Question

A certain plant runs three shifts per day. Of all the items produced by the plant, 50% of them are produced on the first shift, 30% on the second shift, and 20% on the third shift. Of all the items produced on the first shift, 1% are defective, while 2% of the items produced on the second shift and 3% of the items produced on the third shift are defective. An item is sampled at random from the day's production, and it turns out to be defective. What is the probability that it was manufactured during the first shift

Answer 1

Answer:

0.2941 = 29.41% probability that it was manufactured during the first shift.

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: Defective

Event B: Manufactured during the first shift.

Probability of a defective item:

1% of 50%(first shift)

2% of 30%(second shift)

3% of 20%(third shift).

So

$P(A) = 0.01*0.5 + 0.02*0.3 + 0.03*0.2 = 0.017$

Probability of a defective item being produced on the first shift:

1% of 50%. So

$P(A \cap B) = 0.01*0.5 = 0.005$

What is the probability that it was manufactured during the first shift?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.005}{0.017} = 0.2941$

0.2941 = 29.41% probability that it was manufactured during the first shift.