Question

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 $\mathrm{0.8.}$ 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 $p,$ then $98p$ is equal to .

Answer 1

$p\left(E_1\right)=0.9,p\left(E_2\right)=0.8p\left(E\right)=\frac{p\left(\overline{E_1}\right)\cdot p\left(E_2\right)}{p\left(E_1\right)\cdot P\left(\overline{E_2}\right)+p\left(\overline{E_1}\right)\cdot p\left(E_2\right)+p\left(\overline{E_1}\right)\cdot p\left(\overline{E_2}\right)}\Rightarrow p\left(E\right)=\frac{\left(0.1\right)\left(0.8\right)}{\left(0.9\right)\left(0.2\right)+\left(0.1\right)\left(0.8\right)+\left(0.1\right)\left(0.2\right)}\Rightarrow p=\frac{0.08}{0.28}=\frac{2}{7}\therefore 98p=98\times \frac{2}{7}=28$