A factory is interested in detecting defective items in a bo

A factory is interested in detecting defective items in a box. However, they can’t go through every
box and test every item.
a. Suppose that in a given box of 36 items, 5 of them are defective. If the company samples 6 items from
the box to see if there are any defective, what is the probability that the box gets flagged for defectives?
b. If they instead sampled 10 items from the box, what is the new probability of being flagged?
c. Suppose that in a crate of 20 boxes, there is only one box of defectives (which has the breakdown from
above). If they sample one box from the crate at random and test 8 items from that box, what is the
probability that the only box with defectives gets detected/flagged?

Solution

a)

As

P(at least 1 defective) = 1 - P(no defective)

There are 36C6 ways to choose any item, and 31C6 to choose just good items. Hence,

P(no defective) = (31C6) / (36C6) = 0.378008021

Thus,

P(at least 1 defective) = 0.621991979 [answer]

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b)

As

P(at least 1 defective) = 1 - P(no defective)

There are 36C10 ways to choose any item, and 31C10 to choose just good items. Hence,

P(no defective) = (31C10) / (36C10) = 0.174486461

Thus,

P(at least 1 defective) = 0.825513539 [answer]

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c)

For the defective crate, the probability that it gets flagged, if chosen is

As

P(at least 1 defective) = 1 - P(no defective)

There are 36C8 ways to choose any item, and 31C8 to choose just good items. Hence,

P(no defective) = (31C10) / (36C10) = 0.260695187

Thus,

P(at least 1 defective) = 0.739304813 [answer]

Therefore,

P(defective box gets caught) = P(it is chosen) P(it gets flagged) = (1/20)*0.739304813 = 0.036965241 [answer]