Given:
[tex]\begin{gathered} Number(CDs)=1500 \\ Defective(CDs)=170 \end{gathered}[/tex]To Determine: The probability that the entire batch will be accepted if 6 CDs were randomly selected
Solution
[tex]Non-defective(CDs)=1500-170=1330[/tex]Please note accepted CDs is the same as non-defective CDs.
If 6 were radomly selected without replacement, let us calculate the probability from the first to the sixth
[tex]\begin{gathered} Recall: \\ P(A)=\frac{n(A)}{n(S)} \\ So, \\ P(accepted)=\frac{n(Accepted)}{n(Total)} \end{gathered}[/tex]For the first accepted CD's
[tex]P(accepted1)=\frac{1330}{1500}=\frac{133}{150}[/tex]For the second accepted CD's without replacement
[tex]P(accepted2)=\frac{1330-1}{1500-1}=\frac{1329}{1499}=0.886591[/tex]For the third CD's
[tex]P(accepted3)=\frac{1329-1}{1499-1}=\frac{1328}{1498}=0.886515[/tex]For the fourth Cd's
[tex]P(accepted4)=\frac{1328-1}{1498-1}=\frac{1327}{1497}=0.8864395[/tex]For the fifth Cd's
[tex]P(accepted5)=\frac{1327-1}{1497-1}=\frac{1326}{1496}=0.886364[/tex]For the sixth CD's
[tex]P(accepted6)=\frac{1326-1}{1496-1}=\frac{1325}{1495}=0.866288[/tex]Therefore, the probability that the entire batch will be accepted would be
[tex]P(accepted)=\frac{1330}{1500}\times\frac{1329}{1499}\times\frac{1328}{1428}\times\frac{1327}{1427}\times\frac{1326}{1426}\times\frac{1325}{1425}[/tex][tex]P(accepted)=0.485295\approx0.4853(nearest-4-decimal-place)[/tex]Hence, the probability that the entire batch will be accepted is approximately 0.4853
