In a manufacturing plant, three machines, A, B, and C, produce 37%, 36%, and 27%, respectively, of the total production. The company's quality-control department has determined that 1.5% of the items produced by Machine A, 2% of the items produced by Machine B, and 2.5% of the items produced by Machine C are defective. If an item is selected at random and found to be defective, what is the probability that it was produced by Machine B? (Round your answer to three decimal places.)

It will be similar to the previous problem.

http://www.jiskha.com/display.cgi?id=1498184495

The steps are to use law of total probability and then Bayes theorem.

I'll start, and let you finish.

Events:
A=part was produced by machine A
B=part was produced by machine B
C=part was produced by machine C
D=part was defective

We're given:
P(A)=0.37
P(B)=0.36
P(C)=0.27
P(D|A)=0.015 (defective given produced by A)
P(D|B)=0.02
P(D|C)=0.025

Also, by definition of conditional probability:
P(X|Y)=P(X∩Y)P(Y)

Then by the law of total probability:
P(D)=P(D∩A)+P(D∩B)+P(D∩C)
=P(D|A)P(A)+P(D|B)P(B)+P(D|C)*P(C)
= ? (all necessary probabilities are known)

Using Bayes Theorem,
P(B|D) (produced by B given defective)
=P(B∩D)P(D)
=P(D∩B)P(D)
=[P(D|B)/P(B)]*P(D) [from P(D|B)=P(D∩B)P(B)]
=? (all necessary probabilities are known or calculated before)

To solve this problem, we can use Bayes' Theorem. Bayes' Theorem helps us update our probability estimates when new information is given.

Let's define the following events:
- A: The item is produced by Machine A.
- B: The item is produced by Machine B.
- C: The item is produced by Machine C.
- D: The item is defective.

We are given the following probabilities:
P(A) = 0.37 (Machine A produces 37% of the total production)
P(B) = 0.36 (Machine B produces 36% of the total production)
P(C) = 0.27 (Machine C produces 27% of the total production)
P(D|A) = 0.015 (1.5% of items produced by Machine A are defective)
P(D|B) = 0.02 (2% of items produced by Machine B are defective)
P(D|C) = 0.025 (2.5% of items produced by Machine C are defective)

We want to find P(B|D) - the probability that the item was produced by Machine B, given that it is defective.

According to Bayes' Theorem, we have:

P(B|D) = (P(D|B) * P(B)) / (P(D|A) * P(A) + P(D|B) * P(B) + P(D|C) * P(C))

Plugging in the values, we get:

P(B|D) = (0.02 * 0.36) / (0.015 * 0.37 + 0.02 * 0.36 + 0.025 * 0.27)

Calculating this expression, we find that P(B|D) ≈ 0.342.

Therefore, the probability that the defective item was produced by Machine B is approximately 0.342 or 34.2%.