58
26.
These questions can be solved algebraically, or with the Venn diagram below.
a.
1
1
(
)
1
(
)
P A
P A
′
=
−
= 1 – .12 = .88.
b.
The addition rule says
)
(
)
(
(
)
(
)
P
B
P A
P B
A
B
A
P
∪
=
+
−
∩
. Solving for the intersection (“and”)
probability, you get
1
2
1
2
1
2
)
(
)
(
)
(
)
(
A
P A
P A
P A
P A
A
∩
=
+
−
∪
= .12 + .07 – .13 = .06.
c.
A Venn diagram shows that
)
(
)
(
)
(
B
P A
P A
P A
B
′
∩
=
−
∩
. Applying that here with
1
2
A
A
A
=
∩
and
B
=
A
3
, you get
1
2
3
1
2
1
2
3
([
)
(
(
]
)
)
P
A
P A
A
A
A
P A
A
A
′
∩
−
∩
∩
=
∩
∩
=
.06 – .01 = .05.
d.
The event “at most two defects” is the complement of “all three defects,” so the answer is just 1 –
1
2
3
(
)
P A
A
A
∩
∩
= 1 – .01 = .99.
27.
There are 10 equally likely outcomes: {A, B} {A, Co} {A, Cr} {A,F} {B, Co} {B, Cr} {B, F} {Co, Cr}
{Co, F} and {Cr, F}.
a.
P
({A, B}) =
1
10
= .1.
b.
P
(at least one
C
) =
P
({A, Co} or {A, Cr} or {B, Co} or {B, Cr} or {Co, Cr} or {Co, F} or {Cr, F}) =
7
10
= .7.
c.
Replacing each person with his/her years of experience,
P
(at least 15 years) =
P
({3, 14} or {6, 10} or
{6, 14} or {7, 10} or {7, 14} or {10, 14}) =
6
10
= .6.
28.
Recall there are 27 equally likely outcomes.
a.
P
(all the same station) =
P
((1,1,1) or (2,2,2) or (3,3,3)) =
9
1
27
3
=
.
b.
P
(at most 2 are assigned to the same station) = 1 –
P
(all 3 are the same) = 1 –
1
9
=
8
9
.
c.
P
(all different stations) =
P
((1,2,3) or (1,3,2) or (2,1,3) or (2,3,1) or (3,1,2) or (3,2,1))
=
9
2
27
6
=
.
.04
.05
.02
.00
.01
.01
.01
.86
A
1
A
2
A
3
Probability and Statistics for Engineers - Homework #
3
- Chapter
2
:
26, 32, and 38