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If $12$ identical balls are to be placed randomly in $3$ identical boxes, then the probability that one of the boxes contains exactly $3$ balls is
$\frac{4}{{19}}$
$\frac{{55}}{3}{\left( {\frac{2}{3}} \right)^{11}}$
$\frac{{\left( {428} \right){}^{12}{C_3}}}{{{3^{11}}}}$
$\frac{5}{{19}}$
Solution
Number of elements in sample spa ce $=\mathrm{n}(\mathrm{s})$
$=3^{12}$
If $E$ corresponds to the event that one of the boxes contains exactly $3$ balls then $\mathrm{n}(\mathrm{E})=$
${\,^{12}}{{\rm{C}}_3} \times {\,^3}{{\rm{C}}_1} \times \left[ {{2^9} – 2{\rm{x}}{\,^9}{{\rm{C}}_3}} \right] + {\,^2}{{\rm{C}}_6} \times {\,^3}{{\rm{C}}_1} \times {\,^6}{{\rm{C}}_3}$
required probability $ = \frac{{{\rm{n}}({\rm{E}})}}{{{\rm{n}}({\rm{s}})}} = \frac{{\left( {428} \right){\,^{12}}{C_3}}}{{{3^{11}}}}$
