Solutions to Chapter 2 Exercises
June 13, 2012
Exercise 2.1
Notice for n=1 die, there are 6 possible outcomes of the sum (1-6).
Now for n=2 dice, there are 11 possible outcomes of the sum (2-12).
Similarly for n=3 dice, you can obtain a sum from 3-18, which is 16 possible
outcomes.
This leads to the general case for n dice, there are 6n-n+1 cases. This is equivalent to 5n+1.
Exercise 2.2
(a)
(b)
If the coin is heads, then the outcome could be -1, -2, -3, -4, -5, or -6.
If the coin is tails, then the outcome could be 1, 2, 3, 4, 5, 6.
These 12 options are the total possible outcomes of the experiment.
If you repeat the experiment 2 times, then the sum of the dice could range
from -12 to
12 (or -6n to 6n).
Similarly for n=3 times, the values of -18 to 18 could be obtained.
Thus for n times, there are 12n+1 outcomes which range from -6n to 6n.
Exercise 2.3
(a)
Consider the x-y plane. The sample space will be all possible ordered pairs
where x is in [0,1) and y is in [0,2). This will look like a rectangle with the
bottom left corner at the origin with a width of 1 and height of 2, as shown in
the gure below. Note the lines x=1 and y=2 form the top and right side lines of
the rectangle and are not included in the sample space, so they are represented
with dashed lines.
(b)
1
On the x-y plane with the rectangle drawn from part (a), draw the line x=y.
The area below this line but still inside the rectangle will satisfy x > y . From
the graph below, you will see this area is 1/4 of the total area. Thus the fraction
of points in the space satisfying x > y is 1/4.
(c)
Zero, the line x=y has zero area and therefore the fraction of the area of the
line relative to the area of the rectangle is zero.
y
x = y
4
3
2
x
1
1 2 3
4
Exercise 2.4
(a)
The gure below shows the unit circle. The area to the right of the dark line
will be the area x>1/2.
The area of a chord of angle θ in degrees and radius, R, is given by the equation:
R2
A=
2
θπ
− sinθ
180
Here the circle has radius R=1. The angle θ will be the angle from one intersection of the line x = 1/2 and the circle, to the origin, then to the second
intersection of the line x = 1/2 and the circle. For x > 1/2 we can solve for the
angle to be 120o .
A=
1 120π
− sin 120o
2 180
2
≈ 0.614
y
x
1
x = --2
(b)
This space will look like a doughnut. The fraction
p of points is the area of the
whole circle minus the area of a circle of radius 1/2. Since A = π R2 , we get
π - π2 = π2 ≈ 1.571.
(c)
The constraint x+y > 1/2 can be written in the form of the area above the line
y = -x + 1/2. To solve for the intersections between this line and the unit circle,
notice y2 = x2 -x+1/4, plugging this in for y2 in the equation of the circle, we
can solve for the x coordinates of intersection to be 0.9114 and -0.4114. Using
this we can nd θ of this chord to be ≈ 138.59o . From the equation in part (a),
we obtain A ≈ 0.8787.
(d)
Zero, a line will have zero area and therefore the fraction of the area of the line
relative to the area of the circle is zero.
Exercise 2.5
(a)
The sample space will contain 50 outcomes because there are 5 options for
n and 10 options for k, therefore the pair (n,k) will have 50 options.
(b)
Let n=0, then there aren't any values of k such that n > k. Let n=1, if k=0
this is true, so this is one outcome. Similarly, n=2 will create two outcomes;
n=3 will create 3 more; lastly n=4 creates 4 more. In all this is 10 out of 50, so
1/5 of the outcomes satisfy n > k.
(c)
By a similar procedure as in Part (b), if n=0, then there are 9 choices of k
such that n < k. For n=2, there are 8, etc. Thus the total number of outcomes
3
for which n < k is 9+8+7+6+5=35. So the fraction of outcomes in the sample
space satisfying n < k is 35/50 or 7/10.
(d)
For n=k, we see there are 5 solutions (n=k=0; n=k=1; n=k=2; n=k=3;
n=k=4). Thus the fraction here is 5/50 or 1/10. Note since n > k, n < k, and
n=k covers the whole space, then they must add up to 1, so this could've been
determined from
1-Part (b) - Part (c) = 1 - 2/10 - 7/10 = 1/10
Exercise 2.6
(a) Example 2.1: Yes, ipping a fair coin should land on heads and tails equally
often.
Example 2.2: Yes, you would expect each of the six sides to appear equally often.
Example 2.3: Yes, each of the 36 outcomes are equally as likely.
Example 2.4: No, the probability of ζ 1 is 1/2, but the probability of ζ 2 is 1/4.
Furthermore the probability of ζ n is (1/2)n . These will not be equally likely.
Example 2.5: Yes, the real numbers selected by the random number generator
should in theory be equally likely to be selected.
(b) If you are given four Aces and one King face down and told to select one
card, then the outcome will be either an Ace or a King. Selecting an Ace will
occur more often than selecting a King, so the two outcomes will not occur
equally often.
Exercise2.7
Given M events A1 , A2 , . . . , AM : Ai ∩ Aj = φ ∀ i 6= j prove that
Pr
M
[
!
Ai
=
i=1
M
X
P r (Ai ) .
i=1
We shall prove this using induction. For the case M=2
Pr
2
[
!
Ai
=
i=1
2
X
P r (Ai ) .
i=1
This we know to be true from the axioms of the probability. Let us assume that
the proposition is true for M=k.
Pr
k
[
!
Ai
=
i=1
k
X
i=1
4
P r (Ai )
We need to prove that this is true for M=k+1. Dene an auxiliary event B :
B=
k
[
!
Ai
.
i=1
Then the event
k+1
[
!
Ai
= (B ∪ Ak+1 )
i=1
Pr
k+1
[
!
Ai
= P r (B ∪ Ak+1 ) .
i=1
Since the propostion is true for M=2 we can rewrite the above equation as
Pr
k+1
[
!
Ai
= P r (B) + P r (Ak+1 ) .
i=1
Since the proposition is true for M=k we can rewrite this as
=
k
X
P r (Ai ) + P r (Ak+1 )
i=1
=
k+1
X
P r (Ai ) .
i=1
Hence the propostion is true for M=k+1. And by induction the proposition is
true for all M .
Exercise 2.8
First, note that the sets A ∪ B and B can be rewritten as
A ∪ B = {A ∪ B} ∩ {B ∪ B} = {A ∩ B} ∪ B
B = B ∩ {A ∪ A} = {A ∩ B} ∪ {A ∩ B}.
Hence, A ∪ B can be expressed as the union of three mutually exclusive sets.
A ∪ B = {B ∩ A} ∪ {A ∩ B} ∪ {A ∩ B}.
Next, rewrite B ∩ A as
B ∩ A = {B ∩ A} ∪ {A ∩ A} = A ∩ {A ∪ B} = A ∩ {A ∩ B}.
Likewise, A ∩ B = B ∩ {A ∩ B}. Therefore A ∪ B can be rewritten as the
following union of three mutually exclusive sets:
A ∪ B = {A ∩ (A ∩ B)} ∪ {A ∩ B} ∪ {B ∩ (A ∩ B)}.
5
Hence
P r (A ∪ B) = P r A ∩ (A ∩ B) + P r (A ∩ B) + P r B ∩ (A ∩ B) .
Next, write A as the union of the following two mutually exclusive sets
A = {A ∩ {A ∩ B}} ∪ {A ∩ B}.
Hence,
P r (A) = P r A ∩ {A ∩ B} + P r (A ∩ B)
and
P r A ∩ {A ∩ B} = P r (A) − P r (A ∩ B) .
Likewise,
P r B ∩ {A ∩ B} = P r (B) − P r (A ∩ B) .
Finally,
P r (A ∪ B)
= P r A ∩ (A ∩ B) + P r (A ∩ B) + P r B ∩ (A ∩ B)
=
(P r (A) − P r (A ∩ B)) + P r (A ∩ B) + P r (B) − P r (A ∩ B)
= P r (A) + P r (B) − P r (A ∩ B) .
