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Twolffpiggott · Answer 1 · 2020-09-28T05:41:30+0000

Answer:

The probability that a box containing 3 defectives will be shipped is 0.6696.

The probability that a box containing only 1 defective will be sent back for screening is 0.12

Explanation:

Consider the provided information.

Boxes of 25 items are readied for shipment, and a sample of 3 items is tested for defectives.

Part (A) What is the probability that a box containing 3 defectives will be shipped?

Let x represents the number of defective item among the 3 item.

According to hyper geometric distribution:

$h(x; N, n, k) = \frac{\binom{k}{x} \binom{N-k}{n-x}}{\binom{N}{n}}$

Where n is randomly selected without replacement from a population of N items. k items can be classified as successes, and N - k items can be classified as failures.

Therefore, N = 25, n = 3 and k=3 in above formula.

The probability that box deliver 3 defective item.

P(No defective item among selected 3)=P(x=0)

$P(x=0) = \frac{\binom{3}{0} \binom{25-3}{3-0}}{\binom{25}{3}}$

P(x=0) =0.6696

Therefore, the probability that a box containing 3 defectives will be shipped is 0.6696.

Part (B) What is the probability that a box containing only 1 defective will be sent back for screening?

Substitute, N = 25, n = 3 and k=1 in above formula.

The probability that a box containing only 1 defective will be sent back for screening

$P(x=1) = \frac{\binom{1}{1} \binom{25-1}{3-1}}{\binom{25}{3}}$

P(x=1) =0.12

Hence, the probability that a box containing only 1 defective will be sent back for screening is 0.12

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