Our professor provides the answer to this 3 PART QUESTION, but I\'m still having

Question

Our professor provides the answer to this 3 PART QUESTION, but I'm still having trouble getting there. The correct answers are: 0.578; Yes; and 0.607. ONLY RESPOND IF YOU CAN EXPLAIN THE THREE ANSWERS (i.e. 0.578; Yes; 0.607).

Here's the question.

You run a manufacturing operation that requires a stock of replacement parts for a critical component. The component manufacturer notifies you and other customers that 25% of the parts they produced in the previous quarter are defective, and you have 16 of those parts in your stock. If you sample three parts at random, what is the probability that one or more of those three parts will be defective? Will you obtain a different answer if you know that exactly 25% of the parts in your stock (that is, 4 out of 16) are defective? If so, recalculate the probability that one or more of the three parts that you sample will be defective.

Again, the answers should be 0.578; Yes; and 0.607, but it's not clear how.

Explanation / Answer

a) X - number of defective in 3 parts

X is binomial with parameter n = 3 and p = 0.25

P(X >= 1) = 1 - P(X =0)

= 1 - (1 -p)^n

= 1 - (1-0.25)^3

= 1 - 0.75^3

= 0.578125

= 0.578

b)

yes ,

if we know exactlythat 4 are defective we can apply hypergeometric distribution directly

c)

total 16 , 4 defective , 12 defective

3 selected ,

P(X>=1) = 1 - P(X =0)

= 1 - 12C3/16C3 {note that we have to choose 3 not defective from 12 , and 3 total from 16 }

= 1 - (12*11*10)/(16*15*14)

= 1 - 0.39285714

= 0.6071428

= 0.607

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