HW3 Solution
Problem 1
From lecture, you can assess these subjective probabilities via the indirect method where
you vary the probability of one of the lotteries until the individual is indifferent between
both lotteries.
This is the table that is given to you.
In one of the lotteries the assessee
wins if the wind speed
exceeds
a threshold
q
.
When the assessee is indifferent, we get
P
(Wind
> q
) =
p
. Therefore, we have the following:
1.
q
= 10
→
p
= 0
.
95
⇒
P
(Wind
≤
q
) = 0
.
05
2.
q
= 20
→
p
= 0
.
65
⇒
P
(Wind
≤
q
) = 0
.
35
3.
q
= 25
→
p
= 0
.
55
⇒
P
(Wind
≤
q
) = 0
.
45
4.
q
= 35
→
p
= 0
.
45
⇒
P
(Wind
≤
q
) = 0
.
55
Below is the graph:
1
Problem 2
You are given the following probabilities:
1. If we have no ROS, we get flooding if
•
Levees fail, call it event L: D or C or (A and B)
⇔
D or C or (A

B and B) or
•
Pumps fail, call it event P: F or (G and H)
P
(
Flood

wind >
30
,
¬
ROS
) =
P
(
L
∪
P

wind >
30
,
¬
ROS
)
P
(
L
) =
P
(
D
∪
C
∪
(
A
∩
B
))
=
P
(
D
) +
P
(
C
) +
P
(
A
∩
B
)

P
(
D
)
P
(
C
)

P
(
D
)
P
(
A
∩
B
)

P
(
C
)
P
(
A
∩
B
)
=
.
01 +
.
1 +
.
24

.
01(
.
1)

.
01(
.
24)

.
1(
.
24) +
.
01(
.
1)(
.
24)
= 0
.
32284
P
(
P
) =
P
(
F
∪
(
G
∩
H
))
=
P
(
F
) +
P
(
G
∩
H
)

P
(
F
∩
G
∩
H
)
=
P
(
F
) +
P
(
G
)
P
(
H
)

P
(
F
)
P
(
G
)
P
(
H
)
=
.
8 +
.
7(
.
1)

.
7(
.
1)(
.
8)
= 0
.
814
Therefore,
P
(
L
∪
P