Exercise 2.9
P r(A ∪ B ∪ C)
= P r((A ∪ B) ∪ C)
= P r(A ∪ B) + P r(C) − P r((A ∪ B) ∩ C)
= P r(A) + P r(B) − P r(A ∩ B) + P r(C) − P r((A ∩ C) ∪ (B ∩ C))
= P r(A) + P r(B) + P r(C) − P r(A ∩ B) − P r((A ∩ C) ∪ (B ∩ C))
= P r(A) + P r(B) + P r(C) − P r(A ∩ B)
− (P r((A ∩ C) + P r(B ∩ C)) − P r(A ∩ C ∩ B ∩ C))
= P r(A) + P r(B) + P r(C) − P r(A ∩ B) − P r(A ∩ C) − P r(B ∩ C) + P r(A ∩ B ∩ C)
Exercise 2.10
Since A ⊂ B and A ∩ B = A, B = A ∪ {B ∩ A}. Since A and B ∩ A are mutually
exclusive, we have
P r(B) = P r(A) + P r(B ∩ A) .
Hence, by (1) and considering P r(B ∩ A) ≥ 0, P r(A) ≤ P r(B).
6
(1)
Exercise 2.11
M
[
Pr
!
M
X
≤
i=1
P r(Ai )
i=1
We shall prove this by induction. For k = 1 this reduces to
P r(A1 ) ≤ P r(A1 )
which is obviously true.For k = 2 we have
P r(A1 ∪ A2 ) = P r(A1 ) + P r(A2 ) − P r(A1 ∩ A2 )
by the axioms of probability.
⇒ P r(A1 ∪ A2 ) ≤ P r(A1 ) + P r(A2 )
The equality holding when the events are mutually exclusive. Assume that the
propostion is true for M = k. Then we have
Pr
!
k
[
≤
Ai
k
X
i=1
P r(Ai )
i=1
We shall prove that this is true for M = k + 1. Let
B=
k
[
Ai
i=1
Then the following holds
P r(B) ≤
k
X
P r(Ai )
(2)
i=1
Since the proposition is true for M = 2
P r(B ∪ Ak+1 ) ≤ P r(B) + P r(Ak+1 )
(3)
Using (2) in (3) we have
P r(B ∪ Ak+1 ) ≤
k
X
P r(Ai ) + P r(Ak+1 )
i=1
k+1
[
P r(
Ai ) ≤
i=1
P r(
k
X
P r(Ai ) + P r(Ak+1 )
i=1
k+1
[
Ai ) ≤
i=1
k+1
X
P r(Ai )
i=1
Thus the proposition is true for M = k + 1 and by the principle of induction it
is true for all nite M .
7
Exercise 2.12
(a)
Yes, because every event have a positive probability, the Pr(S) = 1 since the
sum of the four outcomes is 1, and since the set is nite the probability of event
(H,H) or (H,T) will be the sum of the two probabilities.
(b)
Pr(rst toss is heads) = Pr(H,H) + Pr(H,T) = 3/8 + 1/4 = 5/8
(c)
Pr(second toss is heads) = Pr(H,H) + Pr(T,H) = 3/8 + 1/4 = 5/8
Exercise 2.13
(a)
Yes, again because every event have a positive probability, the Pr(S) = 1
since the sum of the four outcomes is 1, and since the set is nite the probability
of event (H,H) or (H,T) will be the sum of the two probabilities.
(b)
Pr(rst toss is heads) = Pr(H,H) + Pr(H,T) = 25/64 + 15/64 = 40/64 =
5/8
(c)
Pr(second toss is heads) = Pr(H,H) + Pr(T,H) = 25/64 + 15/64 = 40/64
= 5/8
Exercise 2.14
Yes. Axiom 2.1 is satised since each event has a nonnegative probability.
Axiom 2.2 is satised because the sum of all the outcomes is 1, thus Pr(S) =
1. Lastly Axiom 2.3 is satised because the probability of rolling a 1 or a 2 is
going to be the sum of each of their probabilities.
Exercise 2.15
Assume S is the sample space for a given experiment.
Axiom 2.1: For an event A in S ,
P (A) = lim
n→∞
nA
.
n
Since nA ≥ 0, and n > 0, P (A) ≥ 0.
Axiom 2.2: S is the sample space for the experiment. Since S must happen
with each run of the experiment, nS = n. Hence
P (S) = lim
n→∞
8
nS
=1.
n
Axiom 2.3a: Suppose A ∩ B = 0. For an experiment that is run n times, assume
the event A ∪ B occurs n0 times, while A occurs nA times and B occurs nB
times. Then we have n0 = nA + nB . Hence
n0
nA + nB
nA
nB
= lim
= lim
+ lim
= P (A) + P (B) .
n→∞ n
n→∞
n→∞
n→∞
n
n
n
P (A ∪ B) = lim
Axiom 3.b: For an experiment that is run n times, assume the event Ai occurs
nAi times, i = 1, 2, · · · . Dene event C = A1 P
∪ A2 · · · ∪ Ai · · · . Since any two
events are mutually exclusive, event C occurs ∞
i=1 nAi times. Hence,
P(
∞
[
i=1
P∞
Ai ) = lim
n→∞
i=1
n
nA i
=
∞
X
∞
X
nA i
=
P (Ai ) .
n→∞ n
i=1
i=1
lim
Exercise 2.16
(a) One possible probability assignment is:
P r(H, H, H) = 1/64,
P r(H, H, M ) = P r(H, M, H) = P r(M, H, H) = 3/64,
P r(H, M, M ) = P r(M, H, M ) = P r(M, M, H) = 9/64,
P r(M, M, M ) = 27/64.
Note that the probability of hitting the target of the rst toss is
P r(hit) = P r(H, M, M )+P r(H, H, M )+P r(H, M, H)+P r(H, H, H) = 9/64+3/64+3/64+1/64 = 1/4.
Similar calculations would lead to the probability of a hit on the 2nd or on
the 3rd throw would also be 1/4.
(b) This assignment is not unique. A dierent assignment which leads to a
probability of hit of 1/4 on any throw is as follows:
P r(H, H, H) = 1/128,
P r(H, H, M ) = P r(H, M, H) = P r(M, H, H) = 3/64,
P r(H, M, M ) = P r(M, H, M ) = P r(M, M, H) = 19/128,
P r(M, M, M ) = 13/32.
9
Exercise 2.17
In order to use the classical approach in which each case is equally as likely,
the spinner should be divided into many pieces of equal size. Once in equally
sized pieces, determine how many of these pieces are needed to create the areas
labeled 1, 2, and 3 (making the pieces smaller if necessary). Now each piece is
equally likely and this is a more reasonable probability assignment.
Exercise 2.18
(a) The possible outcomes include: T, HT, HHT, HHHT,HHHH There are 5
outcomes, and you would not expect each to be equally probable.
(b) In order to use the classical approach, each outcome must be equally as
likely. Since there are four coin tosses and each has two possible outcomes,
then consider the 16 dierent possible combinations of ips as the outcomes for
the classical approach {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH,
HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}.
Now each outcome has the probability of occurring of 1/16.
(c) There are 8 dierent combinations that begin with tails, so Pr(T) will be
8/16. Similarly, there are 4 combinations that start o with the sequence HT,
so Pr(HT) = 4/16. This leads to the probability assignment to be as follows:
P r(T ) =
1
1
1
1
1
; P r(HT ) = ; P r(HHT ) = ; P r(HHHT ) =
; P r(HHHH) =
2
4
8
16
16
Exercise 2.19
Extending the solution to Problem 2.18, let n = the number of ips until a tail
occurs (i.e. n=1 corresponds to ipping a tail on the rst toss)
The number of outcomes in the classical approach will be 2n .
As we saw, half of these outcomes must start with T, and we can see as n
increases by 1 the probability decreases by 1/2.
This leads to the probability assignment to be as follows:
P r(T ) =
1
1
1
1
1
; P r(HT ) = ; P r(HHT ) = ; P r(HHHT ) =
; P r(HHHHT ) =
; ...
2
4
8
16
32
Or
P r(T ) =
1
1
1
1
1
; P r(HT ) = 2 ; P r(HHT ) = 3 ; P r(HHHT ) = 4 ; P r(HHHHT ) = 5 ; ...
