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Question Number 99194 by bemath last updated on 19/Jun/20
Answered by bramlex last updated on 19/Jun/20
$${suppose}\:{probability}\:{win}\:{or}\:{draw}\:{or}\:{lose}\:{are} \\ $$$${same}\:{is}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${not}\:{to}\:{lose}\:{in}\:{those}\:{three}\:{matches}\: \\ $$$${case}\left(\mathrm{1}\right)\:\left(\mathrm{3}{w}\right)\Rightarrow\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} =\:\frac{\mathrm{1}}{\mathrm{27}} \\ $$$${case}\left(\mathrm{2}\right)\left(\mathrm{2}{w},\mathrm{1}{d}\right)\Rightarrow{C}\left(\mathrm{3},\mathrm{2}\right)\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} =\:\frac{\mathrm{1}}{\mathrm{9}} \\ $$$${case}\left(\mathrm{3}\right)\left(\mathrm{1}{w},\mathrm{2}{d}\right)\Rightarrow{C}\left(\mathrm{3},\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} =\:\frac{\mathrm{1}}{\mathrm{9}} \\ $$$${case}\left(\mathrm{4}\right)\left(\mathrm{3}{d}\right)\Rightarrow\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{27}} \\ $$$${now}\:{p}\:=\:\mathrm{2}×\left(\frac{\mathrm{1}}{\mathrm{27}}+\frac{\mathrm{1}}{\mathrm{9}}\right)\:=\:\frac{\mathrm{8}}{\mathrm{27}}\:\blacksquare \\ $$