A stimulating problem, combining probabilities with discrete mathematics.
The question requires the probability of successively picking two balls of same colour be the same before and after adding 14 balls of a new colour.
Assume there are n balls with k of each colour, i.e. n/k (∈N) represents the initial number of colours.
Then probability of picking two balls of the same colour is therefore Psame=(k/n)*(k-1)/(n-1).
Probability of picking two balls of the same colour is therefore obtained by multiplying p1 by the number of colours, i.e.
After adding 14 balls of the same (new) colour:
Total number of balls = n+14
Pnew = 14/(n+14)*13/(n+13)
Probability of getting two balls of the same colour
Since question specifies P0=P1, we have
(k-1)/(n-1) = 14/(n+14)*13/(n+13) + (n/k)(k/(n+14)(k-1)/(n+13))
(k-1)/(n-1) = ( n(k-1)+13*14 )/[(n+13)(n+14)]
Cross-multiplying and equating to zero yields the integer equation to be solved for zeroes for integer values of n and k:
f(n,k)=(k-1)(n+14)(n+13) - (n-1)(n*k-n+13*14) = 0
The easiest solution strategy is to assume integer values of k and solve for n such that n is also an integer and such that k|n (k divides n).
This can be easily solved to get (n,k) couplets of
(13,5) but 5 does not divide 13,
(26,6) but 6 does not divide 26
(91,7) yes, 7 divides 91.
n=91, k=7, i.e.
initially there are 91 balls with 7 each of the 13 colours.
You should proceed to check that the solution satisfies the given requirements.