Three machines turn out all the products in a factory, with the first machine producing 20% of the products, the second machine 15%, and the third machine 65%. The first machine produces defective products 17% of the time, the second machine 8% of the time and the third machine 5% of the time. What is the probability that a non-defective product came from the second machine?

Question

Three machines turn out all the products in a factory, with the first machine producing 20% of the products, the second machine 15%, and the third machine 65%. The first machine produces defective products 17% of the time, the second machine 8% of the time and the third machine 5% of the time. What is the probability that a non-defective product came from the second machine?

Answer 1

To find the probability that a non-defective product came from the second machine, we need to calculate the conditional probability.

Let's denote the events as follows:
A: Product comes from the second machine.
B: Product is non-defective.

We are given the following probabilities:
P(A) = 15% = 0.15 (since the second machine produces 15% of the products)
P(B|A) = 92% = 0.92 (since the second machine produces defective products 8% of the time, the probability of a non-defective product is 100% - 8% = 92%)

We want to find P(A|B), the probability that a product comes from the second machine given that it is non-defective.

Using Bayes' theorem, we can calculate P(A|B) as follows:
P(A|B) = P(A) * P(B|A) / P(B)

Now let's calculate P(B), the probability of a non-defective product.
P(B) = P(non-defective product) = 1 - P(defective product)

For the first machine, the probability of a defective product is 17% = 0.17.
For the second machine, the probability of a defective product is 8% = 0.08.
For the third machine, the probability of a defective product is 5% = 0.05.

To calculate P(B), we need to consider the probabilities for all three machines. Since each machine produces a certain percentage of the products, we have to calculate the weighted average of the probabilities of defective products.

P(B) = (P(B|A1) * P(A1)) + (P(B|A2) * P(A2)) + (P(B|A3) * P(A3))
= (1 - 0.17) * 0.20 + (1 - 0.08) * 0.15 + (1 - 0.05) * 0.65
= 0.83 * 0.20 + 0.92 * 0.15 + 0.95 * 0.65
= 0.166 + 0.138 + 0.617
= 0.921

Now we can substitute the values into the equation for P(A|B):
P(A|B) = P(A) * P(B|A) / P(B)
= 0.15 * 0.92 / 0.921
= 0.138 / 0.921
= 0.15

Therefore, the probability that a non-defective product came from the second machine is 0.15, or 15%.

Answer 2

prob(non-defective, second machine)

= (.92)(.15) = .138