Question 1181553: A company buys 20% of its electronic chips from supplier New and the rests from other suppliers. Assume that the company purchases a large number of chips from each supplier.
It is known that 1/10 (one tenth) of supplier New’s chips are defective while the other suppliers have only 1/50 (one fiftieth) chips defective. The company selects 20 chips at random from a batch and tests them. One chip is found to be defective.
What is the probability that this batch of chips came from supplier New?
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem using Bayes' Theorem:
**1. Define Events:**
* N: The batch of chips came from supplier New.
* O: The batch of chips came from other suppliers.
* D: A randomly selected chip is defective.
**2. Given Probabilities:**
* P(N) = 0.20 (20% of chips are from New)
* P(O) = 1 - P(N) = 0.80 (80% of chips are from other suppliers)
* P(D|N) = 1/10 = 0.10 (Probability of a defective chip given it's from New)
* P(D|O) = 1/50 = 0.02 (Probability of a defective chip given it's from other suppliers)
**3. What We Want:**
We want to find P(N|D), the probability that the batch came from supplier New *given* that one chip is found to be defective.
**4. Bayes' Theorem:**
Bayes' Theorem states:
P(N|D) = [P(D|N) * P(N)] / P(D)
We need to find P(D), the overall probability of a defective chip. We can use the law of total probability:
P(D) = P(D|N) * P(N) + P(D|O) * P(O)
P(D) = (0.10 * 0.20) + (0.02 * 0.80)
P(D) = 0.02 + 0.016
P(D) = 0.036
**5. Apply Bayes' Theorem:**
P(N|D) = (0.10 * 0.20) / 0.036
P(N|D) = 0.02 / 0.036
P(N|D) ≈ 0.556
**Answer:**
The probability that the batch of chips came from supplier New, given that one chip is defective, is approximately 0.556 or 55.6%.
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