Question

three machines - a, b, and c - are used to produce a large quantity of identical parts at a factory. machine a produces 60% of the parts, while machines b and c produce 30% and 10% of the parts, respectively. historical records indicate that 10% of the parts produced by machine a are defective, compared with 30% for machine b and 40% for machine c. suppose we randomly select a part produced by one of these three machines at random.
(a) draw a tree diagram to model this chance process. complete the tree diagram by dragging the probabilities and placing them in the boxes.
(b) whats the probability that the part is defective?
(round to 2 decimal places. leave your answer in decimal form.)
(c) if the part is inspected and found to be defective, whats the probability it was produced by machine c?
(round to 3 decimal places. leave your answer in decimal form.)

Explanation:

Step1: Calculate probability of defective part

Use the law of total probability. Let \(P(A) = 0.6\), \(P(B)=0.3\), \(P(C) = 0.1\). Let \(P(D|A)=0.1\), \(P(D|B) = 0.3\), \(P(D|C)=0.4\).

The formula for total probability \(P(D)=P(A)P(D|A)+P(B)P(D|B)+P(C)P(D|C)\)

Substitute the values:

\(P(D)=(0.6\times0.1)+(0.3\times0.3)+(0.1\times0.4)\)

\(P(D)=0.06 + 0.09+0.04\)

\(P(D)=0.19\)

Step2: Calculate probability part from machine C given defective

Use Bayes' theorem \(P(C|D)=\frac{P(C)P(D|C)}{P(D)}\)

We know \(P(C) = 0.1\), \(P(D|C)=0.4\), \(P(D) = 0.19\)

Substitute the values:

\(P(C|D)=\frac{0.1\times0.4}{0.19}=\frac{0.04}{0.19}\approx0.211\)

Answer:

(b) \(0.19\)

(c) \(0.211\)