Given that a loss exceeds $8,$ what is the probability that it exceeds $16?$

I. $\frac{{\displaystyle \underset{8}{\overset{20}{}}}0\mathrm{.}005(20-x)dx}{{\displaystyle \underset{16}{\overset{20}{}}}0\mathrm{.}005(20-x)dx}$

II$.\frac{0\mathrm{.}005(20-16)dx}{{\displaystyle \underset{8}{\overset{20}{}}}0\mathrm{.}005(20-x)dx}$

III. $\frac{0\mathrm{.}005(20-8)}{0\mathrm{.}005(20-16)}$

IV$.\frac{{\displaystyle \underset{16}{\overset{20}{}}}0\mathrm{.}005(20-x)dx}{{\displaystyle \underset{8}{\overset{20}{}}}0\mathrm{.}005(20-x)dx}$

V$.\frac{0\mathrm{.}005(20-16)}{0\mathrm{.}005(20-8)}$

IV

III

v

II