Question

a bag contains 5 white, 7 red,4 black balls. four balls are drawn one by one with replacement. what is the probability that none is white

Answer 1

$$\begin{gathered} \mathrm{Total}\mathrm{no}.\mathrm{of}\mathrm{balls}=16 \\ \mathrm{total}\mathrm{no}.\mathrm{of}\mathrm{non}-\mathrm{white}\mathrm{balls}=11 \\ \mathrm{Probability}\mathrm{of}\mathrm{not}\mathrm{getting}\mathrm{a}\mathrm{white}\mathrm{ball}\mathrm{in}\mathrm{one}\mathrm{step}=\frac{11}{16} \\ \mathrm{Since},\mathrm{the}\mathrm{ball}\mathrm{is}\mathrm{put}\mathrm{back}(\mathrm{ie},\mathrm{re}-\mathrm{placed}),\mathrm{the}\mathrm{probability}\mathrm{of}\mathrm{not}\mathrm{getting}\mathrm{a}\mathrm{white}\mathrm{ball}\mathrm{will}\mathrm{remain}\mathrm{same}\mathrm{every}\mathrm{time}. \\ \mathrm{So},\mathrm{probability}\mathrm{of}\mathrm{drawing}\mathrm{non}-\mathrm{white}\mathrm{balls}4\mathrm{times}={\left(\frac{11}{16}\right)}^4=\frac{14641}{65536} \end{gathered}$$