Respuesta :
Answer:
38.71% probability it was manufactured by machine A.
29.03% probability it was manufactured by machine B.
32.26% probability it was manufactured by machine C.
Step-by-step explanation:
We have these following probabilities:
A 20% probability that the chip was fabricated by machine A.
A 30% probability that the chip was fabricated by machine B.
A 50% probability that the chip was fabricated by machine C.
A 6% probability that a chip fabricated by machine A was defective.
A 3% probability that a chip fabricated by machine B was defective.
A 2% probability that a chip fabricated by machine C was defective.
The question can be formulated as:
What is the probability of B happening, knowing that A has happened.
It can be calculated by the following formula
[tex]P = \frac{P(B).P(A/B)}{P(A)}[/tex]
Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.
What are the probabilities that it was manufactured by machines A, B, and C?
Machine A
What is the probability that the widget was manufactures by machine A, given that it is defective?
P(B) is the probability that it was manufactures by machine A. So [tex]P(B) = 0.20[/tex]
P(A/B) is the probability that a widget is defective, given that it is manufactured by machine A. So [tex]P(A/B) = 0.06[/tex]
P(A) is the probability that a widget is defective. This is the sum of 6% of 20%, 3% of 30% and 2% of 50%. So
[tex]P(A) = 0.06*0.20 + 0.03*0.30 + 0.02*0.5 = 0.031[/tex]
So
[tex]P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.2*0.06}{0.031} = 0.3871[/tex]
38.71% probability it was manufactured by machine A.
Machine B
What is the probability that the widget was manufactures by machine B, given that it is defective?
P(B) is the probability that it was manufactures by machine B. So [tex]P(B) = 0.30[/tex]
P(A/B) is the probability that a widget is defective, given that it is manufactured by machine B. So [tex]P(A/B) = 0.03[/tex]
[tex]P(A) = 0.06*0.20 + 0.03*0.30 + 0.02*0.5 = 0.031[/tex]
So
[tex]P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.3*0.03}{0.031} = 0.2903[/tex]
29.03% probability it was manufactured by machine B.
Machine C
What is the probability that the widget was manufactures by machine C, given that it is defective?
P(B) is the probability that it was manufactures by machine C. So [tex]P(B) = 0.50[/tex]
P(A/B) is the probability that a widget is defective, given that it is manufactured by machine C. So [tex]P(A/B) = 0.02[/tex]
[tex]P(A) = 0.06*0.20 + 0.03*0.30 + 0.02*0.5 = 0.031[/tex]
So
[tex]P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.5*0.02}{0.031} = 0.3226[/tex]
32.26% probability it was manufactured by machine C.
