Respuesta :
Answer:
The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.
Step-by-step explanation:
Using Bayes' Theorem
P(A|B) = [tex]\frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|a)P(a)}[/tex]
where
P(B|A) is probability of event B given event A
P(B|a) is probability of event B not given event A
and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.
For this problem,
Let P(B1) = Probability of machine B1 = 0.3
P(B2) = Probability of machine B2 = 0.2
P(B3) = Probability of machine B3 = 0.5
Let P(D) = Probability of a defective product
P(N) = Probability of a Non-defective product
P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003
P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006
P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010
Likewise,
P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997
P(N|B2) be probability of a non-defective product produced by machine 2 = 1 - P(D|B2) = 1 - 0.006 = 0.994
P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990
For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)
P(B1|N) =[tex]\frac{P(N|B1)P(B1)}{P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)}[/tex] = [tex]\frac{(0.297)(0.3)}{(0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)}[/tex] = 0.1138
Hence the probability that a non-defective product is produced by machine B1 is 11.38%.
