In a workshop, there are five machines and the probability of any one of them to be out of service on a day is $\frac{1}{4}$. 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:
Correct Option: 1
Required probability $=$ when no machine has fault $+$ when only one machine has fault $+$ when only two machines have fault.
$={ }^{5} C_{0}\left(\frac{3}{4}\right)^{5}+{ }^{5} C_{1}\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^{4}+{ }^{5} C_{2}\left(\frac{1}{4}\right)^{2}\left(\frac{3}{4}\right)^{3}$
$=\frac{243}{1024}+\frac{405}{1024}+\frac{270}{1024}=\frac{918}{1024}=\frac{459}{512}=\frac{27 \times 17}{64 \times 8}$
$=\left(\frac{3}{4}\right)^{3} \times k=\left(\frac{3}{4}\right)^{3} \times \frac{17}{8}$
$\therefore \quad k=\frac{17}{8}$
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