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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=((2×(1/6))/1)=(1/3)

$${p}=\frac{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{6}}}{\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Commented by Mingma last updated on 08/Nov/23

Perfect ��

Answered by mr W last updated on 08/Nov/23

a: b= 1: 3, 4, ..., 3n ⇒ 3n−2 2: 6, 7, ..., 3n ⇒ 3n−5 3: 9, 10, ..., 3n ⇒ 3n−8 ... n: 3n ⇒ 1 Σ=1+4+...+(3n−2)=((n(3n−1))/2) p=(((n(3n−1))/2)/C_2 ^(3n) )=((n(3n−1)×2)/(2×3n(3n−1)))=(1/3)

$${a}:\:\:\:\:{b}= \\ $$$$\mathrm{1}:\:\:\:\:\mathrm{3},\:\mathrm{4},\:...,\:\mathrm{3}{n}\:\Rightarrow\:\mathrm{3}{n}−\mathrm{2} \\ $$$$\mathrm{2}:\:\:\:\:\mathrm{6},\:\mathrm{7},\:...,\:\mathrm{3}{n}\:\Rightarrow\:\mathrm{3}{n}−\mathrm{5} \\ $$$$\mathrm{3}:\:\:\:\:\mathrm{9},\:\mathrm{10},\:...,\:\mathrm{3}{n}\:\Rightarrow\:\mathrm{3}{n}−\mathrm{8} \\ $$$$... \\ $$$${n}:\:\mathrm{3}{n}\:\Rightarrow\:\mathrm{1} \\ $$$$\Sigma=\mathrm{1}+\mathrm{4}+...+\left(\mathrm{3}{n}−\mathrm{2}\right)=\frac{{n}\left(\mathrm{3}{n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$${p}=\frac{\frac{{n}\left(\mathrm{3}{n}−\mathrm{1}\right)}{\mathrm{2}}}{{C}_{\mathrm{2}} ^{\mathrm{3}{n}} }=\frac{{n}\left(\mathrm{3}{n}−\mathrm{1}\right)×\mathrm{2}}{\mathrm{2}×\mathrm{3}{n}\left(\mathrm{3}{n}−\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Commented by MM42 last updated on 08/Nov/23

Sir W ” b > 3a ”

$${Sir}\:\:{W} \\ $$$$\:''\:{b}\:>\:\mathrm{3}{a}\:''\: \\ $$

Commented by mr W last updated on 08/Nov/23

you are right sir! i purposely took b≥3a in order to get p=(1/3) independently from n.

$${you}\:{are}\:{right}\:{sir}! \\ $$$${i}\:{purposely}\:{took}\:{b}\geqslant\mathrm{3}{a}\:{in}\:{order}\:{to}\:{get} \\ $$$${p}=\frac{\mathrm{1}}{\mathrm{3}}\:{independently}\:{from}\:{n}. \\ $$

Answered by MM42 last updated on 08/Nov/23

b> 3a a=1→b=4,5,6,..,3n ⇒m_1 =3n−3 a=2→b=7,8,9,...,3n⇒m_2 =3n−6 ⋮ a=n−1→b=3n−2,3n−1,3n⇒m_(n−1) =3 n(A)=3+6+...+(3n−3)=((3n(n−1))/2) n(S)= (((3n)),(( 2)) )=((3n(3n−1))/2) ⇒ p(A)=((n−1)/(3n−1)) ✓ for example : n=3 : 1,2,..,9⇒n(S)= ((9),(2) )=36 A={{1,4},{1,5},...,{1,9},{2,7},{2,8},{2,9}}⇒n(A)=9 ⇒p(A)=(1/4) ✓

$${b}>\:\mathrm{3}{a} \\ $$$${a}=\mathrm{1}\rightarrow{b}=\mathrm{4},\mathrm{5},\mathrm{6},..,\mathrm{3}{n}\:\Rightarrow{m}_{\mathrm{1}} =\mathrm{3}{n}−\mathrm{3} \\ $$$${a}=\mathrm{2}\rightarrow{b}=\mathrm{7},\mathrm{8},\mathrm{9},...,\mathrm{3}{n}\Rightarrow{m}_{\mathrm{2}} =\mathrm{3}{n}−\mathrm{6} \\ $$$$\vdots \\ $$$${a}={n}−\mathrm{1}\rightarrow{b}=\mathrm{3}{n}−\mathrm{2},\mathrm{3}{n}−\mathrm{1},\mathrm{3}{n}\Rightarrow{m}_{{n}−\mathrm{1}} =\mathrm{3} \\ $$$${n}\left({A}\right)=\mathrm{3}+\mathrm{6}+...+\left(\mathrm{3}{n}−\mathrm{3}\right)=\frac{\mathrm{3}{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$${n}\left({S}\right)=\begin{pmatrix}{\mathrm{3}{n}}\\{\:\:\mathrm{2}}\end{pmatrix}=\frac{\mathrm{3}{n}\left(\mathrm{3}{n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\Rightarrow\:{p}\left({A}\right)=\frac{{n}−\mathrm{1}}{\mathrm{3}{n}−\mathrm{1}}\:\:\checkmark \\ $$$${for}\:{example}\:: \\ $$$${n}=\mathrm{3}\::\:\mathrm{1},\mathrm{2},..,\mathrm{9}\Rightarrow{n}\left({S}\right)=\begin{pmatrix}{\mathrm{9}}\\{\mathrm{2}}\end{pmatrix}=\mathrm{36} \\ $$$${A}=\left\{\left\{\mathrm{1},\mathrm{4}\right\},\left\{\mathrm{1},\mathrm{5}\right\},...,\left\{\mathrm{1},\mathrm{9}\right\},\left\{\mathrm{2},\mathrm{7}\right\},\left\{\mathrm{2},\mathrm{8}\right\},\left\{\mathrm{2},\mathrm{9}\right\}\right\}\Rightarrow{n}\left({A}\right)=\mathrm{9} \\ $$$$\Rightarrow{p}\left({A}\right)=\frac{\mathrm{1}}{\mathrm{4}}\:\checkmark\: \\ $$

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