Math Problem Statement

  1. A factory produces items from two machines: Machine A and Machine B. Machine

A produces 60% of the total items, while Machine B produces 40%. The probability

that an item produced by Machine A is defective is P(D|A) = 0.03. The probability

that an item produced by Machine B is defective is P(D|B) = 0.05.

(a) What is the probability that a randomly selected product be defective, P(D)?

(b) If a randomly selected item from the production line is defective, calculate the

probability that it was produced by Machine A, P(A|D).

Solution

Let's break down the problem step by step.

Given Information:

  • Machine A produces 60% of the items → P(A)=0.60P(A) = 0.60
  • Machine B produces 40% of the items → P(B)=0.40P(B) = 0.40
  • The probability that an item produced by Machine A is defective → P(DA)=0.03P(D|A) = 0.03
  • The probability that an item produced by Machine B is defective → P(DB)=0.05P(D|B) = 0.05

We are asked to calculate the following:

Part (a): Probability that a randomly selected product is defective, P(D)P(D)

We will use Total Probability Theorem for this. The probability that a randomly selected item is defective can be found by considering both machines:

P(D)=P(DA)P(A)+P(DB)P(B)P(D) = P(D|A)P(A) + P(D|B)P(B)

Substituting the given values:

P(D)=(0.03×0.60)+(0.05×0.40)P(D) = (0.03 \times 0.60) + (0.05 \times 0.40) P(D)=0.018+0.02=0.038P(D) = 0.018 + 0.02 = 0.038

So, the probability that a randomly selected product is defective is 0.038\boxed{0.038}, or 3.8%.


Part (b): Probability that an item was produced by Machine A, given that it is defective, P(AD)P(A|D)

We will use Bayes' Theorem to find this conditional probability. Bayes' Theorem states:

P(AD)=P(DA)P(A)P(D)P(A|D) = \frac{P(D|A)P(A)}{P(D)}

We already know the values:

  • P(DA)=0.03P(D|A) = 0.03
  • P(A)=0.60P(A) = 0.60
  • P(D)=0.038P(D) = 0.038

Substituting these values into Bayes' Theorem:

P(AD)=(0.03×0.60)0.038P(A|D) = \frac{(0.03 \times 0.60)}{0.038} P(AD)=0.0180.0380.4737P(A|D) = \frac{0.018}{0.038} \approx 0.4737

So, the probability that a defective item was produced by Machine A is approximately 0.4737\boxed{0.4737}, or about 47.37%.


Summary:

  • (a) The probability that a randomly selected product is defective is 0.038.
  • (b) The probability that a defective product came from Machine A is approximately 0.4737.

Would you like more details or clarification on any part? Here are some related questions you can explore:

  1. What happens if the probability of defects changes for either machine?
  2. How does increasing Machine A's defect rate affect P(AD)P(A|D)?
  3. What if the distribution between the machines changes (e.g., 50% from Machine A and 50% from Machine B)?
  4. How do you calculate the expected number of defective items if 1000 products are produced?
  5. How would this problem be approached if more than two machines were involved?

Tip: When using Bayes' Theorem, always check if you have the full set of conditional probabilities, priors, and total probability before calculating.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Probability
Conditional Probability
Bayes' Theorem
Total Probability Theorem

Formulas

P(D) = P(D|A)P(A) + P(D|B)P(B)
P(A|D) = (P(D|A)P(A)) / P(D)

Theorems

Total Probability Theorem
Bayes' Theorem

Suitable Grade Level

Grades 10-12