Saved Bookmarks
| 1. |
Fourteen numbered balls (1, 2, 3, …, 14) are divided in 3 groups randomly. Find the probability that the sum of the numbers on the balls, in each group, is odd. |
|
Answer» Solution :Each group should have odd numbered BALLS. Case I: Two groups have three odd numbered balls and third has only one odd numbered ball. Number of such cases = `(7!)/((3!)^(2)xx1!xx2!)xx3^(7)`(as even number of balls can go in any group) Case II: Two groups have one odd numbered ball each and the third group has five odd numbered balls. Number of such cases = `(7!)/((1!)^(2)xx5! xx 2!)xx3^(7)` Total number of cases = Number of ways in which three non-empty group can be formed `=(3^(14)-.^(3)C_(1)2^(14) + .^(3)C_(2))/(3!)` `therefore` Required PROBABILITY = `((7!)/(3!)^(2)+(7!)/(5! xx 2!))/(3^(14) - .^(3)C_(1)2^(14)+.^(3)C_(2))xx3!xx3^(7)` |
|