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With one method of a procedure called acceptance sampling, a sample of items is randomly selected without replacement and the entire batch is accepted if every item in the sample is okay. A company just manufactured 796 CDs, and 205 are defective. If 3 of these CDs are randomly selected for testing, what is the probability that the entire batch will be accepted?

Verified Answer :

meerkat18

The probability that the first CD in the sample is not defective is 591/796
The probability that the second CD in the sample is not defective is 590/795
The probability that the third CD in the sample is not defective is 589/794

P(sample has 0 defective CDs)=591×590× 589796×795×794=
you can calculate

JeanaShupp

Answer: 0.405224

Step-by-step explanation:

Given :  A company just manufactured 796 CDs, and 205 are defective.

Then, the probability that a CD is defective : \(p=\dfrac{205}{796}=0.257537688442\approx0.26\)

Sample size : n= 3

Binomial probability formula :

\(P(x)=^nC_xp^x(1-p)^{n-x}\)

By using binomial probability formula, the probability that none 3 CDs is  defective:-

\(P(3)=^3C_3(0.26)^0(1-0.26)^{3}\\\\=(1)(0.74)^3\ \ [\text{Since}^nC_n=1]\\\\=0.405224\)

Hence, the the probability that the entire batch will be accepted= 0.405224

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