1
2
2
2
2
2
10
Exercise 2.20
(a) For three dice, the minimum roll is 3 and the maximum roll is 18. Now
3+18
= 10.5. Since 10.5 is not a possible sum, this means 10 and 11 are both
2
equally probable. The probability of the sum adding up to be 10 is the number
of ways you can add three numbers to equal 10 divided by the total number of
outcomes. The total number of outcomes is 63 . The number of ways to add
three numbers to get ten is 24.
So Pr(sum = 10) = Pr(sum = 11) = 1/9 ≈ 0.111.
(b) For four dice, the minimum roll is 4 and the maximum roll is 24. Similarly
4+24
= 14. Thus 14 is the most likely sum. Following the same procedure in
2
part (a), we obtain:
Pr(sum = 14) = 44/1296 ≈ 0.034.
Exercise 2.21
Axiom 2.1:
P r(A | B) =
P r(A, B)
≥0
P r(B)
since both P r(A, B) and P r(B) are greater than zero.
Axiom 2.2:
P r(S | B) =
P r(S, B)
P r(B)
=
=1
P r(B)
P r(B)
Axiom 2.3:
P r(A ∪ B | C)
=
=
=
P r((A ∪ B) ∩ C)
P r((A ∩ C) ∪ (B ∩ C))
=
P r(C)
P r(C)
P r(A ∩ C) P r(B ∩ C) P r(A ∩ B ∩ C)
+
−
P r(C)
P r(C)
P r(C)
P r(A | C) + P r(B | C) − P r(A ∩ B | C)
Exercise 2.22
Show that if P r(B | A) = P r(B) then
(a) P r(A, B) = P r(A)P r(B)
P r(B | A)
=
P r(B | A)
=
P r(A, B)
P r(A)
P r(B)
(4)
(5)
Equating (4) and (5) we get
P r(B)
=
P r(A, B)
=
11
P r(A, B)
P r(A)
P r(A)P r(B)
(6)
(7)
(b) P r(A | B) = P r(A)
P r(A | B)
=
P r(A, B)
P r(B)
(8)
Since (5) ⇒ (7) we get using (7)
P r(A | B)
P r(A | B)
P r(A)P r(B)
P r(B)
= P r(A)
=
(9)
(10)
Exercise 2.23
Let Di = ith diode chosen is defective.
P r(D1 ∩ D2 ) = 1 − P r(D1 ∩ D2 ) = 1 − P r(D1 )P r(D2 | D1 )
P r(D1 ) =
25
30
P r(D2 | D1 ) =
24
29
(after the rst diode is selected and found to be not defective, there are 29
diodes remaining of which 5 are defective and 24 are not)
P r(D1 ∩ D2 ) = 1 −
9
25 24
·
=
30 29
29
Exercise 2.24
(a)
P r(1st = red, 2nd = blue) = P r(1st = red)P r(2nd = blue | 1st = red)
3
P r(1st = red) =
12
5
P r(2nd = blue | 1st = red) =
11
3 5
5
P r(1st = red, 2nd = blue) =
·
=
.
12 11
44
(b) P r(2nd = white) =
(c) Same as part (b).
4
12
= 31 .
12
Exercise 2.25
(a)
P r(5)
=
P r(5)
=
P r(5, 5)
=
1
6
5
6
5 5
25
· =
6 6
36
(b) Pr(sum = 7)
The sum of 7 can occur in the following 6 possible ways.
sum = {(1;6), (2;5) , (3;4) , (4;5) , (5;2) , (6;1) }. And there are a total of 36
outcomes in the sample space.
P r(sum = 7) =
1
6
=
36
6
(c) A= { (3;5) , (5;3) }
P r(A) =
2
1
=
36
18
(d)
P r(A = 5)
=
P r(B = (5 | 4))
=
P r(A, B)
=
1
6
2
6
1
1 2
· =
6 6
18
Alternatively the desired event is given by the following set of outcomes
X ={ (5;4) , (5,5) }
P r(X)
=
P r(5)
=
P r(5, 5)
=
2
1
=
36
18
(e)
13
1
6
1 1
1
· =
6 6
36
(f)
P r(A = 6)
=
P r(B = 6)
=
P r(A ∪ B)
=
=
=
1
6
1
6
P r(A) + P r(B) − P r(A ∩ B)
1 1 1 1
+ − ·
6 6 6 6
11
36
Exercise 2.26
(a)
P r ({1st ∈ (2, 3, 4)} ∩ {2nd ∈ (2, 3, 4)})
= P r (1st ∈ (2, 3, 4)) × P r (2nd ∈ (2, 3, 4))
1
3 3
= × = .
6 6
4
(b) The possible combinations of the two die can be (6, 4), (5, 3), (4, 2), (3, 1).
Hence,
P r (1st − 2nd = 2) = 4 ×
1 1
1
× = .
6 6
9
(c) Since one roll is 6, all possible combinations of two die are (1, 6), (2, 6), (3, 6),
(4, 6), (5, 6), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2), (6, 1). Since only two combinations
satisfy the requirement, (4, 6) and (6, 4), we have
P r (sum = 10 | one roll = 6) =
2
.
11
(d) Similarly, all possible combinations of two die are (1, 5), (2, 5), (3, 5), (4, 5),
(5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6). Since four combinations satisfy
the requirement, (5, 2), (5, 3), (2, 5) and (3, 5), we have
P r (sum ∈ (7,8) | one roll = 5) =
4
.
11
(e) Since the sum is 7, all possible combinations of two die are (4, 3), (5, 2),
(6, 1), (3, 4), (2, 5), (1, 6). Hence, we have
P r (one roll = 4 | sum = 7) =
14
2
1
= .
6
3
Exercise 2.27
Determine if Pr(A|B)>Pr(B|A) is always true, sometimes true, or never true,
given that Pr(A)>Pr(B).
P r(A|B) > P r(B|A)
P r(A ∩ B)
P r(B ∩ A)
>
P r(B)
P r(A)
Since all probabilities are greater than one, we have:
P r(A)P r(A ∩ B) > P r(B)P r(B ∩ A)
And since Pr(A∩B) = Pr(B∩A), we obtain:
P r(A) > P r(B)
This is given to be true, so Pr(A|B)>Pr(B|A) is always true.
**It should be noted that this is assuming Pr(B)6=0
Exercise 2.28
Prove that for any two events A and B, Pr(A∩B) ≤ Pr(A) ≤ Pr(A∪B).
Looking at the rst inequality and since any probability is ≤ 1, we know
Pr(A∩B) = Pr(B|A) Pr(A) ≤ 1*Pr(A) = Pr(A)
This proves the rst inequality.
Now Pr(A∪B) = Pr(A) + Pr(B) - Pr(A∩B)
By the rst inequality proved, we see that Pr(B) ≥ P(A∩B) so Pr(B) P(A∩B) ≥ 0.
This implies Pr(A∪B) ≥ Pr(A)
Which is equivalent to Pr(A) ≤ Pr(A∪B) as desired. Exercise 2.29
Prove Theorem 2.4 that a combined experiment, E = E1 xE2 xE3 x...xEm consisting of experiments Ei each with ni outcomes, i=1, 2, 3,...,m, has a total
number of possible outcomes given by n = n1 n2 n3 ...nm =
m
Y
i=1
15
= ni
Proof by induction:
Let Nm = total number of outcomes in E1 xE2 xE3 x...xEm
Base Cases:
For m=1: E = E1 which has n1 outcomes. Therefore N1 =n1 .
For m=2: E = E1 xE2 which has n1 possible outcomes for the rst part and n2
possible outcomes for the second part. So E will have n1 n2 possible outcomes.
Therefore N2 =n1 n2 = N1 n2 .
Induction Hypothesis
Assume Nm = Nm−1 nm .
Induction Step
Need to show Nm+1 = nm+1 Nm
From our induction hypothesis we know Nm =Nm−1 nm and Nm+1 will have
the all of the possible outcomes of Nm times the possible outcome of Em+1 ,
which is nm+1 . Therefore Nm+1 = Nm nm+1 as required.
Furthermore Nm = Nm−1 nm and Nm−1 = Nm−2 nm−1 etc.
Therefore, Nm = n1 n2 n3 ...nm =
m
Y
= ni
i=1
Exercise 2.30
Prove Theorem 2.8 that given a set of n distinct elements, the number of ways
to partition the set into m groups where the i th group has ni elements is given
by the multinomical coecient,
n
n1 , n2 , ..., nm
=
n!
n1 !n2 !...nm !
