Two machines A and B produce respectively 60% and 40% of the total number of items of a factory. The percentages of defective output for A and B are 3% and 2% respectively. Suppose an item is selected at random, and is found to be defective. The probability that the item was produced by B, is equal to
.4*.02=.008 Pr reject and B
.6*.03=.018 Pr reject and A
What is .008/(.008+.018)=8/26 ?
To find the probability that the defective item was produced by machine B, we can use Bayes' theorem.
Let's define the following events:
A: Item produced by machine A
B: Item produced by machine B
D: Defective item
We want to find P(B|D) (the probability that the item was produced by B given that it is defective).
Using Bayes' theorem, we have:
P(B|D) = (P(D|B) * P(B)) / P(D)
P(D|B) is the probability of selecting a defective item given that it was produced by B. This is given as 2%.
P(B) is the probability of a randomly selected item being produced by B. This is given as 40%.
P(D) is the probability of selecting a defective item. We can find this by using the law of total probability:
P(D) = P(D|A) * P(A) + P(D|B) * P(B)
P(D|A) is the probability of selecting a defective item given that it was produced by A. This is given as 3%.
P(A) is the probability of a randomly selected item being produced by A. This is given as 60%.
Substituting the given values, we can calculate P(D):
P(D) = 0.03 * 0.6 + 0.02 * 0.4
Now we can substitute all the values into the Bayes' theorem equation:
P(B|D) = (0.02 * 0.4) / (0.03 * 0.6 + 0.02 * 0.4)
Simplifying the equation:
P(B|D) = 0.008 / (0.018 + 0.008)
P(B|D) = 0.008 / 0.026
P(B|D) ≈ 0.3077
Therefore, the probability that the defective item was produced by machine B is approximately 0.3077 or 30.77%.
To find the probability that the item was produced by machine B given that it is defective, we can use Bayes' theorem.
Let's define the events:
A: The item is produced by machine A.
B: The item is produced by machine B.
D: The item is defective.
We need to find P(B | D), the probability that the item was produced by B given that it is defective.
According to Bayes' theorem:
P(B | D) = (P(D | B) * P(B)) / P(D)
We are given:
P(A) = 0.6 (machine A produces 60% of the total items)
P(B) = 0.4 (machine B produces 40% of the total items)
P(D | A) = 0.03 (defective output for machine A is 3%)
P(D | B) = 0.02 (defective output for machine B is 2%)
Let's calculate P(D) using the Law of Total Probability:
P(D) = P(D | A) * P(A) + P(D | B) * P(B)
= 0.03 * 0.6 + 0.02 * 0.4
= 0.018 + 0.008
= 0.026
Now, substituting these values in Bayes' theorem:
P(B | D) = (P(D | B) * P(B)) / P(D)
= (0.02 * 0.4) / 0.026
= 0.008 / 0.026
= 0.3077
Therefore, the probability that the defective item was produced by machine B is approximately 0.3077 or 30.77%.