An electric instrument consists of two units. Each unit must function independently for the instrument to operate. The probability that the first unit functions is $0.9$ and that of the second unit is $0.8$. The instrument is switched on and it fails to operate. If the probability that only the first unit failed and second unit is functioning is $\mathrm{p}$, then $98 \mathrm{p}$ is equal to
$\mathrm{I}_{1}=$ first unit is functioning
$\mathrm{I}_{2}=$ second unit is functioning
$\mathrm{P}\left(\mathrm{I}_{1}\right)=0.9, \mathrm{P}\left(\mathrm{I}_{2}\right)=0.8$
$\mathrm{P}\left(\overline{\mathrm{I}}_{1}\right)=0.1, \mathrm{P}\left(\overline{\mathrm{I}}_{2}\right)=0.2$
$\mathrm{P}=\frac{0.8 \times 0.1}{0.1 \times 0.2+0.9 \times 0.2+0.1 \times 0.8}=\frac{8}{28}$
$98 \mathrm{P}=\frac{8}{28} \times 98=28$