Respuesta :
Answer:
0.2027 = 20.27% probability that a non-defective product came from the second machine
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Non-defective product.
Event B: From the second machine.
Probability of a non-defective product:
100-13 = 87% of 30%(first machine)
100-9 = 91% of 20%(second machine)
100-9 = 91% of 50%(third machine).
So
[tex]P(A) = 0.87*0.3 + 0.91*0.2 + 0.91*0.5 = 0.898[/tex]
Non-defective and from the second machine:
91% of 20%. So
[tex]P(A \cap B) = 0.91*0.2 = 0.182[/tex]
What is the probability that a non-defective product came from the second machine
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.182}{0.898} = 0.2027[/tex]
0.2027 = 20.27% probability that a non-defective product came from the second machine
