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