Recurrence relation
A barrel contains 2n balls , numbered 1 to 2n. Choose three balls at random , one after the other , and with the balls replaced after each draw. What is the probability that the three element sequence obtained has the properties that the smallest element is odd and that only the smallest element , if any , is repeated???
NOTE : refer to combinatorics of An EXCURSION IN MATHEMATICS
The answer is 0.5.
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1 solution
The total no of possible outcomes is N = 2n×2n×2n =8 n ^3
To find favourable outcomes
Let a be any odd integer such that a is greater than equal to 1 and less than equal to 2n-1
Now proceed Youll get 4n ^3