Stat/Math 414
Solutions # 5
1.411
.
a) Denote this event by
A
. Let
S
be the space of all unordered samples without
replacement of size 3 from 12 bulbs. We denote by
N
the size of the outcome space
S, N
(
A
) the number of all favorable outcomes.
N
=
±
12
3
¶
, N
(
A
) =
±
3
3
¶
= 1;
P
(
A
) =
N
(
A
)
N
=
1
220
.
b) Denote this event by
A
.
B=“There were 2 defective bulbs in the ﬁrst 4 tests”.
C=“The 5th tested bulb is defective”.
A
=
B
∩
C, P
(
A
) =
P
(
B
∩
C
) =
P
(
C

B
)
P
(
B
)
P
(
B
) =
(
3
2
)(
9
2
)
(
12
4
)
=
24
110
;
P
(
C

B
) =
3

2
12

4
=
1
8
.
Hence
P
(
A
) =
24
110
·
1
8
=
3
110
.
c) Let this event be
A
.
B=“There were 2 defective bulbs in the ﬁrst 9 tests”.
C=“The 10th tested bulb is defective.
A
=
B
∩
C, P
(
B
) =
(
3
2
)(
9
9

2
)
(
12
9
)
=
27
55
;
P
(
C

B
) =
1
3
.
P
(
A
) =
27
55
·
1
3
=
9
55
.
1.414
.
N
= 52
6
;
N
(
A
) = 52
·
51
·
50
·
49
·
48
·
47.
P
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 '08
 staff
 Math, Conditional Probability, prior probabilities, sour milk

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