Question

Twelve balls are distributed among three boxes. The probability that the first box contains three balls is

• A. $\frac{110}{9}{\left(\frac{2}{3}\right)}^{10}$
• B. $\frac{9}{110}{\left(\frac{2}{3}\right)}^{10}$
• C. $\frac{{}^{12}{C}_3}{{12}^3}\times 2^9$
• D. $\frac{{}^{12}{C}_3}{3^{12}}$

Answer 1

The correct option is A $\frac{110}{9}{\left(\frac{2}{3}\right)}^{10}$

Since each ball can be placed in any one of 3 boxes, therefore there are 3 ways in which a ball
can be placed in any one of the three boxes.
n(S) = $3^{12}$
choose any 3 balls and place them in first bag, ways of doing this = ${}^{12}{C}_3$
remaining 9 balls can be placed in other two bag in $2^9$ ways

Probability = ${}^{12}{C}_3\times 2^9$/ $3^{12}$