In a workshop, there are five machines and the probability of any one of them to be out of service on a day is ¼. If the probability that at most two machines will be out of service on the same day is ${\left(\frac{3}{4}\right)}^{3}k$, then k is equal to

Given, probability of any machine to be out of service $=\frac{1}{4}$
So, probability of no machine is out of service $=1-\frac{1}{4}=\frac{3}{4}$

Required probability = when no machine is out of service + when only one machine is out of service + when only two machines are out of service.

$={}^5{C}_0{\left(\frac{3}{4}\right)}^5+{}^5{C}_1\left(\frac{1}{4}\right){\left(\frac{3}{4}\right)}^4+{}^5{C}_2{\left(\frac{1}{4}\right)}^2{\left(\frac{3}{4}\right)}^3$

$={\left(\frac{3}{4}\right)}^3\left[\frac{9}{16}+5\left(\frac{3}{16}\right)+10\left(\frac{1}{16}\right)\right]$

$={\left(\frac{3}{4}\right)}^3\left(\frac{9+15+10}{16}\right)={\left(\frac{3}{4}\right)}^3\left(\frac{34}{16}\right)={\left(\frac{3}{4}\right)}^3\left(\frac{17}{8}\right)={\left(\frac{3}{4}\right)}^{3}k$

$\therefore k=\frac{17}{8}$