Assignment.3 solution
February, 12, 2010
MATH 203
1
Solution :
Defining the events as
A
i
=
{
the ith draw is defective.
}
So we are looking for finding the probability of
P
(
A
c
1
∩
A
c
2
∩
A
3
∩
A
4
)
=
P
(
A
c
1
)
P
(
A
c
2

A
c
1
)
P
(
A
3

A
c
1
∩
A
c
2
)
P
(
A
4

A
c
1
∩
A
c
2
∩
A
3
)
=
15
20
.
14
19
.
5
18
.
4
17
= 0
.
03611971
3.66
Solution :
Since A, B, and C are all mutually exclusive, we know that
P
(
A
∩
B
) =
P
(
A
∩
C
) =
P
(
B
∩
C
) = 0
(a)
P
(
A
∪
B
) =
P
(
A
) +
P
(
B
)

P
(
A
∩
B
) =
.
30 +
.
55

0 =
.
85
(b)
P
(
A
∩
B
) = 0
(c)
P
(
A

B
) =
P
(
A
∩
B
)
P
(
B
)
=
0
.
55
= 0
(d)
P
(
B
∪
C
) =
P
(
B
) +
P
(
C
)

P
(
B
∩
C
) =
.
55 +
.
15

0 =
.
70
(e) No, B and C are not independent events. If B and C are independent
events, then
P
(
B

C
) =
P
(
B
). From the problem, we know P(B) = .55.
P
(
B

C
) =
P
(
B
∩
C
)
P
(
C
)
=
0
.
15
= 0
.
Thus, since
P
(
B

C
)
6
=
P
(
B
), events B and C are not independent.
3.72
Solution :
Define the following events, and from the start we can assume that everything
is conditional on neuroblastoma so
B:
{
Child undergoes surgery
}
B:
{
Surgery is successful in curing the disease
}
From the problem, we know
P
(
B
) =
.
20 and
P
(
C

B
) =
.
95
.
We also know that
P
(
C

B
) =
P
(
C
∩
B
)
P
(
B
)
for any events B and C.
Thus,
P
(
C
∩
B
) =
P
(
C
∩
B
)
P
(
B
)
=
.
95(
.
20) =
.
19
1
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Assignment.2 solution
February, 3, 2010
MATH 203
3.79
Solution :
Define the following events:
R
i
:
{
Fish
i
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 Spring '08
 Dr.JoseCorrea
 Statistics, Conditional Probability, Probability, Probability theory

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