Math Problem Statement

A MACHINE PRODUCES ELECTRONIC COMPONENTS THAT ARE PACKAGED INTO PACKS OF 10. THE PROBABILITY THAT A COMPONENT IS DEFECTIVE IS 0.003, INDEPENDEDNTLY OF ALL OTHER COMPONENTS. FIND THE PROBABILITY THAT AT LEAST ONE OF THE COMPONENTS IN THE PACK IS DEFECTIVE.

tHE MANUFACTURER USES THE FOLLOWING QUALITY CONTROL PROCEDURE TO CHECK LARGE BATCHES OF BOXES:

A PACK OF 10 IS SELECTED AT RANDOM FROM THE BATCH. IF THE PACK CONTAINS ATLEAST ONE DEFECTIVE COMPONENT THEN ANOTHER PACK IS SELECTED FROMT HE SAME BATCH. IF THAT PACK CONTAINS AT LEAST ONE DEFECTIVE COMPONENT, THEN THE WHOLE BATCH IS REJECTED; OTHERWISE THE WHOLE BATCH IS ACCEPTED.

Find the probability that a batch is rejected and suggest a reason why the assumption of independence might not hold.

Solution

Part 1: Probability that at least one component in a pack is defective

We are given:

  • The probability that a single component is defective is p=0.003p = 0.003.
  • Each pack contains 10 components.

Let XX be the number of defective components in a pack. This is a binomial distribution because each component has two possible outcomes (defective or not), and the probability of each outcome is independent: XBinomial(n=10,p=0.003)X \sim \text{Binomial}(n = 10, p = 0.003)

The probability that none of the components in the pack are defective is the probability of 0 defective components: P(X=0)=(1p)10=(0.997)10P(X = 0) = (1 - p)^{10} = (0.997)^{10}

Now, the probability that at least one component is defective is the complement of this: P(at least 1 defective)=1P(X=0)=1(0.997)10P(\text{at least 1 defective}) = 1 - P(X = 0) = 1 - (0.997)^{10}

Let's calculate this probability.

P(at least 1 defective)=1(0.997)10P(\text{at least 1 defective}) = 1 - (0.997)^{10}

Part 2: Probability that a batch is rejected

According to the quality control procedure:

  1. A pack is selected at random from the batch.
  2. If the pack contains at least one defective component, another pack is selected.
  3. If the second pack also contains at least one defective component, the batch is rejected.

Let P(defective in 1 pack)=1(0.997)10P(\text{defective in 1 pack}) = 1 - (0.997)^{10}.

Now, the probability that a batch is rejected is the probability that both packs have at least one defective component. Since the process of selecting packs is independent, the probability of both packs having at least one defective component is: P(batch rejected)=P(defective in 1 pack)×P(defective in 1 pack)P(\text{batch rejected}) = P(\text{defective in 1 pack}) \times P(\text{defective in 1 pack})

Thus: P(batch rejected)=(1(0.997)10)2P(\text{batch rejected}) = \left( 1 - (0.997)^{10} \right)^2

Part 3: Why independence might not hold

The assumption of independence may not hold in practice due to several factors:

  1. Production defects: If there is a manufacturing issue, multiple components produced in the same time period or batch might be defective, leading to correlation among defects within a pack.
  2. Material quality: If poor materials are used for a portion of production, this could increase the chance of multiple defective components being produced together, violating the independence assumption.
  3. Environmental factors: Factors such as machine calibration, temperature, or operator errors might affect multiple components simultaneously.

Would you like me to calculate the exact values for you or do you need further clarifications?


Follow-up Questions:

  1. What is the probability that a batch is accepted?
  2. How does the probability of rejection change if the defect rate increases?
  3. What is the expected number of defective components in a pack?
  4. How would the analysis change if components within a pack were not independent?
  5. How could a manufacturer reduce the probability of rejecting an entire batch?

Tip:

Independence assumptions simplify calculations but may not reflect real-world dependencies, so consider alternative statistical models when dealing with correlated events.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Probability
Binomial Distribution

Formulas

P(X = 0) = (1 - p)^{10} for 0 defective components
P(at least 1 defective) = 1 - (0.997)^{10}
P(batch rejected) = (1 - (0.997)^{10})^2

Theorems

Binomial Distribution Theorem
Complement Rule

Suitable Grade Level

Grades 10-12