Question

In a hospital, there are 20 kidney dialysis machines and the chance of any one of them to be out of service during a day is 0.02. Determine the probability that exactly 3 machines will be out of service on the same day.

Answer 1

$$\begin{gathered} \mathrm{Let}X\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{machines}\mathrm{out}\mathrm{of}\mathrm{service}\mathrm{during}\mathrm{a}\mathrm{day}. \\ \mathrm{Then},X\mathrm{follows}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}n=20 \\ \text{Let}p\mathrm{be}\mathrm{the}\mathrm{probability}\mathrm{of}\mathrm{any}\mathrm{machine}\mathrm{out}\mathrm{of}\mathrm{service}\mathrm{during}\mathrm{a}\mathrm{day}. \\ \therefore p=0.02\mathrm{and}q=0.98 \\ \mathrm{Hence},\mathrm{the}\mathrm{distribution}\mathrm{is}\mathrm{given}\mathrm{by} \\ P(\mathit{\text{X}}=r)={}^{20}C_r{\left(0.02\right)}^r{\left(0.98\right)}^{20-r},r=0,1,2.....20 \\ \therefore P(\mathrm{exactlly}3\mathrm{machines}\mathrm{will}\mathrm{be}\mathrm{out}\mathrm{of}\mathrm{the}\mathrm{service}\mathrm{on}\mathrm{the}\mathrm{same}\mathrm{day})=P(X=3) \\ ={}^{20}C_3{\left(0.02\right)}^3{\left(0.98\right)}^{20-3} \\ =1140(0.000008)(0.7093) \\ =0.006469 \end{gathered}$$