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Question Number 99141 by babar last updated on 18/Jun/20

$$\mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{in}\mathrm{a}\mathrm{random}$$$$\mathrm{arrangement}\mathrm{of}\mathrm{the}\mathrm{letters}\mathrm{of}\mathrm{the}\mathrm{word}$$$${}^{\prime }{\mathrm{UNIVERSITY}}^{\prime }\mathrm{the}\mathrm{two}{\mathrm{I}}^{\prime }\mathrm{do}\mathrm{not}\mathrm{come}$$$$\mathrm{together}\mathrm{is}$$

Answered by bramlex last updated on 19/Jun/20

$$n\left(S\right)=\frac{10!}{2!};n\left(A\right)=9\times 8!$$$$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{9\times 8!}{\left(\frac{10!}{2!}\right)}=\frac{2\times 9}{10\times 9}=\frac{1}{5}$$

Commented by bemath last updated on 19/Jun/20

$$\mathrm{it}\mathrm{should}\mathrm{be}\mathrm{P}\left({\mathrm{A}}^{\mathrm{c}}\right)=1-\mathrm{P}\left(\mathrm{A}\right)=\frac{4}{5}$$

Commented by bramlex last updated on 19/Jun/20

$$ohyes.$$

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