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NIMCET 2019 Question Paper with solutions
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© examsnet.com
Question : 44
Total: 120
A computer producing factory has only two plaats
T
1
and
T
2
- Plant
T
1
produces 20% and plant T2 produces 80% of the total computer produced. 7% of the computers produced in the factory turn out to be defective. It is known that P(computer turns out to the defective given that it is produced in plant
T
1
) = 10P (computer turns out to be defective given that it is produced in plant
T
2
). A computer produced in the factory is randomly selected and its does not turn out to the defective. Then the probability that it is produced in plant
T
2
is
36
73
47
79
78
93
75
83
Validate
Solution:
Let
T
1
and
T
2
be the event that computer is produced by plant
T
1
and
T
2
respectively. Again, let D and ND be the event that computer produced is defective and non-defective respectively
P
(
T
1
)
=
20
100
,
P
(
T
2
)
=
80
100
,
P
(
D
)
=
7
100
Again let
P
(
D
T
1
)
=
10
‌
‌
P
(
D
T
2
)
=
k
∴
‌
‌
P
(
D
T
1
)
=
k
and
P
(
D
T
2
)
=
k
Now,
P
(
D
)
=
P
(
T
1
)
P
(
D
T
1
)
+
P
(
T
2
)
P
(
D
T
2
)
⇒
7
100
=
20
100
×
k
+
80
100
×
k
10
⇒
7
=
20
k
+
8
k
⇒
k
=
1
4
Now, required probability
=
P
(
T
2
/
N
D
)
=
‌
‌
P
(
T
z
)
P
(
N
D
/
T
2
)
P
(
T
1
)
P
(
N
D
/
T
1
)
+
P
(
T
2
)
P
(
N
D
/
T
2
)
=
‌
‌
‌
‌
80
100
×
(
1
−
‌
‌
1
40
)
‌
‌
20
100
(
1
−
‌
‌
1
4
)
+
‌
‌
80
100
(
1
−
‌
‌
1
40
)
=
‌
‌
4
×
‌
‌
39
40
3
4
+
4
×
39
40
=
4
×
39
30
+
4
×
39
=
156
186
=
78
93
© examsnet.com
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