Proof by induction:
Base Cases:
For m=1, then ni = n1 = n. We know n can be partitioned only one way to a
group of n elements. This agrees with
n
n1
=
n!
n!
=
=1
n1 !
n!
Induction Hypothesis
Assume the following holds for some positive integer m
n
n1 , n2 , ..., nm
=
16
n!
n1 !n2 !...nm !
Induction Step
Need to show
n
n1 , n2 , ..., nm , nm+1
=
n!
n1 !n2 !...nm !nm+1 !
We know
n!
n!
1
=
n1 !n2 !...nm !nm+1 !
n1 !n2 !...nm ! nm+1 !
From our induction hypothesis we obtain
n!
=
n1 !n2 !...nm !nm+1 !
n
1
n1 , n2 , ..., nm nm+1 !
Now it can be shown that
n!
=
n1 !n2 !...nm !nm+1 !
n
as desired. n1 , n2 , ..., nm , nm+1
Exercise 2.31
(a)
n
k
=
n!
k!(n − k)!
n!
(n − k)!(n − (n − k))!
n!
n
=
n−k
(n − k)!(k)!
n
n
=
n−k
k
n
n−k
=
(b)
17
n
k
+
n
k+1
=
=
=
=
=
=
n!
n!
+
k!(n − k)! (k + 1)!(n − k − 1)!
n!
1
1
+
k!(n − k − 1)! n − k k + 1
n!
k+1+n−k
k!(n − k − 1)! (n − k)(k + 1)
n!
1+n
k!(n − k − 1)! (n − k)(k + 1)
(n + 1)!
(k + 1)!(n − k)!
n+1
k+1
(c)
n X
n
= 2n
k
k=0
Consider the binomial expansion of (p + q)n . We have
n X
n
(p + q) =
· pk · q (n−k)
k
n
k=0
Put p = q = 1 . Then we get
2n =
n X
n
k
k=0
(d)
n X
n
(−1)k = 0
k
k=0
Consider the binomial expansion of (p + q)n . We have
n X
n
(p + q) =
· pk · q (n−k)
k
n
k=0
Put p = −1, q = 1 . Then we get
n X
n
(−1)k = 0
k
k=0
18
(e)
n X
n
k = n2n−1
k
k=0
Once again consider the binomial expansion (p + q)n .
n X
n
(p + q) =
· pk · q (n−k)
k
n
k=0
Dierentiate both sides with respect to p.
d
d
n
(p + q) =
dp
dp
n−1
n(p + q)
=
!
n X
n
k
(n−k)
·p ·q
k
k=0
n X
k=0
n
k
· k · pk−1 · q (n−k)
(11)
Once again put p = q = 1. We get the following relation.
n2n−1 =
n X
n
·k
k
k=0
The above relation can be rewritten as follows.
n2
n−1
n X
n
n
=
· k + 0.
k
0
k=1
n
X
n
·k
n2n−1 =
k
k=1
(f)
n X
n
k=0
k
· k · (−1)k = 0
For this we proceed in the same way as in (e) but we substitute p = −1, q = 1
in (11). This gives us
n X
n
· k · pk−1 · q (n−k) = 0
k
k=0
n X
n
· k · (−1)k−1 = 0
k
k=0
Multiplying by -1 through out gives the required identity.
n X
n
· k · (−1)k = 0
k
k=0
19
Exercise 2.32
(a)
P r (two Aces)
P r (two of any kind)
3
4 3 48
=
· P r A, A, A = 3 ·
·
·
= 0.0130.
2
52 51 50
= 13 · P r (two Aces) = 0.1694.
(b)
P r (three Aces)
=
P r (three of any kind)
=
4 3 2
·
·
= 0.000181.
52 51 50
13 · P r (three Aces) = 0.00235.
P r (A, A, A) =
(c)
P r (three Hearts)
P r (three of any suit)
13 12 11
·
·
= 0.0129.
52 51 50
4 · P r (three Hearts) = 0.0518.
= P r (H, H, H) =
=
(d)
P r (straight)
= P r (2, 3, 4) + P r (3, 4, 5) + . . . + P r (Q, K, A)
4 4 4
= 11 · P r (2, 3, 4) = 11 ·
·
·
= 0.003092.
52 51 50
Exercise 2.33
(a)
P r (1 Heart)
=
13 · P r H, H, H, . . . , H
=
13 39 38
28
13 ·
·
·
...
=
52 51 50
40
(b)
20
13
1
39
12
·
52
13
= 0.08006
We can choose anywhere between 7 to 13 cards of a given suit to satisfy
the given condition. All these events are mutually exclusive. Let Ai denote the
event of having i Hearts. Following a procedure similar to part (a), it is found
that
13
39
·
i
P r (Ai ) =
13−i
52
13
.
Therefore
P r (at least 7 Hearts) = P r
!
13
[
Ai
13
X
=
i=7
= Pr
!
13
[
P r(Ai )
i=7
Ai
13
X
=
i=7
i=7
13
i
39
13−i
52
13
·
.
The probability of at least 7 cards from any suit is simply 4 times the probability
of at least 7 Hearts. Hence
P r (at least 7 cards from any suit) =
13 4 X 13
39
·
= 0.0403.
52
i
13 − i
13 i=7
(c)
P r (no Hearts)
=
P r H, H, H, . . . , H
=
39 38 37
27
·
·
...
=
52 51 50
40
39
13
52
13
= 0.01279
Exercise 2.34
(a) Four letter sequences means each letter has 26 options, so the total number
of combinations is 26*26*26*26 = 264 = 456,976
(b) Four letter sequences without repeating letters, so each next letter has one
less options, so the total number of combinations is 26*25*24*23 = 358,800
(c) The number of ways 26 letters can be ordered is 26! ≈ 4.03*1026
Exercise 2.35
(a) There are 2n distinct words.
(b) Method 1:
P r(2 ones) = P r({110} ∪ {101} ∪ {011}) = P r(110) + P r(101) + P r(011) =
21
3
.
8
Method 2:
2 3−2
3
1
1
3
P r(2 ones) =
·
·
= .
2
2
2
8
Exercise 2.36
(a) There are mn distinct words.
(b) For this case there are 43 distinct words.
P r(2 pulses of level 2) =
2 4−2
4
1
2
8
·
·
=
.
2
3
3
27
Exercise 2.37
(a) For some even interger n, then there will be n/2 0s and n/2 1s. The number
of ways they can be arranged will be
n!
( n2 )!( n2 )!
(b) Want to nd the smallest n such that the expression in part (a) is greater
than or equal to 100. If n=8, we get 70 dierent characters. For n=10 there are
252 dierent characters. So the minimum length of codeword would need to be
10 bits.
Exercise 2.38
(a) Since the rst digit of an area code is not a zero, then this position has 9
options. Now the second digit of the area code has 2 options (1 or 0). Now the
third digit of the area code cannot be a 1 if the second digit is 1, so look at two
cases:
Case 1: The second digit is a 0. Now the number of combinations is 9*1*10
= 90
Case 2: The second digit is a 1. The number of combinations is 9*1*9 = 81
In all there will be 171 valid area codes.
(b) For a seven digit phone number, the rst digit cannot be a 0, and the second
cannot be a 0 or 1. This leaves the number of combinations to be 9*8*105 =
7,200,000.
(c) The number of ten digit phone numbers is just the total possible combinations of the area codes and seven digit phone numbers, so 171*72*105 =
1,231,200,000.
22
Exercise 2.39
(a)
P r (3 heads) =
9 1 3 1 9−3
21
( ) ( )
=
= 0.1641.
3 2 2
128
(b)
P r (at least 3 heads)
9 X
9 1 i 1 9−i
=
( )( )
i 2 2
i=3
2 X
233
9 1 i 1 9−i
=
= 0.9102 .
= 1−
( )( )
2
2
256
i
i=0
(c)
P r (at least 3 heads and at least 2 tails)
=
=
P r (3 ≤ number of heads ≤ 7)
7 X
9 1 i 1 9−i
57
=
= 0.8906 .
( )( )
i
2
2
64
i=3
Exercise 2.40
(a) The number of possible blackjack hands is
52
2
= 1326.
