Question Number 152683 by rexford last updated on 31/Aug/21
$${The}\:{probability}\:{that}\:{athlete}\:{will}\:{win}\:{a}\:{race}\:{is}\:\frac{\mathrm{1}}{\mathrm{6}}\:{and}\:{that} \\ $$$${he}\:{will}\:{be}\:{second}\:{and}\:{third}\:{are}\:\frac{\mathrm{1}}{\mathrm{4}}\:{and}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${respectively}.{what}\:{is}\:{the}\:{probability}\:{that}\:{he}\:{will}\:{not}\:{be}\:{first} \\ $$$${in}\:{the}\:{first}\:{three}\:{place}! \\ $$$${Please},{help}\:{me}\:{out} \\ $$
Answered by Olaf_Thorendsen last updated on 31/Aug/21
$$\mathrm{P}\left(\mathrm{X}\leqslant\mathrm{3}\right)\:=\:\mathrm{P}\left(\mathrm{X}=\mathrm{1}\right)+\mathrm{P}\left(\mathrm{X}=\mathrm{2}\right)+\mathrm{P}\left(\mathrm{X}=\mathrm{3}\right) \\ $$$$\mathrm{P}\left(\mathrm{X}\leqslant\mathrm{3}\right)\:=\:\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{3}}\:=\:\frac{\mathrm{2}+\mathrm{3}+\mathrm{4}}{\mathrm{12}}\:=\:\frac{\mathrm{5}}{\mathrm{6}} \\ $$