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I had a quiz and one of the question is : If I have 25 items among them 10 are defective. The question is I pick 5 out of 25 and test them, what is the chance of 3 of them are defective.

The correct answer is using hypergeometric distribution. However my question is if I assume all 5 items are tested at the same time ( that means independent from each other) is it possible that it will become binomial distribution ?

Another question is if I pick randomly 5 items out of 25 and randomly number them as item 1,2,3,4,5. What is the probability of defective for item 1? My TA said that because we pick 5 out of 25 without replacement then the probability of defective of one item in the batch of 5 is no longer $\frac{10}{25}$ so I should use hyper geometric distribution in this case. I am quite confused since picking give you no new information how can the probability of defective of one random item changes? Can someone help me to clarify?

Thank you

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The only way to get a binomial distribution is to assume that the sampling is done with replacement. If the sampling is without replacement, the distribution of the number of defectives is hypergeometric.

For your second question, if you pick $5$ items and number them, the probability the first is defective is $\frac{10}{25}$. The probability the second is defective is also $\frac{10}{25}$. And so on. Many people find this unintuitive at first.

Of course the conditional probability the second is defective, given the first is, is not $\frac{10}{25}$. When we do sampling without replacement, we lose independence.

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The use of binomial and hypergeometric hinges on the concept of whether the samples are replaced or not. If it is replaced, then the probability of picking the second of n itmes remains the same as the first through n times that are picked while without replacement, the probability keeps reducing everytime you pick an item, in other words, the denominator is not any more the same 25. This is the key difference between the binomial and hypergeometric.

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