Respuesta :
Answer:
40.68% probability that the lamp was manufactured in factory C
Step-by-step explanation:
Conditional probability formula:
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.
Noting that A and B here are general notations, not related to the factories in the question.
We have these following probabilities:
35% probability that a lamp is selected from factory A.
If a lamp is selected from factory A, 1.5% probability that it is defective.
35% probability that a lamp is selected from factory B.
If a lamp is selected from factory A, 1% probability that it is defective.
30% probability that a lamp is selected from factory C.
If a lamp is selected from factory A, 2% probability that it is defective.
If a randomly selected lamp is defective, what is the probability that the lamp was manufactured in factory C?
Event A: Defective lamp
Event B: From factory C.
[tex]P(A) = 0.35*0.015 + 0.35*0.01 + 0.3*0.02 = 0.01475[/tex]
Intersection:
30% probability that a lamp is selected from factory C.
If a lamp is selected from factory A, 2% probability that it is defective.
Then
[tex]P(A \cap B) = 0.3*0.02 = 0.006[/tex]
Finally:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.006}{0.01475} = 0.4068[/tex]
40.68% probability that the lamp was manufactured in factory C
