Given that

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 $p.$ 

Let $E_1=$ Event of first unit is working 

$E_2=$ Event of second unit is working

$P\left(E_1\right)=0.9 P\left(E_2\right)=0.8$

We know that $P\left(\overline{E}\right) = 1 - P\left(E\right)$

$P\left({\overline{E}}_1\right)=0.1 P\left({\overline{E}}_2\right)=0.2$

By using Baye's theorem

Required probability

$p = \frac{P\left(E_2\right)\times P\left(\overline{E_1}\right)}{P\left(\overline{E_2}\right)\times P\left(\overline{E_1}\right)+P\left(E_1\right)\times P\left(\overline{E_2}\right)+P\left(E_2\right)\times P\left(\overline{E_1}\right)}$

$p=\frac{0.8\times 0.1}{0.1\times 0.2+0.9\times 0.2+0.1\times 0.8}=\frac{\left({\displaystyle \frac{8}{100}}\right)}{{\displaystyle \frac{\left(2 +18 +8\right)}{100}}}= \frac{8}{28}$

So, $98P=98\times \frac{8}{28}=28$