STAT 100A HWII Solution
Problem 1:
If we flip a fair coin
n
times independently, what is the probability that we observe
k
heads?
k
= 0
,
1
, ..., n
. Please explain your answer.
A: The probability is
(
n
k
)
/
2
n
. The reason is that all the 2
n
sequences are equally likely, and the
number of sequences with exactly
k
heads is
(
n
k
)
, i.e., among the
n
flips, we choose
k
flips to be
heads, and the rest of the
n

k
flips to be tails. The number of such choices is
(
n
k
)
.
Problem 2:
Prove the following two identities:
(1)
P
(
¯
A

B
) = 1

P
(
A

B
).
A:
P
(
¯
A

B
) =
P
(
¯
A
∩
B
)
P
(
B
)
=
P
(
B
)

P
(
A
∩
B
)
P
(
B
)
= 1

P
(
A

B
)
.
(2)
P
(
A
∪
B

C
) =
P
(
A

C
) +
P
(
B

C
)

P
(
A
∩
B

C
).
A:
P
(
A
∪
B

C
)
=
P
((
A
∪
B
)
∩
C
)
P
(
C
)
=
P
((
A
∩
C
)
∪
(
B
∩
C
))
P
(
C
)
=
P
(
A
∩
C
) +
P
(
B
∩
C
)

P
((
A
∩
C
)
∩
(
B
∩
C
))
P
(
C
)
=
P
(
A

C
) +
P
(
B

C
)

P
(
A
∩
B

C
)
Problem 3:
Independence. If
P
(
A

B
) =
P
(
A
), prove
(1)
P
(
A
∩
B
) =
P
(
A
)
P
(
B
).
A:
P
(
A

B
) =
P
(
A
∩
B
)
/P
(
B
) =
P
(
A
), so
P
(
A
∩
B
) =
P
(
A
)
P
(
B
).
(2)
P
(
B

A
) =
P
(
B
).
A:
P
(
B

A
) =
P
(
B
∩
A
)
/P
(
A
) =
P
(
A
)
P
(
B
)
/P
(
A
) =
P
(
B
).
Problem 4:
Prove the following two identities (
C
stands for cause,
E
stands for effect).
(1) Rule of total probability:
P
(
E
) =
P
(
E

C
)
P
(
C
) +
P
(
E

¯
C
)
P
(
¯
C
).
A:
P
(
E
) =
P
(
E
∩
C
) +
P
(
E
∩
¯
C
) =
P
(
C
)
P
(
E

C
) +
P
(
¯
C
)
P
(
E

¯
C
).
(2) Bayes rule:
P
(
C

E
) =
P
(
C
)
P
(
E

C
)
/
[
P
(
C
)
P
(
E

C
) +
P
(
¯
C
)
P
(
E

¯
C
].
A:
P
(
C

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

C
)
/
[
P
(
C
)
P
(
E

C
) +
P
(
¯
C
)
P
(
E

¯
C
].
Problem 5:
Suppose 1% of the population is inflicted with a particular disease. For a medical test,
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '07
 Wu
 Conditional Probability, Probability, Probability theory, Bayesian probability, randomly selected person, STAT 100A HWII

Click to edit the document details