(b) If we don't make a distinction between suits, then there are thirteen dierent
cards of which to choose
two plus you could have any of 13 dierent pairs.
So there will be 13
2 + 13 = 91 dierent hands.
(c) If we consider 10's, Jacks, Queens, and Kings to all be the same, then there
would only be 10 dierent cardsof which to choose two plus the possible number
of pairs. Thus there will be 10
2 + 10 = 55 dierent hands.
Exercise 2.41
(a) Pr(hand worth 18 or more) = Pr(A ∩ 7, 8, 9, 10, J, Q, or K) + Pr(10, J,
Q, K ∩ 8 or 9) + Pr(10, J, Q, K ∩ 10, J, Q,K) + Pr(9 ∩ 9)
=
4
52
28
52
+
16
52
=
8
52
+
16
52
492
≈ 0.182
2704
23
15
52
+
4
52
3
52
(b) Pr(12 ≤ sum ≤ 17) = 1 - Pr(hand worth 11 or less) - Pr(hand worth 18 or
more)
So to solve this we'll nd Pr(hand worth 11 or less) = Pr(2 ∩ 2) + Pr(2 ∩
3, 4, 5, 6, 7, 8, or 9) + Pr(3 ∩ 3) + Pr(3 ∩ 4, 5, 6, 7, or 8) + Pr(4 ∩ 4) + Pr(4
∩ 5, 6, 7) + Pr(5 ∩ 5) + Pr(5 ∩ 6)
Noticing Pr(2 ∩ 2) = Pr(3 ∩ 3) = Pr(4 ∩ 4) = Pr(5 ∩ 5), we get
3
4
28
4
20
4
12
4
4
4
+
+
+
+
=4
52
52
52
52
52
52
52
52
52
52
304
Now Pr(hand worth 11 or less) = 2704
This leads to Pr(12 ≤ sum ≤ 17) = 1 -
492
2704
-
304
2704
=
1908
2704
≈ 0.706
Exercise 2.42
You have 16 points. The dealer has 7 + x points.
Pr(dealer has more points than you) = Pr(dealer's points ≥ 17)
In order for the dealer to have 17 or more points, the dealer would need a 10,
J, Q, K, or A.
Since one 10 is already on the table, then there are 3+4+4+4+4=19 cards
that would give the dealer more points than you. The probability of the dealer
getting one of these cards is 19/49 ≈ 0.3878.
Exercise 2.43
(a) A poker hand is 5 out of 52 cards, so there are
52
5
= 2,598,960 possibilities.
(b) There will be 48 poker hands containing all four aces.
Thus Pr(being dealt four aces) = 48/2,598,960 ≈ 1.85*10−5 .
Now Pr(being dealt four of a kind) = 13 * Pr(being dealt four aces) =
624/2,598,960
≈ 2.40*10−4
(c) For a poker hand with 3 aces and 2 other cards, there will be 43 48
=
2
4512 dierent
poker hands.
Thus Pr(being dealt three aces) = 4512/2,598,960 ≈ 0.0017.
Now Pr(being dealt three of a kind) = 13 * Pr(being dealt three aces) =
58656/2,598,960
≈ 0.0226.
24
Exercise 2.44
(a) The number
can divide 20 students into 4 teams of 5 is:
10of ways he 10
20
15
5
5
5 ≈ 1.17*10
(b) Let T stand for Tommy and B for Bobby.
Pr(T and B are on the same team) = Pr(T and B on Team 1) + Pr(T and
B on Team 2) + Pr(T and B on Team 3) + Pr(T and B on Team 4) = 4*Pr(T
and B on Team 1)
Now Pr(T and B on Team 1) = Pr(T on Team 1) * Pr(B on Team 1 | T on
Team 1) = (5/20)(4/19)
So Pr(T and B are on the same team) = 4(5/20)(4/19) = 80/380 ≈ 0.21.
(c) Let F stand for Frank.
Pr(T not with F ∩ B not with F) = Pr(T and B are on the same team ∩
not with F) + Pr(T and B on dierent teams ∩ not with F)
4
15
15
10
5
+ 1
≈ 0.614
=4
20
19
18
19
18
Exercise 2.45
P r(defective) = P r(defective | A) · P r(A) + P r(defective | B) · P r(B)
1
0.15
= (0.15) ·
+ (0.05) ·
= 0.137.
1.15
1.15
Exercise 2.46
(a) Pr(0 received) = Pr(0 transmitted)*Pr(0 received | 0 transmitted) +
Pr(1 transmitted)*Pr(0 received | 1 transmitted) = (1/2)(0.95) + (1/2)(0.10)
= 0.525
Pr(1 received) = Pr(0 transmitted)*Pr(1 received | 0 transmitted) +
Pr(1 transmitted)*Pr(1 received | 1 transmitted) = (1/2)(0.05) + (1/2)(0.90)
= 0.475
Also you could've noticed that Pr(1 received) = 1 - Pr(0 received) = 0.475
(b) Pr(1 transmitted | 0 received)
=
=
Pr(1 transmitted ∩0 received)
Pr(0 received)
Pr(1 transmitted)*Pr(0 received |1 transmittted)
Pr(0 received)
25
=
0.5 ∗ 0.1
≈ 0.095
0.525
Following the same approach we see Pr(0 transmitted | 1 received)
=
0.5 ∗ 0.05
≈ 0.0526
0.475
(c) Pr(error) is the sum of the two answers in part (b) ≈ 0.148.
Exercise 2.47
(a) Pr(E received) = Pr(E received|0 transmitted) Pr(0 transmitted) + Pr(E
received|1 transmitted) Pr(1 transmitted) = (0.09)(0.5) + (0.16)(0.5) = 0.125
(b) Pr(0 transmitted|E received) =
Pr(E received|0 transmitted) Pr(0 transmitted)
Pr(E received)
=
(0.09)(0.5)
≈ 0.2601
(0.125)
(c) Pr(error) = Pr(0 received|1 transmitted) + Pr(1 received|0 transmitted) =
0.04 + 0.01 = 0.05
Exercise 2.48
(a) If 7 H were observed, what is the probability the coin ipped was the fair
coin?
P r(f aircoin|7H) =
P r(f aircoin ∩ 7H)
P r(f aircoin ∩ 7H)
=
P r(7H)
P r(f aircoin ∩ 7H) + P r(unf aircoin ∩ 7H)
We see that
10 7 3
1
1
10
1
1
7
P r(f aircoin ∩ 7H) =
=
2 210
2
7
2
2
P r(unf aircoin ∩ 7H) =
7 3
1
10
3
1
2
7
4
4
P r(f aircoin|7H) ≈ 0.319.
26
(b) If 3 H were observed, what is the probability the coin ipped was the fair
coin?
Following the same method above:
P r(f aircoin|3H) =
P r(f aircoin ∩ 3H)
P r(f aircoin ∩ 3H)
=
P r(3H)
P r(f aircoin ∩ 3H) + P r(unf aircoin ∩ 3H)
We see that
10 3 7
1
10
1
1
1
3
=
P r(f aircoin ∩ 3H) =
10
2 2
2
2
2
3
3 7
1
10
3
1
P r(unf aircoin ∩ 3H) =
2
3
4
4
P r(f aircoin|3H) ≈ 0.974.
Exercise 2.49
We know that Pr(B) + Pr(G) = 1. And Pr(B|P) + Pr(G|P) = 1.
Also Pr(P|B) =
P r(B∩P )
P r(B)
and Pr(P|G) =
P r(G∩P )
P r(G)
Now Pr(P) = Pr(B ∩ P) + Pr(G ∩ P) = Pr(B) Pr(P|B) + Pr(G) Pr(P|G)
Lastly, Pr(B|P) =
P r(B∩P )
P r(P )
and Pr(G|P) =
P r(G∩P )
P r(P )
Using these equations the table can be completed. Note the blue answers
are the ones that needed to be determined and are rounded to the nearest three
digits if necessary.
