Question Number 99141 by babar last updated on 18/Jun/20
$$\mathrm{The}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{in}\:\mathrm{a}\:\mathrm{random} \\ $$$$\mathrm{arrangement}\:\mathrm{of}\:\mathrm{the}\:\mathrm{letters}\:\mathrm{of}\:\mathrm{the}\:\mathrm{word} \\ $$$$'\mathrm{UNIVERSITY}'\:\mathrm{the}\:\mathrm{two}\:\mathrm{I}'\:\mathrm{do}\:\mathrm{not}\:\mathrm{come} \\ $$$$\mathrm{together}\:\mathrm{is} \\ $$
Answered by bramlex last updated on 19/Jun/20
$${n}\left({S}\right)=\frac{\mathrm{10}!}{\mathrm{2}!}\:;\:{n}\left({A}\right)\:=\:\mathrm{9}×\mathrm{8}! \\ $$$${P}\left({A}\right)\:=\:\frac{{n}\left({A}\right)}{{n}\left({S}\right)}\:=\:\frac{\mathrm{9}×\mathrm{8}!}{\left(\frac{\mathrm{10}!}{\mathrm{2}!}\right)}\:=\:\frac{\mathrm{2}×\mathrm{9}}{\mathrm{10}×\mathrm{9}}\:=\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$
Commented by bemath last updated on 19/Jun/20
$$\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{P}\left(\mathrm{A}^{\mathrm{c}} \right)\:=\:\mathrm{1}−\mathrm{P}\left(\mathrm{A}\right)\:=\:\frac{\mathrm{4}}{\mathrm{5}} \\ $$
Commented by bramlex last updated on 19/Jun/20
$${oh}\:{yes}.\: \\ $$