A box of manufactured items contains 12 items of which 4 are defective. If 3 items...

A box of manufactured items contains 12 items of which 4 are defective. If 3 items are drawn at random without replacement, what is the probability that:

1. The first one is defective and rest are good

2. Exactly one of three is defective

Expert Solution

Out of 12 items, 4 are defective and 8 are good.

1:

Since draws are done without replacement so after first defective we have 8 good items out of 11 items. Likewise after second good item, we have 7 good items out of 10 items. So the probability that the first one is defective and rest are good is

P(the first one is defective and rest are good) = (4/12) * (8/11) * (7/10) = 0.1697

2:

Here we need to use hypergeometric distribution. The probability that exactly one of three is defective is

orchestra answered 2 years ago

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