Chapter 2:
Probability
9
Instructor’s Solutions Manual
2.5
a.
A
S
A
B
B
A
B
A
B
A
=
∩
=
∪
∩
=
∩
∪
∩
)
(
)
(
)
(
.
b.
A
A
B
B
B
A
B
B
A
B
=
∩
=
∩
∪
∩
=
∩
∪
)
(
)
(
)
(
)
(
.
c.
=
∩
∩
=
∩
∩
∩
)
(
)
(
)
(
B
B
A
B
A
B
A
0
/
.
The result follows from part a.
d.
)
(
)
(
B
B
A
B
A
B
∩
∩
=
∩
∩
=
0
/
.
The result follows from part b.
2.6
A
= {(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (1,4), (2,4), (3,4), (4,4), (5,4), (6,4), (1,6), (2,6),
(3,6), (4,6), (5,6), (6,6)}
C
= {(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)}
A
∩
B
= {(2,2), (4,2), (6,2), (2,4), (4,4), (6,4), (2,6), (4,6), (6,6)}
B
A
∩
= {(1,2), (3,2), (5,2), (1,4), (3,4), (5,4), (1,6), (3,6), (5,6)}
B
A
∪
= everything but {(1,2), (1,4), (1,6), (3,2), (3,4), (3,6), (5,2), (5,4), (5,6)}
A
C
A
=
∩
2.7
A
= {two males} = {M
1
, M
2
), (M
1
,M
3
), (M
2
,M
3
)}
B
= {at least one female} = {(M
1
,W
1
), (M
2
,W
1
), (M
3
,W
1
), (M
1
,W
2
), (M
2
,W
2
), (M
3
,W
2
),
{W
1
,W
2
)}
B
= {no females} = A
S
B
A
=
∪
=
∩
B
A
0
/
A
B
A
=
∩
2.8
a.
36 + 6 = 42
b.
33
c.
18
2.9
S
= {A+, B+, AB+, O+, A-, B-, AB-, O-}
2.10
a.
S
= {A, B, AB, O}
b.
P
({A}) = 0.41,
P
({B}) = 0.10,
P
({AB}) = 0.04,
P
({O}) = 0.45.
c.
P
({A} or {B}) =
P
({A}) +
P
({B}) = 0.51, since the events are mutually exclusive.
2.11
a.
Since
P
(
S
) =
P
(E
1
) + … +
P
(E
5
) = 1, 1 = .15 + .15 + .40 + 3
P
(E
5
).
So,
P
(E
5
) = .10 and
P
(E
4
) = .20.
b.
Obviously,
P
(E
3
) +
P
(E
4
) +
P
(E
5
) = .6.
Thus, they are all equal to .2
2.12
a.
Let L = {left tern}, R = {right turn}, C = {continues straight}.
b.
P
(vehicle turns) =
P
(L) +
P
(R) = 1/3 + 1/3 = 2/3.
2.13
a.
Denote the events as very likely (VL), somewhat likely (SL), unlikely (U), other (O).
b.
Not equally likely:
P
(VL) = .24,
P
(SL) = .24,
P
(U) = .40,
P
(O) = .12.
c.
P
(at least SL) =
P
(SL) +
P
(VL) = .48.
2.14
a.
P
(needs glasses) = .44 + .14 = .48
b.
P
(needs glasses but doesn’t use them) = .14
c.
P
(uses glasses) = .44 + .02 = .46
2.15
a.
Since the events are M.E.,
P
(
S
) =
P
(E
1
) + … +
P
(E
4
) = 1.
So,
P
(E
2
) = 1 – .01 – .09 –
.81 = .09.
b.
P
(at least one hit) =
P
(E
1
) +
P
(E
2
) +
P
(E
3
) = .19.