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Question Number 142745 by iloveisrael last updated on 05/Jun/21

Commented by MJS_new last updated on 05/Jun/21

$$\mathrm{vr}\times \mathrm{th}7\mathrm{gr}\mathrm{bl}\theta \mathrm{p}\mathrm{njk}\mathrm{\Pi }\mathrm{y}\notin \mathrm{m}\mathrm{trf}ϝ\mathrm{sh}5\mathrm{jk}[\mathrm{♠}>bkl$$

Commented by iloveisrael last updated on 05/Jun/21

$$hahahaha.....great$$

Commented by iloveisrael last updated on 05/Jun/21

$$$$maybe in this forum there are members who understand Japanese

Commented by mr W last updated on 05/Jun/21

$$inaboxthereare3nwhiteballsand$$$$2nredballs.threeballsaretaken.$$$$theprobabilitythattwoballsarewhite$$$$andoneballisredis{P}_n.find\underset{n\to \mathrm{\infty }}{\lim }{P}_n.$$

Commented by iloveisrael last updated on 05/Jun/21

$$ouwnice.thanks$$

Commented by MJS_new last updated on 05/Jun/21

$$\frac{3n}{3n+2n}\times \frac{3n-1}{3n+2n-1}\times \frac{2n}{3n+2n-2}\times 3=$$$$=\frac{18(3n^2-n)}{5(25n^2-15n+2)}\Rightarrow \mathrm{limit}\mathrm{with}n\to \mathrm{\infty }\mathrm{is}\frac{54}{125}=43.2\mathrm{\%}$$

Commented by EDWIN88 last updated on 05/Jun/21

$$\mathrm{white}\mathrm{balls}=3\mathrm{n}$$$$\mathrm{red}\mathrm{balls}=2\mathrm{n}$$$${\mathrm{P}}_{\mathrm{n}}=\frac{{\mathrm{C}}_2^{3\mathrm{n}}.{\mathrm{C}}_1^{2\mathrm{n}}}{{\mathrm{C}}_3^{5\mathrm{n}}}=\frac{\frac{3\mathrm{n}(3\mathrm{n}-1)}{\cancel{2.1}}.\frac{\cancel{2n}}{1}}{\frac{5\mathrm{n}(5\mathrm{n}-1)(5\mathrm{n}-2)}{3.2.1}}$$$$=\frac{{18\mathrm{n}}^2(3\mathrm{n}-1)}{5\mathrm{n}(5\mathrm{n}-1)(5\mathrm{n}-2)}$$$$\text{Double subscripts: use braces to clarify}$$$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{54(1-\frac{1}{3\mathrm{n}})}{125(1-\frac{1}{5\mathrm{n}})(1-\frac{2}{5\mathrm{n}})}=\frac{54}{125}◻$$

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