Year in School
2nd grade
4th grade
6th grade
8th grade
Pr(P)
0.278
0.37
0.50
0.74
Pr(B)
0.45
0.60
0.52
0.641
Pr(G)
0.55
0.40
0.48
0.359
27
Pr(P|B)
0.25
0.35
0.50
0.75
Pr(P|G)
0.30
0.40
0.50
0.722
Pr(B|P)
0.405
0.568
0.52
0.65
Pr(G|P)
0.595
0.432
0.48
0.35
Exercise 2.50
For the probability assignment in Exercise 2.12,
Pr(H on rst toss) = Pr(H,H) + Pr(H,T) = 3/8 + 1/4 = 5/8,
Pr(H on second toss) = Pr(H,H) + Pr(T,H) = 3/8 + 1/4 = 5/8,
Pr(T on rst toss) = Pr(T,T) + Pr(T,H) = 1/8+1/4 =3/8,
Pr(T on second toss) = Pr(T,T) + Pr(H,T) = 1/8+1/4=3/8.
Therefore, Pr(H on rst toss)Pr(H on second toss) = (5/8)(5/8) = 25/64. This is
not equal to Pr(H,H)=3/8. So this probability assignment does not correspond
to independent coin tosses. Going through a similar set of calculations for
the distribution specied in Exercise 2.13, it is also found that this does not
correspond to independent tosses.
Exercise 2.51
(a) Let p = P r (Ace is drawn) = 4/52 = 1/13.
P r (1st Ace on 5th selection) = (1 − p)4 p =
12 4 1
·
= 0.0558.
13 13
(b)
P r (at least 5 cards drawn before Ace) = (1 − p) =
5
12 5
= 0.6702.
13
(c)
48
·
52
48
P r (at least 5 cards drawn before Ace) =
52
P r (1st Ace drawn on 5th selection) =
47
·
51
47
·
51
46
·
50
46
·
50
45
·
49
45
·
49
4
= 0.0599 .
48
44
·
= 0.6588 .
48
Exercise 2.52
(a)
13
If the cards are replaced the probability of drawing a club is p = 52
= 14 . We
need to nd the probability that we draw 2 clubs in 7 trials and a club on the
7
8th trial. This is given by
· p2 (1 − p)5 · p.
2
P r(3rd club drawn on 8th selection) =
7
2
1 3
· ( )3 ( )5 = 0.0779.
4 4
(b)
We need to nd the probability that we draw either 0, 1 or 2 clubs in 8 trials.
This follows the binomial distribution and this probability is given by
2 X
1 3
729
8
P r(at least 8 cards drawn before 3rd club) =
·( )i ( )8−i =
= 0.7119.
i
4
4
1024
i=0
28
(c)
P r(3rd club drawn on 8th selection)
= P r(2 clubs drawn in 7 selections)P r(8th card = club | 2 clubs in rst 7 selections).
13 39
P r(2 clubs in 7 selections) =
2
5
52
7
= 0.3357.
2
P r(8th card = club | 2 clubs in rst 7 selections) =
.
45
13 39
2
· 2 525 = 0.0149.
P r(3rd club drawn on 8th selection) =
45
7
P r(at least 8 cards drawn before 3rd club)
=
=
P r(less than 3 clubs in 7 selections)
2
X
P r(k clubs in 7 selections)
k=0
=
2
X
13
k
39
7−k
52
7
k=0
= 0.7677.
Exercise 2.53
n
1
P r(all nversions erased) =
.
2
Exercise 2.54
(i) (I, M ) is not permissible. If A and B are independent, then P r(A, B) =
P r(A)P r(B). By assumption, P r(A) 6= 0 and P r(B) 6= 0 and hence for independent events A and B it follows that P r(A, B) 6= 0 and therefore they are not
mutuallly exclusive.
(ii) (I, N M ) is permissible. For example let A = {throw a 1 on rst roll of a die}
and B = {throw a 2 on second roll of a die}. These two events are independent,
but not mutually exclusive.
(iii) (N I, M ) is permissible. For example let A = {throw a 1 on rst roll of a die}
and B = {sum of two rolls of a die is 8}. These two events are not independent,
but they are mutually exclusive.
(iv) (N I, N M ) is permissible. This is the most general case. For example let
A = {throw a 2 on rst roll of a die} and B = {sum of two rolls of a die is 8}.
These two events are not independent, nor are they mutually exclusive.
29
Exercise 2.55
(a) This is true.
Proof:
Since A and B are independent we have Pr(A ∩ B) = Pr(A) Pr(B)
Want to show Pr(A ∩ B̄ ) = Pr(A) Pr(B̄ )
We know Pr(A ∩ B) + Pr(A ∩ B̄ ) = Pr(A)
Thus Pr(A ∩ B̄ ) = Pr(A) - Pr(A ∩ B)
So Pr(A ∩ B̄ ) = Pr(A) - Pr(A) Pr(B) = Pr(A) 1-Pr(B)
And 1- Pr(B) = Pr(B̄ )
Therefore Pr(A ∩ B̄ ) = Pr(A) Pr(B̄ ) as required. A is independent of B̄ (b) This is true.
Proof:
Want to show Pr(Ā ∩ B̄ ) = Pr(Ā) Pr(B̄ )
We know Pr(Ā) = 1- Pr(A) and similarly Pr(B̄ ) = 1- Pr(B)
Thus Pr(Ā) Pr(B̄ ) = 1 - Pr(B) - Pr(A) - Pr(A)Pr(B)
= 1 - Pr(B) - Pr(A) - Pr(A ∩ B)
= 1 - Pr(A ∪ B)
= Pr(Ā ∩ B̄ ) as required. Ā is independent of B̄ Exercise 2.56
Pr(A → D) = Pr((L1 ∩ L2 ∩ L4 ) ∪ (L1 ∩ L5 ) ∪ (L3 ∩ L4 ))
= Pr(L1 ∩ L2 ∩ L4 ) + Pr(L1 ∩ L5 ) + Pr(L3 ∩ L4 ) - Pr(L1 ∩ L2 ∩ L4 ∩ L5 )
- Pr(L1 ∩ L2 ∩ L3 ∩ L4 ) - Pr(L1 ∩ L3 ∩ L4 ∩ L5 ) + Pr(L1 ∩ L2 ∩ L3 ∩ L4 ∩ L5 )
= p3 + 2p2 - 4p4 + p5 .
Exercise 2.57
(a) P r(X = k) =
(b)
n
X
n
k
pk (1 − p)
P r(X = k) =
k=0
n−k
.
n X
n k
n−k
n
p (1 − p)
= (p + (1 − p)) = 1n = 1.
k
k=0
(c) Binomial
30
Exercise 2.58
(a) Since
∞
X
PX (k) = 1 ,
k=0
i.e.,
∞
X
c
0.37k = 1 .
k=0
= 1, which gives c = 0.63.
Hence
(b) Similarly,
c
1−0.37
∞
X
c
0.82k = 1 .
k=1
= 1 + c, which gives c = 0.2195.
Hence
(c) Similarly,
c
1−0.82
c
24
X
0.41k = 1
k=0
which gives c = 0.5900.
(d) Similarly,
c
15
X
0.91k = 1
k=1
which gives c = 0.1307.
(e) Similarly,
c
6
X
0.412k = 1
k=0
which gives c = 0.8319.
Exercise 2.59
(a) The probability mass function for a binomial random variable is given by
P r(X = k) =
n
k
pk (1 − p)n−k
Tabulating the probabilities for various values of k we get the following
31
k
0
1
2
3
4
5
6
7
8
9
10
Probability
0.10737418240000
0.26843545600000
0.30198988800000
0.20132659200000
0.08808038400000
0.02642411520000
0.00550502400000
0.00078643200000
0.00007372800000
0.00000409600000
0.00000010240000
Refer to the gure below for the plot.
Probablity Mass Function of Binomial & Poisson Random Variables
0.35
Binomial n=10, p=0.2
Poisson n=10, np= 0.2
0.3
Pr(X=k) −>
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
k−>
6
7
8
9
10
(b) The probability mass function for a Poisson distribution is given by
P r(X = k) = e−α
αk
k!
Tabulating the probabilities for various values of k we get the following
32
k
0
1
2
3
4
5
6
7
8
9
10
Probability
0.13533528323661
0.27067056647323
0.27067056647323
0.18044704431548
0.09022352215774
0.03608940886310
0.01202980295437
0.00343708655839
0.00085927163960
0.00019094925324
0.00003818985065
Refer to Figure ?? for the plot.
