CBSE BOARD XII, asked by yashRonaldo2292, 18 days ago

The factory produces 30% of the defective printers, if a sample of 20 printers are selected at random then compute the random variable

Answers

Answered by singhrajputriya180
0

Answer:

Let D,E

1

,E

2

and E

3

denote the events that the item is defective, machine A is chosen, machine B is chosen and machine C is chosen, respectively.

⇒ P(E

1

)=30%=

100

30

=0.3

⇒ P(E

2

)=25%=

100

25

=0.25

⇒ P(E

3

)=100%−(30+25)%=45%=

100

45

=0.45

Now,

⇒ P(

E

1

D

)=1%=

100

1

=0.01

⇒ P(

E

2

D

)=1.2%=

100

1.2

=0.012

⇒ P(

E

3

D

)=2%=

100

2

=0.02

Probability of being produced by B,

⇒ P(

D

E

2

)=

P(E

2

)×P(

E

2

D

)+P(E

1

)×P(

E

1

D

)+P(E

3

)×p(

E

2

D

)

P(E

2

)×P(

E

2

D

)

=

(0.25×0.012)+(0.3×0.01)+(0.45×0.02)

0.25×0.012

=

0.003+0.003+0.009

0.04

(3+3+9)×10

−3

3×10

−3

=

15

3

=0.2

Answered by vvvvvedAnt
0

Explanation:

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