Answer:
0.545 = 54.5% probability that it came from machine A
Step-by-step explanation:
Bayes Theorem:
Two events, A and B.
[tex]P(B|A) = \frac{P(B)*P(A|B)}{P(A)}[/tex]
In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.
In this question:
Event A: Defective.
Event B: Coming from machine A.
Machine A is responsible for 30%
This means that [tex]P(B) = 0.3[/tex]
10% of the output from machine A is defective
This means that [tex]P(B|A) = 0.1[/tex]
Probability of being defective:
Machine A is responsible for 30%. Of those, 10% are defective.
Machine B is responsible for 20%. Of those, 5% are defective.
Machine C is responsible for 100 - (30+20) = 50%. Of those, 3% are defective. Then
[tex]P(A) = 0.3*0.1 + 0.2*0.05 + 0.5*0.03 = 0.055[/tex]
Finally:
[tex]P(B|A) = \frac{P(B)*P(A|B)}{P(A)} = \frac{0.3*0.1}{0.055} = 0.545[/tex]
0.545 = 54.5% probability that it came from machine A