If 12 identical balls are to be placed in 3 i | vedclass

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Question

$x + y = 9$

$P{ = ^9}{C_3}{\left( {\frac{1}{3}} ight)^3}{\left( {\frac{2}{3}} ight)^9}$

$ = \frac{{12 	imes 11 	imes 10}}{3} 	imes \frac{

Vedclass · Accepted Answer

$\frac{{55}}{3}{\left( {\frac{2}{3}} \right)^{11}}$

Vedclass · Answer

$x + y = 9$

$P{ = ^9}{C_3}{\left( {\frac{1}{3}} ight)^3}{\left( {\frac{2}{3}} ight)^9}$

$ = \frac{{12 	imes 11 	imes 10}}{3} 	imes \frac{1}{{{3^3}}} 	imes {\left( {\frac{2}{3}} ight)^9}$

$ = \frac{{220}}{{{3^3}}} 	imes {\left( {\frac{2}{3}} ight)^9}$

$ = \frac{{55}}{3} 	imes \frac{{{2^2}}}{{{3^2}}} 	imes {\left( {\frac{2}{3}} ight)^9}$

$ = \frac{{55}}{3} 	imes {\left( {\frac{2}{3}} ight)^{11}}$

If $12$ identical balls are to be placed in $3$ identical boxes, then the probability that one of the boxes contains exactly $3$ balls is :

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