1 k 4 10−k
P
= 0.0328
(c) Binomial: P r(X ≥ 5) = 1 − P r(X < 5) = 1 − 4k=0 10
k 5 5
P
k
Poisson: P r(X ≥ 5) = 1 − P r(X < 5) = 1 − 4k=0 2k! e−2 = 0.0527
The Poisson approximation is not particularly good for this example.
Exercise 2.60
(a)
λt =
P r(X < 3) =
10calls
1
·
minute = 1
minute
10
2
X
1 −1
1
1
5
e = (1 + 1 + ) =
= 0.9197.
k!
e
2
2e
k=0
(b)
λt =
P r(X < 3) =
2
X
60k
k=0
k!
10calls
minute
· (6 minutes) = 60
e−60 = e−60 (1 + 60 +
3600
) = 1861 · e−60 = 1.627 × 10−23 .
2
Exercise 2.61
(a)
P r(win with one ticket) = p =
(b)
P r(4 winners) =
1
= 6.29 ∗ 10−8 .
50
6
6
6 ∗ 106 4
p (1 − p)6∗10 −4 = 5.8055 ∗ 10−4 .
4
33
(c) The probability that n people win is
P r(n winners) =
6
6 ∗ 106 n
p (1 − p)6∗10 −n .
n
Note that when n = 0, the above probability evaluates to 0.6855. Since this
probability is greater than 1/2, it is impossible for any other number of winers
to be more probable. Hence the most probable number of winning tickets is
zero.
(d) Here, for the Possion approximation, α = np = (501 ) ∗ 6 ∗ 106 = 0.3774.
6
Hence,
P (4) =
0.37744 −0.3774
α4 −α
e =
e
= 5.7955 ∗ 10−4 .
4!
4!
Compared with results in (b), the Possion distribution is an accurate approximation in this example.
For the case of (c),
P (n) =
αn −α
0.3774n −0.3774
e =
e
.
n!
n!
When n = 0, P (n) will be maximum, which is the same as the result in c).
Exercise 2.62
P r(X = 0)
=
P r(X = 1)
=
P r(X = 2)
=
4 3 2
· · =
6 5 4
2 4 3
3· · ·
6 5 4
4 2 1
3· · ·
6 5 4
1
.
5
3
.
5
1
= .
5
=
Exercise 2.63
Let p = P r(success) =
1
10
.
10
· p · (1 − p)9 = 0.3874.
1
P r(≥ 2 successes) = 1 − P r(≤ 1 success) = 1 − P r(0 successes) − P r(1 success).
P r(1 success) =
P r(0 successes) = (1 − p)1 0 = 0.3487.
P r(≥ 2 successes) = 1 − 0.3487 − 0.3874 = 0.2639.
34
Exercise 2.64
(a) If more than 1 error occurs in a 7-bit data block, the decoder will be in
error. Thus, the decoder error probability is
Pe =
7 X
7
(0.03)i (1 − 0.03)7−i = 0.0171 .
i
i=2
(b) Similarly,
Pe =
15 X
15
i=3
i
(0.03)i (1 − 0.03)15−i = 0.0094 .
(c) Similarly,
Pe =
31 X
31
i=4
i
(0.03)i (1 − 0.03)31−i = 0.0133 .
Exercise 2.65
(a) Pr(red) = 18/38 ≈ 0.4737
(b) Let N be the number of plays/bets and the player bets on red until he wins.
Pr(N=1) = 18/38
Pr(N=2) = (20/38)(18/38)
Pr(N=3) = (20/38)2 (18/38)
Pr(N=k) = (20/38)k−1 (18/38)
(c) Now let M be the number of plays/bets and the gambler bets on red until
he wins twice.
Pr(M=2) = (18/38)2
Pr(M=3) = (3 -1) (20/38)(18/38)2
Pr(M=4) = (4-1) (20/38)2 (18/38)2
Pr(M=k) = (k-1) (20/38)k−2 (18/38)2
Exercise 2.66
Let N be the number of cards that are drawn before the second appearance of
an Ace. For example for {2, 5, K, 7, A, 5, 3, J, A,...}, N has a value of N = 8.
Pr(N=1) is the set {A, A} = (4/52)2
Pr(N=2) is the set {A, non-ace, A} or {non-ace, A, A} = 2(48/52)(4/52)2
Pr(N=3) = 3(48/52)2 (4/52)2
Pr(N=k) = k(48/52)k−1 (4/52)2
35
Exercise 2.67
P r(K = k) = P r
k
[
[{M = m} ∩ {N = k − m}]
m=0
=
k
X
P r(M = m, N = k − m)
k+1
k
X
=
P r(M = m)P r(N = k − m)
m=0
=
k
X
(λB )m e−λB (λA )k−m e−λA
m!
(k − m)!
m=0
= e−λB e−λA (λA )k
k
X
(λB )m (λA )−m
(m!)(k − m)!
m=0
Multiplying and dividing by k!
m
k
k!
e−(λB +λA ) (λA )k X
λB
=
k!
m!(k − m)! λA
m=0
=
m
k e−(λB +λA ) (λA )k X k
λB
k!
λA
m
m=0
=
k
e−(λB +λA ) (λA )k
λB
1+
k!
λA
=
e−(λB +λA ) (λA + λB )k
k!
This follows a Poisson random variable.
Exercise 2.68
(a) If the sum of the rst roll is 7, then we have Pr(N=1) = 6/36 = 1/6
Now Pr(N=2) = Pr(sum of rst roll is not 7 ∩ sum of second roll is 7) =
(30/36)(6/36)
Pr(N=k) = (30/36)k−1 (6/36) = (5/6)k−1 (1/6)
(b) Pr(N≥4) = Pr(N=4) + Pr(N=5) + ... = 1 Pr(N=1) + Pr(N=2) +
Pr(N=3)
From Part (a) we know Pr(N=1) = 1/36, Pr(N=2) = 5/36, Pr(N=3) =
25/216
Therefore Pr(N≥4) = 125/216 ≈ 0.579
36
Exercise 2.69
(a)
5
P r (two Aces) =
· P r A, A, A, A, A
2
5
4 3 48 47 46
=
·
·
·
·
·
= 0.03993.
2
52 51 50 49 48
P r (two of any kind) = 13 · P r (two Aces) = 0.5191.
Note that the above calculations also allow that the hand may have two pair
or a full house. Hence to be completely accurate (in the poker sense) we must
subtract these probabilities.
P r (two of a kind)
=
0.5191 − P r (two pair) − P r (full house)
=
0.5191 − 0.04754 − 0.001441 = 0.4701.
Note that Pr(2 pair) is calculated in part (c) and Pr(full house) is calculated in
part (e).
(b)
5
P r (three Aces) =
· P r A, A, A, A, A
3
5
4 3 2 48 47
=
·
·
·
·
·
= 0.001736.
3
52 51 50 49 48
P r (three of any kind) = 13 · P r (three Aces) = 0.02257.
Note that the above calculations also allow that the hand may have a full house
and hence this probability must be subtracted.
P r (three of a kind)
=
0.02257 − P r (full house)
=
0.02257 − 0.001441 = 0.02113.
(c)
P r (two Aces, two Kings)
5
3
=
·
· P r A, A, K, K, A ∪ K
2
2
5
3
4 3 4 3 44
=
·
·
·
·
·
·
= 0.0006095.
52 51 50 49 48
2
2
13
P r (two pair) =
· P r (two Aces, two Kings) = 0.04754.
2
(d)
P r (ve Hearts)
=
P r (ush)
=
13 12 11 10 9
·
·
·
·
= 0.0004952.
52 51 50 49 48
4 · P r (ve Hearts) = 0.001981.
37
Note that the above calculations also allow that the hand may have a straightush and hence this probability must be subtracted.
P r (straight-ush)
=
P r (ush)
=
4·9
.
52 · 51 · 50 · 49 · 48
4 · 9 · (13 · 12 · 11 · 10 − 1)
= 0.001981.
52 · 51 · 50 · 49 · 48
(e)
P r (three Aces, two Kings)
5
· P r (A, A, A, K, K)
3
4 3 2 4 3
·
·
·
·
= 9.235 × 10−6 .
= 10 ·
52 51 50 49 48
P r (full house) = 13 · 12 · P r (three Aces, two Kings) = 0.001441.
