Two machines produce light bulbs: Machine A produces 60% of the bulbs while machine B produces the remaining. Machine A produces defective bulbs at a rate of 5%, while machine B produces defective bulbs at a rate of 3%. If a randomly selected bulb is defective, what is the probability that it was produced by machine A?

This problem can be solved using Bayes' Theorem. Let A be the event that a bulb is produced by Machine A, and B be the event that a bulb is produced by Machine B. Let D be the event that a bulb is defective.

We are given:

• P(A) = 0.60 (Machine A produces 60% of bulbs)
• P(B) = 0.40 (Machine B produces 40% of bulbs)
• P(D|A) = 0.05 (5% of bulbs from Machine A are defective)
• P(D|B) = 0.03 (3% of bulbs from Machine B are defective)

We want to find P(A|D), the probability that a defective bulb was produced by Machine A.

Bayes' Theorem states: P(A|D) = [P(D|A) * P(A)] / P(D)

First, we need to calculate the overall probability of a bulb being defective, P(D). This can be found using the law of total probability: P(D) = P(D|A) * P(A) + P(D|B) * P(B) P(D) = (0.05 * 0.60) + (0.03 * 0.40) P(D) = 0.030 + 0.012 P(D) = 0.042

Now, we can plug this into Bayes' Theorem: P(A|D) = (0.05 * 0.60) / 0.042 P(A|D) = 0.030 / 0.042 P(A|D) = 30 / 42 P(A|D) = 5 / 7

Therefore, the probability that a randomly selected defective bulb was produced by Machine A is 5/7, or approximately 0.714.