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SOLUTIONS MANUAL
, Classical and Quantum Information Theory
SOLUTIONS TO EXERCISES
This manual is not available to students, or to those studying the book by themselves.
Instructors receiving the manual must agree not to share the manual's password, to
make any of the solutions publicly accessible, on the web or otherwise, or to make
any of their own solutions to the exercises publicly accessible.
Difficulty scale : B = basic, M = medium, T = tricky
Chapter 1
E1.1 (B) : Flipping two coins simultaneously, what are the probabilities associated with
the following events:
(a) getting two heads ?
(b) getting one heads and one tails ?
Answer: with heads H = and tails = T , we have p ( H ) = p (T ) = . The two coins
outcomes are independent in all cases. For question (a), the probability of getting
two heads is
p ( H , H ) = p( H ) × p( H ) = () 2 =
For question (b), there are two possible outcomes : ( H , T ) or (T , H ) , each with
probability 1/4. Thus,
[ ]
p ( H , T ) or (T , H ) = p ( H , T ) + p (T , H ) = + =
E1.2 (B) : Rolling three dice, one wins if the outcome is 4-2-1 in any order. What is the
probability to win in the first, the second and the third dice roll ? What is the number
of rolls required to have at least 50% chances to win ?
Answer: This the old “421” game played in French cafés. The probability to win or get
a any given combination of 4-2-1 is p = 3 × () = 0.0138, which answers the first
3
question. The probability to loose in the first roll is then q = 1 − p = 1 − 0.038 = 0.986 .
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, Classical and Quantum Information Theory
SOLUTIONS TO EXERCISES
The probability to loose two times in a row is therefore r = q × q = (0.986) = 0.972 .
2
Thus the probability to win in the second roll is s = 1 − r = 1 − 0.972 = 0.027 , which
answers the second question. For the third roll, we find that the probability to win is
t = 1 − q 3 = 1 − (0.986) 3 = 0.041 . In order to evaluate the number of rolls needed to
win with at least 50% chances, we must solve the equation
u = 1 − q N = 1 − (0.986) N ≥ 0.5 or (0.986) N < 0.5 . With a pocket calculator, or a
computer spreadsheet, or with logarithms ( N ≥ log(0.5) / log(0.986) ) we get N = 50 .
E1.3 (B) : A lotto game has 50 numbered balls, out of which six ones are picked at
random. What is the probability of winning by betting on any six number combination
?
Answer: The number of different ways to randomly pick up 6 numbered balls out of a
group of 50 is :
n = C 50
1
× C 49
1
× C 48
1
× C 47
1
× C 46
1
× C 45
1
= 50 × 49 × 48 × 47 × 46 × 45 = 1.14 × 1010
We must divide this result by the number of ball permutations 6! = 720 to obtain the
number of actual 6-combinations :
n 1.14 × 10
10
m= = = 15.9 × 10 6
6! 720
The same result is obtained by directly using the combinatorial coefficient :
50! 50! 50 × 49 × 48 × 47 × 46 × 45
m = C 506 = = = = 15.9 × 10 6
6!(50 − 6)! 6! 44! 720
www.cambridge.org/desurvire © Cambridge University Press 2009
, Classical and Quantum Information Theory
SOLUTIONS TO EXERCISES
The probability of winning is therefore p( x) = 1 / m , or one out of 16 millions,
approximately.
E1.4 (B) : Three competing car companies A, B and C have market shares of 60%, 30%
and 10%, respectively. The probability for the cars to show some construction defects
are 5% for A, 7% for B and 15% for C. What is the probability for any car bought at
random to show some construction defect ?
Answer: The probability for a any car to show some construction defect is
p(defect ) = p(defect A) p( A) + p(defect B) p ( B) + p(defect C ) p(C ) =
= 0.05 × 0.6 + 0.07 × 0.3 + 0.15 × 0.1 = 0.066
representing a probability of 6.6%.
E1.5 (M) : A bag contains 6 billiard balls numbered from one to 6. If two balls are
picked at random from the bag , what is the probability of getting
(a) two balls with even numbers ?
(b) at least one ball with odd number?
(c) ball #3 in the pick ?
You must propose two different methods to solve the exercise.
Answer : Consider first the events space for pick = ( x, y ) :
⎧(1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,3), (2,4), (2,5), (2,6) ⎫
⎪ ⎪
S = ⎨(3,1), (3,2), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,5), (4,6)⎬
⎪(5,1), (5,2), (5,3), (5,4), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5)⎪
⎩ ⎭
with makes up 30 equiprobable events, with p ( x, y ) = . We just need to count
the number of events matching the criteria of questions (a),(b)(c). Thus for question
(a) :
[ ]
p (2,4) or (2,6) or (4,2) or (4,6) or (6,2) or (6,6) = + + + + +
= =
For question (b), one can count 24 corresponding events where at least one ball has
an odd number, thus p ( x) = = . A smarter method consists in observing
that for x = “at least one odd ball”, the complementary event is x = “no odd ball” =
”two even balls”, thus
p ( x) = 1 − p ( x ) = 1 − =
www.cambridge.org/desurvire © Cambridge University Press 2009
SOLUTIONS MANUAL
, Classical and Quantum Information Theory
SOLUTIONS TO EXERCISES
This manual is not available to students, or to those studying the book by themselves.
