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JEE Advanced 2016 Paper 1
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© examsnet.com
Question : 41
Total: 54
A computer producing factory has only two plants
T
1
and
T
2
Plant
T
1
produces 20% and plant
T
2
produces 80% of the total computers produced. 7% of computers produced in the factory turn outto be defective. It is known that
P(computer turns out to be defective given that is produced in plant
T
1
= 10P(computer turns out to be defective given that it is produced in plant
T
2
where P(E) denotes the probability of an event E. A computer produces in the factory is randomlyselected and it does not turn out to be defective. Then the probabality that it is produced in plant
T
2
is
36
73
47
79
78
93
75
83
Validate
Solution:
P
(
T
1
)
=
20
100
P
(
T
2
)
=
80
100
Let
P
(
D
T
2
)
=
x
P
(
D
T
1
)
=
10
x
P
(
D
)
=
7
100
(given)
P
(
T
1
)
P
(
D
T
1
)
+
P
(
T
2
)
+
P
(
D
T
2
)
=
7
100
20
100
×
10
x
+
80
100
×
x
=
7
100
x
=
1
40
P
(
D
T
2
)
=
1
40
⇒
P
(
−
D
T
2
)
=
39
40
P
(
D
T
1
)
=
10
40
⇒
P
(
−
D
T
1
)
=
30
40
P
(
T
2
−
D
)
=
80
100
×
39
40
20
100
×
30
40
+
80
100
×
39
40
=
78
93
© examsnet.com
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