Question Number 195206 by otchereabdullai@gmail.com last updated on 27/Jul/23
Answered by MM42 last updated on 27/Jul/23
$${p}\left({a}\right)={p}\left({b}\right)={p}\left({c}\right)=\frac{\mathrm{3}}{\mathrm{4}}\:;\:{probability}\:{of}\:{winning}\:{each}\:{race} \\ $$$$\left({i}\right)\:{p}\left({a}'\cap{b}\cap{c}'\right)=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{64}} \\ $$$$\left({ii}\right)\:{p}\left({a}\cap{b}\cap{c}\right)=\frac{\mathrm{27}}{\mathrm{64}} \\ $$$$\left({iii}\right)\:{p}\left(\Sigma{abc}'\right)=\mathrm{3}×\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{9}}{\mathrm{16}}=\frac{\mathrm{27}}{\mathrm{64}} \\ $$
Commented by otchereabdullai@gmail.com last updated on 27/Jul/23
$${thanks}\:{alot}\:{sir} \\ $$