Math Problem Statement

please solve 100% sure

In a certain assembly plant, three machines, B1, B2, and B3, make 30%, 45%, and 25% of the products, respectively. It is known from past experience that 2%, 3%, and 2% of the products made by each machine, respectively, are defective. Now, suppose that a finished product is randomly selected. 1. If a product was chosen randomly and found to be defective, what is the probability that it was made by B1? 2. What is the probability that it is produced by B2 and defective? 3. What is the probability that it is defective?

Solution

Let's go through this problem step-by-step. The question is based on conditional probabilities.

Given Information

  1. Production percentages:

    • Machine B1 makes 30% of the products.
    • Machine B2 makes 45% of the products.
    • Machine B3 makes 25% of the products.
  2. Defect rates:

    • B1 has a defect rate of 2%.
    • B2 has a defect rate of 3%.
    • B3 has a defect rate of 2%.

Let's solve each question.


Question 1: Probability that a defective product was made by B1

We use Bayes' theorem to calculate this conditional probability.

Let:

  • P(B1)=0.30P(B1) = 0.30
  • P(B2)=0.45P(B2) = 0.45
  • P(B3)=0.25P(B3) = 0.25
  • P(DB1)=0.02P(D | B1) = 0.02
  • P(DB2)=0.03P(D | B2) = 0.03
  • P(DB3)=0.02P(D | B3) = 0.02

Total probability of defect, P(D)P(D): P(D)=P(DB1)P(B1)+P(DB2)P(B2)+P(DB3)P(B3)P(D) = P(D | B1) \cdot P(B1) + P(D | B2) \cdot P(B2) + P(D | B3) \cdot P(B3) =(0.02×0.30)+(0.03×0.45)+(0.02×0.25)= (0.02 \times 0.30) + (0.03 \times 0.45) + (0.02 \times 0.25) =0.006+0.0135+0.005=0.0245= 0.006 + 0.0135 + 0.005 = 0.0245

Probability that a defective product was made by B1: P(B1D)=P(DB1)P(B1)P(D)=0.02×0.300.0245=0.0060.02450.245P(B1 | D) = \frac{P(D | B1) \cdot P(B1)}{P(D)} = \frac{0.02 \times 0.30}{0.0245} = \frac{0.006}{0.0245} \approx 0.245

So, the answer for the first question is approximately 0.245.


Question 2: Probability that it is produced by B2 and defective

This is simply the joint probability P(B2 and D)P(B2 \text{ and } D): P(B2 and D)=P(DB2)P(B2)=0.03×0.45=0.0135P(B2 \text{ and } D) = P(D | B2) \cdot P(B2) = 0.03 \times 0.45 = 0.0135

So, the answer for the second question is 0.0135.


Question 3: Probability that it is defective

This is the total probability of a defective product, which we calculated above: P(D)=0.0245P(D) = 0.0245

So, the answer for the third question is 0.0245.


Summary of Answers

  1. Probability that a defective product was made by B1: 0.245
  2. Probability that it is produced by B2 and defective: 0.0135
  3. Probability that it is defective: 0.0245

Would you like further explanation on any of these steps, or do you have any additional questions?

Related Questions

  1. How do you calculate the probability of a product being defective from a specific machine?
  2. What is Bayes' theorem, and how is it applied here?
  3. Can you explain how joint probabilities differ from conditional probabilities?
  4. How would the probabilities change if defect rates of the machines were different?
  5. What other scenarios can conditional probability be applied to in manufacturing?

Tip

Remember, Bayes' theorem is powerful for questions where you are given conditional probabilities and asked to "reverse" them to find the probability of an event given an outcome.

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Math Problem Analysis

Mathematical Concepts

Probability
Conditional Probability
Bayes' Theorem

Formulas

P(D) = P(D | B1) * P(B1) + P(D | B2) * P(B2) + P(D | B3) * P(B3)
P(B1 | D) = P(D | B1) * P(B1) / P(D)
P(B2 and D) = P(D | B2) * P(B2)

Theorems

Bayes' Theorem

Suitable Grade Level

College Level (Introductory Probability)