Instructors receiving the manual must agree not to share the manual's password, to
make any of the solutions publicly accessible, on the web or otherwise, or to make
any of their own solutions to the exercises publicly accessible.
Difficulty scale : B = basic, M = medium, T = tricky
Chapter 1
E1.1 (B) : Flipping two coins simultaneously, what are the probabilities associated with
the following events:
(a) getting two heads ?
(b) getting one heads and one tails ?
Answer: with heads H = and tails = T , we have p ( H ) = p (T ) = . The two coins
outcomes are independent in all cases. For question (a), the probability of getting
two heads is
p ( H , H ) = p( H ) × p( H ) = () 2 =
For question (b), there are two possible outcomes : ( H , T ) or (T , H ) , each with
probability 1/4. Thus,
[ ]
p ( H , T ) or (T , H ) = p ( H , T ) + p (T , H ) = + =
E1.2 (B) : Rolling three dice, one wins if the outcome is 4-2-1 in any order. What is the
probability to win in the first, the second and the third dice roll ? What is the number
of rolls required to have at least 50% chances to win ?
Answer: This the old “421” game played in French cafés. The probability to win or get
a any given combination of 4-2-1 is p = 3 × () = 0.0138, which answers the first
3
question. The probability to loose in the first roll is then q = 1 − p = 1 − 0.038 = 0.986 .
www.cambridge.org/desurvire © Cambridge University Press 2009
, Classical and Quantum Information Theory
SOLUTIONS TO EXERCISES
The probability to loose two times in a row is therefore r = q × q = (0.986) = 0.972 .
2
Thus the probability to win in the second roll is s = 1 − r = 1 − 0.972 = 0.027 , which
answers the second question. For the third roll, we find that the probability to win is
t = 1 − q 3 = 1 − (0.986) 3 = 0.041 . In order to evaluate the number of rolls needed to
win with at least 50% chances, we must solve the equation
u = 1 − q N = 1 − (0.986) N ≥ 0.5 or (0.986) N < 0.5 . With a pocket calculator, or a
computer spreadsheet, or with logarithms ( N ≥ log(0.5) / log(0.986) ) we get N = 50 .
E1.3 (B) : A lotto game has 50 numbered balls, out of which six ones are picked at
random. What is the probability of winning by betting on any six number combination
?
Answer: The number of different ways to randomly pick up 6 numbered balls out of a
group of 50 is :
n = C 50
1
× C 49
1
× C 48
1
× C 47
1
× C 46
1
× C 45
1
= 50 × 49 × 48 × 47 × 46 × 45 = 1.14 × 1010
We must divide this result by the number of ball permutations 6! = 720 to obtain the
number of actual 6-combinations :
n 1.14 × 10
10
m= = = 15.9 × 10 6
6! 720
The same result is obtained by directly using the combinatorial coefficient :
50! 50! 50 × 49 × 48 × 47 × 46 × 45
m = C 506 = = = = 15.9 × 10 6
6!(50 − 6)! 6! 44! 720
www.cambridge.org/desurvire © Cambridge University Press 2009
, Classical and Quantum Information Theory
SOLUTIONS TO EXERCISES
The probability of winning is therefore p( x) = 1 / m , or one out of 16 millions,
approximately.
E1.4 (B) : Three competing car companies A, B and C have market shares of 60%, 30%
and 10%, respectively. The probability for the cars to show some construction defects
are 5% for A, 7% for B and 15% for C. What is the probability for any car bought at
random to show some construction defect ?
Answer: The probability for a any car to show some construction defect is
p(defect ) = p(defect A) p( A) + p(defect B) p ( B) + p(defect C ) p(C ) =
= 0.05 × 0.6 + 0.07 × 0.3 + 0.15 × 0.1 = 0.066
representing a probability of 6.6%.
E1.5 (M) : A bag contains 6 billiard balls numbered from one to 6. If two balls are
picked at random from the bag , what is the probability of getting
(a) two balls with even numbers ?
(b) at least one ball with odd number?
(c) ball #3 in the pick ?
You must propose two different methods to solve the exercise.
Answer : Consider first the events space for pick = ( x, y ) :
⎧(1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,3), (2,4), (2,5), (2,6) ⎫
⎪ ⎪
S = ⎨(3,1), (3,2), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,5), (4,6)⎬
⎪(5,1), (5,2), (5,3), (5,4), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5)⎪
⎩ ⎭
with makes up 30 equiprobable events, with p ( x, y ) = . We just need to count
the number of events matching the criteria of questions (a),(b)(c). Thus for question
(a) :
[ ]
p (2,4) or (2,6) or (4,2) or (4,6) or (6,2) or (6,6) = + + + + +
= =
For question (b), one can count 24 corresponding events where at least one ball has
an odd number, thus p ( x) = = . A smarter method consists in observing
that for x = “at least one odd ball”, the complementary event is x = “no odd ball” =
”two even balls”, thus
p ( x) = 1 − p ( x ) = 1 − =
www.cambridge.org/desurvire © Cambridge University Press 2009
