The correct option (a) (110/9)(2/3)10
Explanation:
12 balls distributed among 3 boxes.
Each ball can be placed in any of 3 boxes.
∴ n = 3 × 3 × ...... 3 = 312.
No. of ways that 3 ball out of 12 can be put on 1st box = 12C3. & remaining 9 balls can be distributed in remaining 2 boxes in
2 × 2 × ---- 2 = 29 ways
This can be done in 12C3 (29) ways.
∴ P(event) = [{12C3(29)}/ (312)]
= [(220 × 29)/(310 × 9)]
= [(110 × 210)/(9 × 310)]
= (110/9)(2/3)10.
