ) + 1 ⋅ P(A 2 ) + 1 ⋅ P(A 3 ) = 0.8784.
1.45) An aircraft is targeted by two independently acting ground-to-air missiles 1 and 2. Each missile hits the aircraft with probability 0.6 if these missiles are not being destroyed before. The aircraft will crash with probability 1 if being hit by at least one missile. On the other hand, the aircraft
defends itself by firing one air-to-air missile each at the approaching ground-to-air missiles. The
air-to-air missiles destroy their respective targets with probablity 0.5.
(1) What is the probability p that the aircraft will crash as a result of this attack?
(2) What is the probability that the aircraft will crash if two air-to-air missiles are fired at each of
the approaching ground-to-air-missiles?
Solution
(1) Let B i be the event that the aircraft will be hit by missile i, and A i be the event that the air-toair missile fired at missile i hits this missile; i = 1, 2. Then
P(B i ) = P(B i A i )P(A i ) + P(B i A i )P(A i ) = 0 ⋅ 0.5 + 0.6 ⋅ 0.50 = 0.3.
Since B 1 and B 2 are independent, the desired probability is
p = P(B 1 ∪ B 2 ) = P(B 1 ) + P(B 2 ) − P(B 1 ∩ B 2 ) = 0.3 + 0.3 − 0.3 ⋅ 0.3 = 0.51.
(2) In this case, each of the approaching ground-to-air-missiles 1 and 2 are destroyed with probability 1 − 0.5 2 = 0.75. Hence, if A i refers now to the effect of two identical air-to-air missiles fired
at missile i, then
P(B i ) = P(B i A i )P(A i ) + P(B i A i )P(A i ) = 0 ⋅ 0.75 + 0.6 ⋅ 0.25 = 0.15.
Hence,
p = P(B 1 ∪ B 2 ) = 0.15 + 0.15 − 0.15 ⋅ 0.15 = 02775.
Get all Chapter’s Instant download by email at etutorsource@gmail.com
Get all Chapter’s Instant download by email at etutorsource@gmail.com
14
SOLUTIONS MANUAL
1.46) The liquid flow in a pipe can be interrupted by two independently operating valves V 1 and
V 2 , which are connected in series (Figure). For interrupting the liquid flow it is sufficient if one
valve closes properly. The probability that an interruption is achieved when necessary is 0.98 for
both valves. On the other hand, liquid flow is only possible if both valves are open. Switching from
'closed' to 'open' is successful with probability 0.99 for each of the valves.
(1) Determine the probability to be able to interrupt the liquid flow if necessary.
(2) What is the probability to be able to resume liquid flow if both valves are closed?
V1
V2
Solution
Let A 1 (A 2 ) be the event that valve V 1 (V 2 ) properly opens when closed, and B 1 (B 2 ) be the
event that valve V 1 (V 2 ) properly closes when open.
(1) The desired probability is
P(B 1 ∪ B 2 ) = P(B 1 ) + P(B 2 ) − P(B 1 ∩ B 2 ) = 0.98 + 0.98 − 0.98 ⋅ 0.98 = 0.9996.
(2) The desired probability is
P(A 1 ∩ A 2 ) = P(A 1 ) ⋅ P(A 2 ) = 0.99 ⋅ 0.99 = 0.9801.
Get all Chapter’s Instant download by email at etutorsource@gmail.com
We Don’t reply in this website, you need to contact by email for all chapters
Instant download. Just send email and get all chapters download.
Get all Chapters Solutions Manual/Test Bank Instant Download by email at
etutorsource@gmail.com
You can also order by WhatsApp
https://api.whatsapp.com/send/?phone=%2B447507735190&text&type=ph
one_number&app_absent=0
Send email or WhatsApp with complete Book title, Edition Number and
Author Name.