Math Problem Statement

An item is produced in three factories AA, BB and CC. Factory AA produces 2 times the number of items produced by factory BB, and the factories BB and CC produces the same number of items. It is known that 8%, 7%, 3% of the items produced by factories AA, BB and CC respectively are defective. All items produced in the three factories are stocked, and an item is selected at random. What is the probability that the selected item is defective? (Enter the answer correct to two decimal places)

Solution

To solve this problem, we need to find the probability that a randomly selected item from the combined production of the three factories is defective. We'll use the law of total probability.

Step 1: Define Variables

Let's denote:

  • NAN_A as the number of items produced by factory AA,
  • NBN_B as the number of items produced by factory BB,
  • NCN_C as the number of items produced by factory CC.

We know that: NA=2NBN_A = 2N_B NB=NCN_B = N_C

Step 2: Define Defect Probabilities

The probabilities that an item produced by each factory is defective are:

  • P(DA)=0.08P(D \mid A) = 0.08 (8% of items from AA are defective),
  • P(DB)=0.07P(D \mid B) = 0.07 (7% of items from BB are defective),
  • P(DC)=0.03P(D \mid C) = 0.03 (3% of items from CC are defective).

Step 3: Calculate the Total Production

Let NN be the total production from all factories: N=NA+NB+NCN = N_A + N_B + N_C

Substituting the relationships between the productions: N=2NB+NB+NB=4NBN = 2N_B + N_B + N_B = 4N_B

Step 4: Calculate the Proportions of Production

The proportions of the total production from each factory are:

  • From AA: NAN=2NB4NB=12\frac{N_A}{N} = \frac{2N_B}{4N_B} = \frac{1}{2},
  • From BB: NBN=NB4NB=14\frac{N_B}{N} = \frac{N_B}{4N_B} = \frac{1}{4},
  • From CC: NCN=NB4NB=14\frac{N_C}{N} = \frac{N_B}{4N_B} = \frac{1}{4}.

Step 5: Use the Law of Total Probability

The total probability that a randomly selected item is defective is given by: P(D)=P(A)P(DA)+P(B)P(DB)+P(C)P(DC)P(D) = P(A)P(D \mid A) + P(B)P(D \mid B) + P(C)P(D \mid C)

Substitute the known values: P(D)=(12×0.08)+(14×0.07)+(14×0.03)P(D) = \left(\frac{1}{2} \times 0.08\right) + \left(\frac{1}{4} \times 0.07\right) + \left(\frac{1}{4} \times 0.03\right)

Calculate each term: P(D)=0.04+0.0175+0.0075P(D) = 0.04 + 0.0175 + 0.0075

Sum these probabilities: P(D)=0.04+0.0175+0.0075=0.065P(D) = 0.04 + 0.0175 + 0.0075 = 0.065

Thus, the probability that a randomly selected item is defective is: 0.07\boxed{0.07}

Would you like more details or have any questions?

Related Questions:

  1. What is the probability that a non-defective item is selected?
  2. How does the probability change if factory AA's defect rate drops to 5%?
  3. If an item is found to be defective, what is the probability that it came from factory BB?
  4. How many defective items are produced by each factory if NB=1000N_B = 1000?
  5. If factory CC doubles its production, how does it affect the overall defect probability?
  6. How would the probability change if the total production of all factories increased by 50%?
  7. What is the expected number of defective items in a batch of 10000 items produced?
  8. How do changes in individual defect rates impact the overall defect probability?

Tip:

When solving probability problems involving multiple sources or conditions, clearly break down the contributions from each source and use the law of total probability for accurate results.

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Math Problem Analysis

Mathematical Concepts

Probability
Law of Total Probability
Defect Probability

Formulas

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Theorems

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Suitable Grade Level

High School