Question

Three machines turn out all the products in a factory, with the first machine producing 30% of the products, the second machine 20%, and the third machine 50%. The first machine produces defective products 13% of the time, the second machine 9% of the time and the third machine 9% of the time. What is the probability that a non-defective product came from the second machine

Answer 1

Answer:

0.2027 = 20.27% probability that a non-defective product came from the second machine

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Non-defective product.

Event B: From the second machine.

Probability of a non-defective product:

100-13 = 87% of 30%(first machine)

100-9 = 91% of 20%(second machine)

100-9 = 91% of 50%(third machine).

So

$P(A) = 0.87*0.3 + 0.91*0.2 + 0.91*0.5 = 0.898$

Non-defective and from the second machine:

91% of 20%. So

$P(A \cap B) = 0.91*0.2 = 0.182$

What is the probability that a non-defective product came from the second machine

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.182}{0.898} = 0.2027$

0.2027 = 20.27% probability that a non-defective product came from the second machine