Answer:
The number of returns is
[tex]=60[/tex]The number of tax returns to be selected is
[tex]=3[/tex]The number of errors is
[tex]=8[/tex]The number of returns with no errors is
[tex]\begin{gathered} =60-8 \\ =52 \end{gathered}[/tex]The number of possible returns that the auditor can pick is
[tex]=^{60}C_3=\frac{60!}{57!3!}=\frac{60\times59\times58}{6}=34220[/tex]The number of those 3 returns with no errors will be
[tex]=^{52}C_3=\frac{52!}{49!3!}=\frac{52\times51\times50}{6}=22100[/tex]Hence,
The number of possible sets of 3 returns with no errors,
as a fraction of the number of possible sets of 3 returns available for auditing is
[tex]\begin{gathered} =\frac{22100}{34220} \\ =0.646 \\ =0.646\times100 \\ =64.6\% \end{gathered}[/tex]Hence,
The final answer is
[tex]\Rightarrow64.6\%[/tex]