For solving above problem I have used classical approach for probability calculation i.e Number of Favorable . Moves/Total number of Move.
Total Number of Moves as per the star and bar method will be n+r-1 Choose r-1 i.e (14 C 2 ).
Number of Favorable moves will be equal to Number of ways of distributing 9 identical balls into 2 different boxes as chosen box will have exactly three balls.
Number of Favorable moves will be 10 C 1.
P=(14 C 2)/(10 C 1)
But as per the solution mentioned answer will be 55/3(2/3)^11.
They have solved this problem as per the binomial distribution. If X is random variable denoting number of balls in that Particular box the answer will be P(X=3)=12C3 p^3(1-p)9.
where p is the probability of selecting that particular box and 1-p will be the probability of selecting other boxes.
Why my approach is not correct. or is there something that I am not getting. Please suggest.