• # question_answer 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                     [JEE MAIN Held on 07-01-2020 Evening] A) $4$                  B) $\frac{17}{4}$ C) $\frac{17}{8}$                       D) $\frac{17}{2}$

 [c] Probability that a machine is faulted $=\frac{1}{4}=P$ Probability that a machine is not faulted $=1-\frac{1}{4}=\frac{3}{4}=q$ $\therefore$    Probability that at most two machine is faulted $=P(X=0)+P(X=1)+P(X=2)$ $\therefore \,\,\,\,{{\,}^{5}}{{C}_{0}}{{\left( \frac{1}{4} \right)}^{0}}\cdot {{\left( \frac{3}{4} \right)}^{5}}{{+}^{5}}{{C}_{1}}{{\left( \frac{1}{4} \right)}^{1}}\cdot {{\left( \frac{3}{4} \right)}^{4}}+{{\,}^{5}}{{C}_{2}}{{\left( \frac{1}{4} \right)}^{2}}\cdot {{\left( \frac{3}{4} \right)}^{3}}$ $={{\left( \frac{3}{4} \right)}^{3}}\cdot k$ $\Rightarrow \,\,{{\left( \frac{3}{4} \right)}^{2}}+5\cdot \frac{1}{4}\cdot \frac{3}{4}+10\cdot {{\left( \frac{1}{4} \right)}^{2}}=k$ $\therefore \,\,\,k=\frac{10+15+9}{16}=\frac{34}{16}=\frac{17}{8}$