Quality Motors has three plants. Plant 1 produces 35% of the car output, plant 2 produces 20% and plant 3 produces the remaining 45%. One percent of the output of plant 1 is defective, 1.8% of the output of plant 2 is defective and 2% of the output of plant 3 is defective. The annual total production of Quality Motors is 1,000,000 cars. A car is chosen at random from the annual output and is found defection. Use Bayes' Rule to find the probability that it came from plant 2.

We are given that Quality Motors has three plants. Plant 1 produces 35% of the car output, plant 2 produces 20% and plant 3 produces the remaining 45%.

One percent of the output of plant 1 is defective, 1.8% of the output of plant 2 is defective and 2% of the output of plant 3 is defective.

Let the Probability that car output is produced by Plant 1 = P(A) = 0.35

Probability that car output is produced by Plant 2 = P(A) = 0.20

Probability that car output is produced by Plant 3 = P(A) = 0.45

Also, let D = event that output is defective

So, Probability that output defective given that it was produced by plant 1 = P(D/A) = 0.01

Probability that output defective given that it was produced by plant 2 = P(D/B) = 0.018

Probability that output defective given that it was produced by plant 3 = P(D/C) = 0.02

Now, a car is chosen at random from the annual output and is found defection so, the probability that it came from plant 2 is = P(B/D)

We will use the concept of Bayes' Theorem here to calculate the above probability.

SO, P(B/D) = [tex]\frac{P(B) \times P(D/B)}{P(A) \times P(D/A) + P(B) \times P(D/B)+P(C) \times P(D/C)}[/tex]

= [tex]\frac{0.20 \times 0.018}{0.35 \times 0.01 + 0.20 \times 0.018+0.45 \times 0.02}[/tex]

= [tex]\frac{0.0036}{0.0161}[/tex]

= 0.224

Hence, the required probability is 0.224.