Math Problem Statement
A factory received a shipment of 16 generators, and the vendor who sold the items knows there are 7 generators in the shipment that are defective. Before the receiving foreman accepts the delivery, he samples the shipment, and if too many of the generators in the sample are defective, he will refuse the shipment.
For each of the following, give your responses as reduced fractions.
If a sample of 7 generators is selected, find the probability that all in the sample are defective.
If a sample of 7 generators is selected, find the probability that none in the sample are defective.
Solution
The problem deals with sampling without replacement from a shipment of 16 generators, of which 7 are defective. We need to find two probabilities: (1) the probability that all 7 generators sampled are defective, and (2) the probability that none of the 7 generators sampled are defective.
Part 1: Probability that all 7 generators in the sample are defective
To calculate this, we need to determine how many ways we can choose 7 defective generators from the 7 defective generators in the shipment, divided by how many ways we can choose any 7 generators from the 16 generators in the shipment.
- Number of ways to choose 7 defective generators from the 7 defective generators:
- Total number of ways to choose 7 generators from the 16 generators:
Thus, the probability that all 7 generators in the sample are defective is: Now, we compute : So, the probability is:
Part 2: Probability that none of the generators in the sample are defective
For this case, we need to calculate the number of ways to choose 7 non-defective generators from the 9 non-defective ones, divided by the total number of ways to choose 7 generators from the 16 generators.
- Number of ways to choose 7 non-defective generators from the 9 non-defective generators:
- Total number of ways to choose 7 generators from the 16 generators is the same as in Part 1:
Thus, the probability that none of the generators in the sample are defective is: Simplifying this fraction:
Final Results:
- The probability that all 7 generators in the sample are defective is:
- The probability that none of the generators in the sample are defective is:
Would you like further details or have any questions?
Here are 5 related questions to explore:
- What is the probability that exactly 1 generator in the sample is defective?
- What is the probability that at least one generator in the sample is defective?
- How does the probability change if only 5 generators are sampled?
- What is the expected number of defective generators in a sample of 7?
- How would the calculation change if the sampling was with replacement?
Tip: When calculating probabilities from large combinations, simplifying fractions helps avoid numerical errors and improves clarity.
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Math Problem Analysis
Mathematical Concepts
Combinatorics
Probability
Formulas
P(all defective) = (C(7, 7) / C(16, 7))
P(none defective) = (C(9, 7) / C(16, 7))
Theorems
Combinatorial Probability
Suitable Grade Level
Grades 10-12
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