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JEE Main 6 Apr 2024 Shift 1 Paper
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Section:
Mathematics
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© examsnet.com
Question : 10 of 90
Marks:
+1
,
-0
A company has two plants A and B to manufacture motorcycles.
60
%
60\%
60%
motorcycles are manufactured at plant
A
A
A
and the remaining are manufactured at plant B.
80
%
80\%
80%
of the motorcycles manufactured at plant
A
A
A
are rated of the standard quality, while
90
%
90\%
90%
of the motorcycles manufactured at plant B are rated of the standard quality. A motorcycle picked up randomly from the total production is found to be of the standard quality. If
p
p
p
is the probability that it was manufactured at plant
B
B
B
, then
126
p
126p
126
p
is
[6 Apr 2024 Shift 1]
54
64
66
56
Validate
Solution:
👈: Video Solution
A
B
Manufactured
60
%
60\%
60%
40
%
40\%
40%
Standard quality
80
%
80\%
80%
90
%
90\%
90%
P
(
P(
P
(
Manufactured at
B
/
B/
B
/
found standard quality
)
=
)=
)
=
?
A : Found S.Q
B : Manufacture B
C : Manufacture A
P
(
E
1
)
=
40
100
\;P(E_1)=\frac{40}{100}
P
(
E
1
)
=
100
40
P
(
E
2
)
=
60
100
\;P(E_2)=\frac{60}{100}
P
(
E
2
)
=
100
60
P
(
A
/
E
1
)
=
90
100
\;P(A/E_1)=\frac{90}{100}
P
(
A
/
E
1
)
=
100
90
P
(
A
/
E
2
)
=
80
100
\;P(A/E_2)=\frac{80}{100}
P
(
A
/
E
2
)
=
100
80
∵
P
(
E
1
/
A
)
=
P
(
A
/
E
1
)
P
(
E
1
)
P
(
A
/
E
1
)
P
(
E
1
)
+
P
(
A
/
E
2
)
P
(
E
2
)
=
3
7
\because P(E_1/A)=\frac{P(A/E_1)P(E_1)}{P(A/E_1)P(E_1)+P(A/E_2)P(E_2)}=\frac{3}{7}
∵
P
(
E
1
/
A
)
=
P
(
A
/
E
1
)
P
(
E
1
)
+
P
(
A
/
E
2
)
P
(
E
2
)
P
(
A
/
E
1
)
P
(
E
1
)
=
7
3
∴
126
P
=
54
\therefore 126P=54
∴
126
P
=
54
© examsnet.com
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