ANSWER
[tex]P=\frac{81}{625}[/tex]EXPLANATION
There are 4 defects out of 10 total computers. This means that there are 6 computers without defects.
The probability that 1 computer selected will not be defective is the total number of non-defective computers divided by the total number of computers:
[tex]P(one-without-defect)=\frac{6}{10}[/tex]Therefore, if a sample of 4 computers is selected, the probability that the sample will not contain a defective computer is:
[tex]\begin{gathered} P=\frac{6}{10}\cdot\frac{6}{10}\cdot\frac{6}{10}\cdot\frac{6}{10}=(\frac{6}{10})^4 \\ P=\frac{81}{625} \end{gathered}[/tex]