Question

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

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

Answer 1

Answer: (A)

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

Answer 2

Success $(p)=\frac{1}{3}$
Failure $(q)=\frac{2}{3}$
Acc. to binomial distribution, we have to find,
$P ( x =3)=12_{ C _{3}} \cdot\left(\frac{1}{3}\right)^{3} \cdot\left(\frac{2}{3}\right)^{9}$
$=\frac{55}{3}\left(\frac{2}{3}\right)^{11}$