Question

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:

• A. $55{\left(\frac{2}{3}\right)}^{10}$
• B. $220{\left(\frac{1}{3}\right)}^{12}$
• C. $22{\left(\frac{1}{3}\right)}^{11}$
• D. $\frac{55}{3}{\left(\frac{2}{3}\right)}^{11}$

Answer 1

The correct option is D $\frac{55}{3}{\left(\frac{2}{3}\right)}^{11}$

$P{=}^{12C_3}{\left(\frac{1}{3}\right)}^3{\left(\frac{2}{3}\right)}^9$

$=\frac{12\times 11\times 10}{6}\times \frac{1}{3^3}\times {\left(\frac{2}{3}\right)}^9$

$=\frac{220}{3^3}{\left(\frac{2}{3}\right)}^9=\frac{55}{3}\times \frac{2^2}{3^2}\times {\left(\frac{2}{3}\right)}^9$