(ii) A factory has two machines. Past records show that the first machine produces 40% of output and the second machine produces 60% of output. Further 4% and 2% of the products produced by the first machine and the second machine were defectives. If a defective item is drawn at random, what is the probability that the defective item was produced by the first machine or second machine?

say we make 1000 total

400 from #1, 16 defective
600 from #2, 12 defective

Tat is a total of 28 defective
16/28 = .57 from #1
12/28 = .43 from #2

To find the probability that the defective item was produced by the first machine or the second machine, we can use Bayes' theorem.

Let's define the following events:
A: Defective item was produced by the first machine.
B: Defective item was produced by the second machine.

We are given the following information:
P(A) = 0.4 (probability of the first machine producing an item)
P(B) = 0.6 (probability of the second machine producing an item)
P(D|A) = 0.04 (probability of an item being defective given that it was produced by the first machine)
P(D|B) = 0.02 (probability of an item being defective given that it was produced by the second machine)

We want to find P(A|D) or P(B|D), which is the probability of an item being produced by the first machine or the second machine given that it is defective.

Bayes' theorem states:
P(A|D) = (P(D|A) * P(A)) / P(D)
P(B|D) = (P(D|B) * P(B)) / P(D)

To find P(D), we can use the law of total probability, which states:
P(D) = P(D|A) * P(A) + P(D|B) * P(B)

Now let's calculate the probabilities:

P(D) = (0.04 * 0.4) + (0.02 * 0.6)
P(D) = 0.016 + 0.012
P(D) = 0.028

Now we can use Bayes' theorem:

P(A|D) = (0.04 * 0.4) / 0.028
P(A|D) = 0.016 / 0.028
P(A|D) ≈ 0.571

Similarly,

P(B|D) = (0.02 * 0.6) / 0.028
P(B|D) = 0.012 / 0.028
P(B|D) ≈ 0.429

Therefore, the probability that the defective item was produced by the first machine is approximately 0.571, and the probability that it was produced by the second machine is approximately 0.429.