Math, asked by sibimathew120, 3 days ago

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​

Answers

Answered by palaksoni5274
0

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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