SOLUTION FOR QUIZ 1, STAT 4382
As usual, you can find very minor mistakes so do everything by yourself and check my
calculations. If you do everything on your own, and do correctly with understanding why
you use this or that solution — you should be ready for Exam 1. Read your text and look
at HWs — this is also a good path to prepare for the exam.
Note: there are no points for the quiz. But during exam you will be asked to grade your
first 4 homeworks from 0 to 40 (each hw gives you 10 points).
Now let us look at your problems.
1. By the total probability Theorem
P
(
B
) =
P
(
BA
) +
P
(
BA
c
)
.
(1)
Using the given probabilities we write
P
(
BA
) =
P
(
A
)
P
(
B

A
) = (
.
4)(
.
8) =
.
32
and
P
(
BA
c
) =
P
(
A
c
)
P
(
B

A
c
) = [1

P
(
A
)]
P
(
B

A
c
) = [1

.
4][
.
4] =
.
24
.
Plug the results in (1) and get
P
(
B
) =
.
32 +
.
24 =
.
56
.
2. We know (we discussed this in class and it was your HW plus a formula in the text)
P
(
AB
) =
P
(
A
) +
P
(
B
)

P
(
A
∪
B
)
.
Given the same event, conditional probabilities have the same properties as regular ones, so
we can write
P
(
AB

C
) =
P
(
A

C
) +
P
(
B

C
)

P
(
A
∪
B

C
)
.
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 Fall '11
 krwaw
 Conditional Probability, Probability theory, Yi, X1, 1 j

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