This preview shows page 1. Sign up to view the full content.
STAT 100A HWIII Solution
Problem 1:
Prove the following two identities (
C
stands for cause,
E
stands for eﬀect), where
{
C
i
,i
= 1
,...,n
}
partition the whole sample space.
(1) Rule of total probability:
P
(
E
) =
∑
n
i
=1
P
(
C
i
)
P
(
E

C
i
).
A:
P
(
E
) =
P
(
∪
n
i
=1
(
E
∩
C
i
)) =
∑
n
i
=1
P
(
E
∩
C
i
) =
∑
n
i
=1
P
(
C
i
)
P
(
E

C
i
).
(2) Bayes rule:
P
(
C
j

E
) =
P
(
C
j
)
P
(
E

C
j
)
/
∑
n
i
=1
P
(
C
i
)
P
(
E

C
i
).
A:
P
(
C
j

E
) =
P
(
C
j
∩
E
)
/P
(
E
) =
P
(
C
j
)
P
(
E

C
j
)
/
∑
n
i
=1
P
(
C
i
)
P
(
E

C
i
).
Problem 2:
Suppose 1% of the population is inﬂicted with a particular disease. For a medical test,
if a person has the disease, then 95% chance the person will be tested positive. If a person does
not have the disease, then 90% chance the person will be tested negative. Using precise notation,
calculate
(1) The probability that a randomly selected person will be tested positive.
A: Let
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '09
 Wu

Click to edit the document details