An electric instrument consists of two units. Each unit must...

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

$P\left({E}_{1}\right) = 0 . 9P\left({E}_{2}\right) = 0 . 8$

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

$P\left(\overset{¯}{E}1\right) = 0 . 1P\left(\overset{¯}{E}2\right) = 0 . 2$

By using Baye's theorem

Required probability

$p = \dfrac{P\left({E}_{2}\right) \times P\left({E}_{1}\right)}{P\left({E}_{2}\right) \times P\left({E}_{1}\right) + P\left({E}_{1}\right) \times P\left({E}_{2}\right) + P\left({E}_{2}\right) \times P\left({E}_{1}\right)}$

$p = \dfrac{0 . 8 \times 0 . 1}{0 . 1 \times 0 . 2 + 0 . 9 \times 0 . 2 + 0 . 1 \times 0 . 8} = \dfrac{\left(\dfrac{8}{100}\right)}{\dfrac{\left(2 + 18 + 8\right)}{100}} = \dfrac{8}{28}$

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