maths
posted by Anonymous .
In an urn, there are several colored balls, with equal numbers of each color. We add 14 balls which are all of the same new color, that is different from those in the urn. It is calculated that the probability of drawing, without replacement, two balls of the same color is the same (when compared before and after the balls are added). How many balls are there in the urn initially?

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.
Initial situation:
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)*(k1)/(n1).
Probability of picking two balls of the same colour is therefore obtained by multiplying p1 by the number of colours, i.e.
P0=(n/k)*Psame=(k1)/(n1)
After adding 14 balls of the same (new) colour:
Total number of balls = n+14
Psame=(k/(n+14)*(k1)/(n+13)
Pnew = 14/(n+14)*13/(n+13)
Probability of getting two balls of the same colour
P1=14/(n+14)*13/(n+13)+ (n/k)*(k/(n+14)*(k1)/(n+13)
Solution
Since question specifies P0=P1, we have
(k1)/(n1) = 14/(n+14)*13/(n+13) + (n/k)(k/(n+14)(k1)/(n+13))
(k1)/(n1) = ( n(k1)+13*14 )/[(n+13)(n+14)]
Crossmultiplying and equating to zero yields the integer equation to be solved for zeroes for integer values of n and k:
f(n,k)=(k1)(n+14)(n+13)  (n1)(n*kn+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 kn (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.
So
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.