=
(f)
P r (10,J,Q,K,A)
=
P r (straight)
=
4 4 4 4 4
·
·
·
·
= 3.283 × 10−6 .
52 51 50 49 48
4 · P r (10,J,Q,K,A) = 2.955 × 10−5 .
Exercise 2.70
The highest number rolled wins. If there is a tie, the defense is the winner.
Let D stand for defense and O for oense.
(a) Both players roll a single die:
Pr(O wins) = Pr(O wins|O rolls 1) Pr(O rolls 1) + Pr(O wins|O rolls 2) Pr(O
rolls 2) + Pr(O wins|O rolls 3) Pr(O rolls 3) + Pr(O wins|O rolls 4) Pr(O rolls
4) + Pr(O wins|O rolls 5) Pr(O rolls 5)+ Pr(O wins|O rolls 6) Pr(O rolls 6)
1 0 1 2 3 4 5
15
Pr(O wins) =
+ + + + +
=
≈ 0.4167.
6 6 6 6 6 6 6
36
Therefore Pr(D wins) = 1 - Pr(O wins) ≈ 0.5833.
(b) Oense rolls two dice while defense rolls one die:
With three dice being rolled, there will be 6*6*6= 216 dierent possible
outcomes. The oense will win if his/her highest roll is higher than the defense's
roll.
Pr(O wins) = Pr(O highest rolls is 1) Pr(O wins|O highest roll is 1) + ...+
Pr(O highest rolls is 6) Pr(O wins|O highest roll is 6)
Pr(O wins) =
1
3 1 5 2 7 3 9 4 11 5
125
∗0+ ∗ + ∗ + ∗ + ∗ + ∗ =
≈ 0.5787.
36
36 6 36 6 36 6 36 6 36 6
216
38
Therefore Pr(D wins) ≈ 0.4213.
(c) Oense rolls three dice while defense rolls one die:
As in Part (b): Pr(O wins) = Pr(O highest rolls is 1) Pr(O wins|O highest
roll is 1) + ...+ Pr(O highest rolls is 6) Pr(O wins|O highest roll is 6)
Pr(O wins) =
1
7 1 19 2 37 3 61 4 91 5
855
∗0+
∗ +
∗ +
∗ +
∗ +
∗ =
≈ 0.6597.
216
216 6 216 6 216 6 216 6 216 6
1296
Therefore Pr(D wins) ≈ 0.3403.
Exercise 2.71
Each player's highest roll is compared with the other player's highest roll, and
their second highest roll is compared with the other player's second highest roll.
The higher number of each comparison wins. If there is a tie, the defense is the
winner.
Let D stand for defense and O for oense.
(a) Both the oense and defense roll two dice:
(i) Pr(O wins both comparisons)
To see the various combinations, consider the following matrix where the
defense rolls (i,j) and the probability of the oense winning both given (i,j) is
in the matrix. Note the probabilities in the matrix need to be multiplied by the
1
factor 36
.
Adding up the elements of the matrix, Pr(O wins both) = ( 293
36 ) Pr(the spe1
cic roll (i,j)) = 293
*
.
Therefore
Pr(O
wins
both)
≈
0.2261.
36
36
(ii) Pr(each wins one), rst solve part (iii).
Now Pr(each wins one) = 1 - Pr(O wins both) - Pr(D wins both) ≈ 0.3271.
(iii) Pr(D wins both comparisons):
Now Pr(D wins both) is represented in the matrix given the defense rolls
(i,j). Note again the probabilities in the matrix need to be multiplied by the
1
factor 36
39
Adding up the elements of the matrix, Pr(D wins both) = 579
36 * Pr(the spe1
cic roll (i,j)) = 579
*
.
Therefore
Pr(D
wins
both)
≈
0.4468.
36
36
(b) The oense rolls 3 dice and the defense rolls two:
Following the same procedure as part (a), we see
(i) Pr(O wins both comparisons) ≈ 0.3717.
Note: Another way to solve this is to rst solve part (iii) then nd:
Pr(O wins max) ≈ 0.4716 and thus Pr(D wins max) ≈ 0.5284.
Pr(O wins 2nd comparison) ≈ 0.6075 and thus Pr(D wins 2nd comparison)
≈ 0.3925.
Now let A = Pr(D wins max) and B = Pr(D wins 2nd)
Pr(A ∪ B) = Pr(A) + Pr(B) - Pr(A ∩ B) = 0.5284 + 0.3925 - 0.2926 ≈
0.6283.
And Pr(Ā ∩ B̄ ) = 1 - Pr(A ∪ B) = Pr(O wins both) ≈ 0.3717.
(ii) Pr(each wins one) ≈ 0.3357.
Using part (i) and (iii) we see Pr(each wins one) = 1 - Pr(O wins both) Pr(D wins both) ≈ 0.3357.
(iii) Pr(D wins both comparisons):
≈ 0.2926.
Pr(D wins both) is represented in the matrix given the defense rolls (i,j).
1
Note the probabilities in the matrix need to be multiplied by the factor 216
Adding up the elements of the matrix, Pr(D wins both) = 2275
216 * Pr(the
1
specic roll (i,j)) = 2275
*
.
Therefore
Pr(D
wins
both)
≈
0.2926.
216
36
Exercise 2.72
(a) Suppose the slots are A, B, C, D. Without loss of generality suppose Player
1 chooses slots A, B, and C. Pr(player 1 wins) = Pr(nobody chooses D) = (3/4)3
≈ 0.4219.
(b) If Players 2, 3, and 4 choose to all hide in the same place, then three boxes
will be empty and their chance of not being chosen is 1/4. Thus Player 1 will
40
win 3/4 of the time.
(c) If Players 2, 3, and 4 choose to all hide in dierent slots, then in order
for Player 1 to win, Player 1 must select the three boxes containing the three
players. There is one combination of boxes out of the possible four, so Player 1
will win 1/4 of the time.
Exercise 2.73
Pr(rst player loses) = 1/5 since one of the ve detonators is live.
Now the Pr(second player loses) = Pr(rst player didn't lose ∩ choosing live
detonator)
This is equivalent to
Pr(rst player didn't lose) * Pr(choosing live detonator | rst player didn't
lose) = (4/5)(1/4) = 1/5
Now Pr(third player loses)
= Pr(rst player didn't lose ∩ second player didn't lose ∩ choosing live
detonator)
= Pr(rst player didn't lose) Pr(second player didn't loose | rst player didn't
lose) Pr(third player choose the detonator | rst and second player didn't)
= (4/5)(3/4)(1/3) = 1/5.
Finally, Pr(fourth player loses) = Pr(rst, second, and third players didn't lose
∩ choosing live detonator) = (4/5)(3/4)(2/3)(1/2) = 1/5.
Therefore this is a fair game as each player is equally likely to lose.
Exercise 2.74
(a) Given Pr(bulb lasts k hours) = (1-a)ak
So if the bulb is S-type: Pr(bulb lasts exactly 200 hours)=(0.01)(0.99200 ) ≈
0.00134
If the bulb is L-type: Pr(bulb lasts exactly 200 hours)=(0.001)(0.999200 ) ≈
0.000819
Using Bayes Rule:
Pr(S-type|200 hours) =
Pr(200 hours|S-type) Pr(S-type)
Pr(200 hours|S-type) Pr(S-type) + Pr(200 hours|L-type) Pr(L-type)
41
0.01 0.99200
= 3
0.01 0.99200 41 0.001 0.999200
4
3
4
≈ 0.8308.
Therefore the unmarked box is most likely the S-type.
(b) Pr(error) = Pr(L-type|200 hours) = 1 - Pr(S-type|200 hours) = 0.1692.
Exercise 2.75
You are better o changing to the unopened door. You're original selection has
a 1/3 chance of containing the prize, and this will not change after the host
opens a door. However, after the host takes away a door that he knows contains
a goat, then the other door has a 2/3 chance of containing the prize.
To view this consider the three cases of where the prize (P) could be in
relation to the goats (G):
Door1 Door2 Door3
Case1
Case2
Case3
P
G
G
G
P
G
G
G
P
Without loss of generality, assume you select Door 1. Now Pr(P) = 1/3
before and after the host opens a door. However, after the host reveals a goat
(from either Door 2 or Door 3) then notice that switching doors will lead to the
prize 2 out of 3 times (in Case 2 and Case 3).
42