Question

Each of the nine balls can be placed with equal probability in any of the three initially empty boxes. Let $P_1$ be the probability that there will be four balls in the first box, three in the second box and two in the third box, and $P _2$ be the probability that there will be three balls in each box then the ratio $\left( P _2 / P _1\right)$ equals

• A. $ \frac{4}{3}$
• B. $\frac{2}{9}$
• C. 8
• D. 2

Answer 1

Answer: (A)

$n ( S )=3^9$
$P_2=\frac{9 ! \cdot 3 !}{(3 !)^3 \cdot 3 !} \cdot \frac{1}{3^9}$
(all boxes are different)
$P_1=\frac{9 !}{4 ! \cdot 3 ! \cdot 2 !} \cdot \frac{1}{3^9} $

$\frac{P_2}{P_1}=\frac{9 !}{(3 !)^3} \cdot \frac{4 ! \cdot 3 ! \cdot 2 !}{9 !}=\frac{4}{3}$