Let 9 distinct balls be distributed among 4 boxes, B1, B2, B3 and B4.

Question

Let 9 distinct balls be distributed among 4 boxes, B1, B2, B3 and B4. If the probability than B3 contains exactly 3 balls is $$k{\left( {{3 \over 4}} \right)^9}$$ then k lies in the set :

Accepted Answer

Required probability = $${{{}^9{C_3}{{.3}^6}} \over {{4^9}}}$$

$$ = {{{}^9{C_3}} \over {27}}.{\left( {{3 \over 4}} \right)^9}$$

$$ = {{28} \over 9}.{\left( {{3 \over 4}} \right)^9} \Rightarrow k = {{28} \over 9}$$

Which satisfies $$\left| {x - 3} \right| < 1$$

Solutions · 1