a computer mouse testing lab tests two kind of mouse m1 and m2. it is given that 12% of the total mouse were found to be defective and 80% of the non defective mouse were type m1 how many m1 mouse in total were tested as a non defective in the lab if around 5600 non defective mouse were of type m2
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Answer:
Let x=P (computer turns out to be defective given that it is produced in plant T
2
),
100
7
=
5
1
×(10x)+
5
4
×x⇒7=200x+80x⇒x=
280
7
Event A = computer is produced in T
2
Event B = computer is not defective
P (produced in T
2
not defective) =
P(B)
P(A∩B)
=
5
1
(1−10x)+
5
4
(1−x)
4/5(1−x)
=
5
1
(
280
280−70
)+
5
4
(
280
273
)
5
4
(
280
273
)
=
210+4×273
4×273
=
105+2×273
2×273
=
651
546
=
93
78
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