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Homework 3 — Solutions
Chapter 2
Problem 7 (a) There are 615 outcomes in the sample space.
(b) There are 315 outcomes without any blue-collar workers, so that
there are 615 − 315 outcomes with at least one blue-collar worker.
(c) If there are no independents, then for each player, there are 4
outcomes, so that there are 415 outcomes altogether.
Problem 8 Suppose that A, B are mutually exclusive events for which P (A) = 0.3
and P (B) = 0.5.
(a) P (either A or B) = P (A) + P (B) = 0.8 because A and B are
mutually exclusive.
(b) P (A occurs but B does not) = 0.3.
(c) P (A ∩ B) = 0.
Problem 9 Let A be the event that a randomly chosen customer carries an Amex
card, and let V be the event that a randomly chosen customer carries a
Visa card. Then P (A or B) = P (A ∪ B) = P (A) + P (B) − P (AB) =
0.24 + 0.61 − 0.11 = 0.74. Hence, 74 percent of all customers carry an
acceptable credit card.
Problem 12 Let E be the event that a randomly chosen student is taking Spanish,
let F be the event that a randomly chosen student is taking French, and
let D be the event that a randomly chosen student is taking German.
1
Note that
P (E) =
P (F ) =
P (D) =
P (EF ) =
P (ED) =
P (F D) =
P (EF D) =
28
100
26
100
16
100
12
100
4
100
6
100
2
100
=
=
=
=
=
=
=
7
25
13
50
4
25
3
25
1
25
3
50
1
50
(a) The probability of a student not being in any of these classes is
1 − P (E ∪ F ∪ D) =1 − (P (E) + P (F ) + P (D)
− P (EF ) − P (ED) − P (F D) + P (EF D))
1
1
28 + 26 + 16 − 12 − 4 − 6 + 2
=1− = .
=1 −
100
2
2
(b) We have
P (only E) = P (E) − P (EF ) − P (ED) + P (EF D) = 0.14
P (only F ) = P (F ) − P (EF ) − P (F D) + P (EF D) = 0.1
P (only D) = P (D) − P (ED) − P (F D) + P (EF D) = 0.08
Hence, P (exactly one language class) = 0.14 + 0.1 + 0.08 = 0.32.
(c) Since fifty students are not taking any of the courses, the proba(502)
49
= 198
,
bility that neither of two randomly picked students is 100
(2)
so that the probability of at least one of them taking a language
49
class is 1 − 198
= 149
.
198
Problem 17 There are 64 · 63 · 62 · 61 · 60 · 59 · 58 · 57 ways of arranging 8 castles
a
Q8 on
2
chess board. Of these, there are 64 · 49 · 36 · 25 · 16 · 9 · 4 · 1 = i=1 i in
2
which none of the rooks can capture any of the others. So the answer
is
Q8 2
i=1 i
.171
64 · 63 · 62 · 61 · 60 · 59 · 58 · 57
Problem 18
2 · 4 · 16
.
52 · 51
Problem 20 Let A be the event that you are dealt a blackjack, and let B be the
event that the dealer is dealt a blackjack.
Then
2 · 4 · 16
52 · 51
4 · 4 · 16 · 3 · 15
P (AB) =
52 · 51 · 50 · 49
P (A ∪ B) = P (A) + P (B) − P (AB) = 0.0948.
P (A) = P (B) =
Hence, then probability that neither you nor the dealer is dealt a blackjack is 1 − P (A ∪ B) = 0.9052.
4
Problem 21 (a) P (1) = 20
= 15 , P (2) =
1
and P (5) = 20
.
8
20
= 25 , P (3) =
5
20
= 14 , P (4) =
2
20
=
1
,
10
1
4
= 12
, P (2) =
(b) There are 48 children altogether, so that P (1) = 48
2·8
1
3·5
5
4·2
1
5
= 3 , P (3) = 48 = 16 , P (4) = 48 = 6 , and P (5) = 48 .
48
Problem 25 Let En be the event that a sum of 5 occurs on the nth roll, and no
sum of 5 or 7 occurs on the first n − 1 rolls. There are 36 outcomes of
a single roll, and four of them give a sum of 5, while 6 of them give a
sum of 7. Hence,
n−1
n−1
26
4
13
1
P (En ) =
=
.
36
36
18
9
A sum of 5 occurs before a sum of 7 precisely if the events En occurs for
some n. Since En and Em are disjoint if n 6= m, the desired probability
is
n−1
∞
∞ X
X
13
1
1
1
1 18
2
P (En ) =
· = ·
= .
13 =
18
9
9 1 − 18
9 5
5
n=1
n=1
3
Problem 27
P (A wins in one move) =
P (A wins in three moves) =
P (A wins in five moves) =
P (A wins in seven moves) =
P (A wins) =
3
10
7
10
7
10
7
10
3
10
6
9
6
·
9
6
·
9
3
7
=
8
40
5 4 3
1
· · · =
8 7 6
12
5 4 3 2 3
1
· · · · · =
8 7 6 5 4
40
7
1
1
7
+
+
+
=
40 12 40
12
·
·
Problem 28 (a) Without replacement:
5
3
P (all three balls are the same color) =
+
6
+
3
19
3
8
3
With replacement:
P (all three balls are the same color) =
5
19
3 3 3
6
8
+
+
19
19
(b) Without replacement:
5
1
P (all three balls are of different colors) =
·
6
·
1
19
3
8
1
With replacement:
P (all three balls are of different colors) = 3! ·
5 6 8
·
·
19 19 19
Problem 32 There are (b + g)! ways to line up the children. There are g · (b + g − 1)!
arrangements with a girl in the ith position. The desired probability is
g(b+g−1)!
g
= b+g
.
(b+g)!
Problem 37 (a) There are 10
selections for the final exam. The number of selec5
tions that allow the student to solve all problems is 75 , so that
( 7)
the desired probability is 105 = 0.08333.
(5)
4
(b) There are 74 · 31 selections that’ll let the student solve exactly
four problems, so that the probability of solving at least four prob(7)+(7)·(3)
lems is 5 104 1 = 21 .
(5)
5