A) \[220{{\left( \frac{1}{3} \right)}^{12}}\]
B) \[22{{\left( \frac{1}{3} \right)}^{11}}\]
C) \[\frac{55}{3}{{\left( \frac{2}{3} \right)}^{11}}\]
D) \[55{{\left( \frac{2}{3} \right)}^{10}}\]
Correct Answer: C
Solution :
Probability is a ratio hence, answer will not be effected when given balls are different. \[n(s)={{3}^{12}}\]since each ball will go in three ways. Event : select 3 balls out of 12 in \[^{12}{{C}_{3}}\] ways. Remaining 9 balls will be placed in other 2 boxes in \[{{2}^{9}}\] ways. \[\therefore \]\[n(E){{=}^{12}}{{C}_{3}}\times {{2}^{9}}\] \[P(E)=n(E0/n(s)\] \[=\frac{^{12}{{C}_{3}}\times {{2}^{9}}}{{{3}^{12}}}\] \[=\frac{220\times {{2}^{9}}}{{{3}^{12}}}=\frac{55}{3}{{\left( \frac{2}{3} \right)}^{11}}\] (Note : Assumption is atleast 1box contains 3 balls but not exactly)
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