In a bin containing 100 parts, ten (10) are defective. If a sample of seven (7) parts is randomly selected without replacement, what is the probability of precisely two defective parts being in the sample?
Question
In a bin containing 100 parts, ten (10) are defective. If a sample of seven (7) parts is randomly selected without replacement, what is the probability of precisely two defective parts being in the sample?
Solution
To solve this problem, we will use the concept of hypergeometric distribution. The hypergeometric distribution model is appropriate for a population consisting of two types of outcomes, and sampling is done without replacement.
Here are the steps:
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Identify the parameters:
- N = Total number of parts = 100
- K = Total number of defective parts = 10
- n = Number of parts in the sample = 7
- k = Number of defective parts in the sample = 2
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Use the formula for hypergeometric distribution: P(X = k) = [C(K, k) * C(N-K, n-k)] / C(N, n)
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Substitute the values into the formula: P(X = 2) = [C(10, 2) * C(100-10, 7-2)] / C(100, 7)
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Calculate the combinations:
- C(10, 2) = 10! / [2!(10-2)!] = 45
- C(90, 5) = 90! / [5!(90-5)!] = 43,949,268
- C(100, 7) = 100! / [7!(100-7)!] = 16,007,560,800
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Substitute the calculated combinations back into the formula: P(X = 2) = (45 * 43,949,268) / 16,007,560,800
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Calculate the probability: P(X = 2) = 0.1234 or 12.34%
So, the probability of precisely two defective parts being in the sample is approximately 12.34%.
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