Question

Three factories produce light bulbs to supply the market. Factory A produces 40%, 20% of the tools are produced in factory B and 40% in factory C. 5% of the bulbs produced in factory A, 3% of the bulbs produced in factory B and 4% of the bulbs produced in factory C are defective.A bulb is selected at random in the market and found to be defective. What is the probability that this bulb was produced by factory A?

Answer 1

Answer:

0.4762 = 47.62% probability that this bulb was produced by factory A

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: Bulb is defective.

Event B: Produced by factory A.

Probability of a bulb being defective.

5% of 40%(factory A)

3% of 20%(factory B)

4% of 40%(factory C). So

$P(A) = 0.05*0.4 + 0.03*0.2 + 0.04*0.4 = 0.042$

Defective and from factory A:

5% of 40%. So

$P(A \cap B) = 0.05*0.4 = 0.02$

What is the probability that this bulb was produced by factory A?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.02}{0.042} = 0.4762$

0.4762 = 47.62% probability that this bulb was produced by factory A