Question Number 199711 by Mingma last updated on 08/Nov/23
Answered by mr W last updated on 08/Nov/23
Commented by mr W last updated on 08/Nov/23
$${p}=\frac{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{6}}}{\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by Mingma last updated on 08/Nov/23
Perfect
Answered by mr W last updated on 08/Nov/23
$${a}:\:\:\:\:{b}= \\ $$$$\mathrm{1}:\:\:\:\:\mathrm{3},\:\mathrm{4},\:…,\:\mathrm{3}{n}\:\Rightarrow\:\mathrm{3}{n}−\mathrm{2} \\ $$$$\mathrm{2}:\:\:\:\:\mathrm{6},\:\mathrm{7},\:…,\:\mathrm{3}{n}\:\Rightarrow\:\mathrm{3}{n}−\mathrm{5} \\ $$$$\mathrm{3}:\:\:\:\:\mathrm{9},\:\mathrm{10},\:…,\:\mathrm{3}{n}\:\Rightarrow\:\mathrm{3}{n}−\mathrm{8} \\ $$$$… \\ $$$${n}:\:\mathrm{3}{n}\:\Rightarrow\:\mathrm{1} \\ $$$$\Sigma=\mathrm{1}+\mathrm{4}+…+\left(\mathrm{3}{n}−\mathrm{2}\right)=\frac{{n}\left(\mathrm{3}{n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$${p}=\frac{\frac{{n}\left(\mathrm{3}{n}−\mathrm{1}\right)}{\mathrm{2}}}{{C}_{\mathrm{2}} ^{\mathrm{3}{n}} }=\frac{{n}\left(\mathrm{3}{n}−\mathrm{1}\right)×\mathrm{2}}{\mathrm{2}×\mathrm{3}{n}\left(\mathrm{3}{n}−\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by MM42 last updated on 08/Nov/23
$${Sir}\:\:{W} \\ $$$$\:''\:{b}\:>\:\mathrm{3}{a}\:''\: \\ $$
Commented by mr W last updated on 08/Nov/23
$${you}\:{are}\:{right}\:{sir}! \\ $$$${i}\:{purposely}\:{took}\:{b}\geqslant\mathrm{3}{a}\:{in}\:{order}\:{to}\:{get} \\ $$$${p}=\frac{\mathrm{1}}{\mathrm{3}}\:{independently}\:{from}\:{n}. \\ $$
Answered by MM42 last updated on 08/Nov/23
$${b}>\:\mathrm{3}{a} \\ $$$${a}=\mathrm{1}\rightarrow{b}=\mathrm{4},\mathrm{5},\mathrm{6},..,\mathrm{3}{n}\:\Rightarrow{m}_{\mathrm{1}} =\mathrm{3}{n}−\mathrm{3} \\ $$$${a}=\mathrm{2}\rightarrow{b}=\mathrm{7},\mathrm{8},\mathrm{9},…,\mathrm{3}{n}\Rightarrow{m}_{\mathrm{2}} =\mathrm{3}{n}−\mathrm{6} \\ $$$$\vdots \\ $$$${a}={n}−\mathrm{1}\rightarrow{b}=\mathrm{3}{n}−\mathrm{2},\mathrm{3}{n}−\mathrm{1},\mathrm{3}{n}\Rightarrow{m}_{{n}−\mathrm{1}} =\mathrm{3} \\ $$$${n}\left({A}\right)=\mathrm{3}+\mathrm{6}+…+\left(\mathrm{3}{n}−\mathrm{3}\right)=\frac{\mathrm{3}{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$${n}\left({S}\right)=\begin{pmatrix}{\mathrm{3}{n}}\\{\:\:\mathrm{2}}\end{pmatrix}=\frac{\mathrm{3}{n}\left(\mathrm{3}{n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\Rightarrow\:{p}\left({A}\right)=\frac{{n}−\mathrm{1}}{\mathrm{3}{n}−\mathrm{1}}\:\:\checkmark \\ $$$${for}\:{example}\:: \\ $$$${n}=\mathrm{3}\::\:\mathrm{1},\mathrm{2},..,\mathrm{9}\Rightarrow{n}\left({S}\right)=\begin{pmatrix}{\mathrm{9}}\\{\mathrm{2}}\end{pmatrix}=\mathrm{36} \\ $$$${A}=\left\{\left\{\mathrm{1},\mathrm{4}\right\},\left\{\mathrm{1},\mathrm{5}\right\},…,\left\{\mathrm{1},\mathrm{9}\right\},\left\{\mathrm{2},\mathrm{7}\right\},\left\{\mathrm{2},\mathrm{8}\right\},\left\{\mathrm{2},\mathrm{9}\right\}\right\}\Rightarrow{n}\left({A}\right)=\mathrm{9} \\ $$$$\Rightarrow{p}\left({A}\right)=\frac{\mathrm{1}}{\mathrm{4}}\:\checkmark\: \\ $$