If 7 different balls are distributed among 4 different boxes, then the probability that the first box…
If 7 different balls are distributed among 4 different boxes, then the probability that the first box contains 3 balls is
$\frac{35}{128}\left(\frac{3}{4}\right)^3$
$\frac{35}{64}\left(\frac{3}{4}\right)^4$
$\frac{7}{8}\left(\frac{3}{4}\right)^7$
$\frac{5}{16}\left(\frac{3}{4}\right)^5$
Solution
Since, total number of ways of distributing of 7 balls in 4 box $=4^7$
Total number of ways such that first box contain 3 balls $={ }^7 C_3 \times 3^4$
So, required probability $=\frac{{ }^7 C_3 \times 3^4}{4^7}=\frac{35}{64}\left(\frac{3}{4}\right)^4$.