Question Number 219527 by mr W last updated on 27/Apr/25
Commented by Nicholas666 last updated on 27/Apr/25
$$\frac{\mathrm{1}}{\mathrm{48019}} \\ $$$$ \\ $$
Commented by mr W last updated on 27/Apr/25
$${can}\:{you}\:{please}\:{show}\:{how}? \\ $$
Answered by mahdipoor last updated on 27/Apr/25
$${all}\:{postion} \\ $$$$=\:\begin{pmatrix}{\mathrm{20}}\\{\mathrm{2}}\end{pmatrix}\:+\begin{pmatrix}{\mathrm{18}}\\{\mathrm{2}}\end{pmatrix}+…+\begin{pmatrix}{\mathrm{2}}\\{\mathrm{2}}\end{pmatrix}\:=\mathrm{715} \\ $$$${in}\:{one}\:\left({or}\:{more}\right)\:{lattery}\:\:{people}\:{from}\:{same}\:{college} \\ $$$$=\:\begin{pmatrix}{\mathrm{20}}\\{\mathrm{1}}\end{pmatrix}×\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{pmatrix}+\begin{pmatrix}{\mathrm{18}}\\{\mathrm{2}}\end{pmatrix}+…+\begin{pmatrix}{\mathrm{2}}\\{\mathrm{2}}\end{pmatrix}\:=\:\mathrm{545} \\ $$$${ln}\:{each}\:{lattery}\:{never}\:{people}\:{fram}\:{same}\:{college} \\ $$$$=\mathrm{715}−\mathrm{545}=\mathrm{170} \\ $$$${p}=\frac{\mathrm{170}}{\mathrm{715}} \\ $$
Commented by mahdipoor last updated on 27/Apr/25
$${it}\:{is}\:{right}? \\ $$
Commented by mr W last updated on 28/Apr/25
$${thanks}\:{sir}!\:{but}\:{the}\:{right}\:{answer} \\ $$$${should}\:{be}\:{what}\:{wuji}\:{sir}\:{has}\:{given} \\ $$$${below}:\:\approx\mathrm{59\%}. \\ $$
Answered by Wuji last updated on 27/Apr/25
$${Total}\:{number}\:{of}\:{ways}\:{to}\:{split}\:\mathrm{20}\:{labeled}\:{people}\:{into}\:\mathrm{10} \\ $$$${unordered}\:{pairs} \\ $$$$\left(\mathrm{20}−\mathrm{1}\right)!!\:=\mathrm{19}!!\:=\frac{\mathrm{20}!}{\mathrm{2}^{\mathrm{10}} \mathrm{10}!}\:=\mathrm{654729075} \\ $$$${let}\:{q}\:{college}\:{out}\:{of}\:\mathrm{10}\:{to}\:{be}\:{bad} \\ $$$$\begin{pmatrix}{\mathrm{10}}\\{\:\:{q}}\end{pmatrix}\:\: \\ $$$$\left.\:\mathrm{20}−\mathrm{2}{q}\:\:\:\Rightarrow\mathrm{2}\left(\mathrm{10}−{q}\right)−\mathrm{1}\right)!!=\left(\mathrm{20}−\mathrm{2}{q}−\mathrm{1}\right)!! \\ $$$$\Rightarrow\begin{pmatrix}{\mathrm{10}}\\{\:\:{q}}\end{pmatrix}\:\left(\mathrm{19}−\mathrm{2}{q}\right)!! \\ $$$${let}\:{A}_{{i}} =\mathrm{2}\:{students}\:{of}\:{college}\:{i}\:{matched}\:{to}\:{each}\:{other} \\ $$$${A}_{{i}} \cup…\cup{A}_{\mathrm{10}} \:=\underset{{k}=\mathrm{0}} {\overset{\mathrm{10}} {\sum}}\left(−\mathrm{1}\right)^{{q}} \:\Sigma\:{s}\subset\underset{\mid{s}\mid={k}} {\:}\left\{\mathrm{1},…,\mathrm{10}\right\}\:\cap_{{i}\in{S}} \:{A}_{{i}} \:=\underset{{q}=\mathrm{0}} {\overset{\mathrm{10}} {\sum}}\left(−\mathrm{1}\right)^{{q}} \begin{pmatrix}{\mathrm{10}}\\{\:\:{q}}\end{pmatrix}\:\left(\mathrm{19}−\mathrm{2}{q}\right)!! \\ $$$${N}_{{good}} \:\:=\underset{{q}=\mathrm{0}} {\overset{\mathrm{10}} {\sum}}\left(−\mathrm{1}\right)^{{q}} \begin{pmatrix}{\mathrm{10}}\\{\:\:{q}}\end{pmatrix}\:\left(\mathrm{19}−\mathrm{2}{q}\right)!!\:=\mathrm{387099936} \\ $$$${P}=\frac{{N}_{{good}} }{\mathrm{19}!!}\:=\frac{\mathrm{387099936}}{\mathrm{654729075}}\:\:\approx\mathrm{0}.\mathrm{59124} \\ $$$$\:\therefore\:{about}\:\mathrm{59}.\mathrm{12\%}\:\:{chance}\:{that}\:{no}\:{two} \\ $$$${students}\:{from}\:{the}\:{same}\:{college}\:{end}\:{up}\:{playing}\:{each}\:{other} \\ $$
Commented by mr W last updated on 28/Apr/25
$${thanks}\:{sir}! \\ $$