Math Problem Statement

In a factory, machine A produces 40% of the output and machine B produces 60%. On the average, 9 items in 1000 produced by A are defective and 1 item in 250 produced by B is defective. An item drawn at random from a day’s output is defective. What is the probability that it was produced by A or B.

Solution

This problem can be solved using Bayes' Theorem.

Let:

  • AA represent the event that the item was produced by machine A.
  • BB represent the event that the item was produced by machine B.
  • DD represent the event that the item is defective.

Given Information:

  • P(A)=0.40P(A) = 0.40 (40% of the output is produced by machine A)
  • P(B)=0.60P(B) = 0.60 (60% of the output is produced by machine B)
  • Defective rates:
    • P(DA)=91000=0.009P(D | A) = \frac{9}{1000} = 0.009 (the probability of an item being defective if it is produced by machine A)
    • P(DB)=1250=0.004P(D | B) = \frac{1}{250} = 0.004 (the probability of an item being defective if it is produced by machine B)

We are tasked with finding:

  • P(AD)P(A | D): the probability that the item was produced by machine A given that it is defective.
  • P(BD)P(B | D): the probability that the item was produced by machine B given that it is defective.

Step 1: Find the total probability that the item is defective, P(D)P(D), using the Law of Total Probability:

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

Substitute the given values:

P(D)=(0.009×0.40)+(0.004×0.60)P(D) = (0.009 \times 0.40) + (0.004 \times 0.60) P(D)=0.0036+0.0024=0.006P(D) = 0.0036 + 0.0024 = 0.006

Step 2: Use Bayes' Theorem to find the conditional probabilities.

(a) Probability that the defective item was produced by machine A:

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

Substitute the known values:

P(AD)=0.009×0.400.006=0.00360.006=0.6P(A | D) = \frac{0.009 \times 0.40}{0.006} = \frac{0.0036}{0.006} = 0.6

(b) Probability that the defective item was produced by machine B:

P(BD)=P(DB)P(B)P(D)P(B | D) = \frac{P(D | B) P(B)}{P(D)}

Substitute the known values:

P(BD)=0.004×0.600.006=0.00240.006=0.4P(B | D) = \frac{0.004 \times 0.60}{0.006} = \frac{0.0024}{0.006} = 0.4

Final Answer:

  • The probability that the defective item was produced by machine A is P(AD)=0.6P(A | D) = 0.6.
  • The probability that the defective item was produced by machine B is P(BD)=0.4P(B | D) = 0.4.

Would you like further details on Bayes' Theorem or the Law of Total Probability?


5 Relative Questions:

  1. How does Bayes' Theorem apply in real-world problems?
  2. What if machine C was added with a different defective rate, how would that affect the calculation?
  3. Can you calculate the overall defect rate if the defect rate for machine A changes?
  4. How can the Law of Total Probability be extended to more machines?
  5. What is the importance of conditional probability in decision making?

Tip:

Always ensure to correctly identify the "prior probabilities" (like P(A)P(A) and P(B)P(B)) and the "likelihoods" (like P(DA)P(D | A) and P(DB)P(D | B)) when applying Bayes' Theorem.

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Math Problem Analysis

Mathematical Concepts

Probability
Bayes' Theorem
Law of Total Probability
Conditional Probability

Formulas

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

Theorems

Bayes' Theorem
Law of Total Probability

Suitable Grade Level

Grades 10-12