Question 1021467
Let's say there are {{{X}}} cherry candies.
Then there are {{{100-X}}} grape candies.
The probability of choosing CC or GG is,
{{{P=P(CC)+P(GG)}}}
{{{P=((X)/100)((X-1)/99)+((100-X)/100)((99-X)/99))}}}
{{{P=(X(X-1)+(100-X)(99-X))/9900}}}

The probability of choosing CG or GC is,
{{{P=P(CG)+P(GC)}}}
{{{P=(X/100)((100-X)/99)+(100-X)/100)(X/99)}}}
{{{P=(2X(100-X))/9900}}}
So then make them equal to each other,
{{{X(X-1)+(100-X)(99-X)=2X(100-X)}}}
{{{X^2-X+9900-199X+X^2=200X-2X^2}}}
{{{4X^2-400X+9900=0}}}
{{{X^2-100X+2475=0}}}
{{{(X-55)(X-45)=0}}}
Two solutions but the question stipulated {{{X>50}}}
So then,
{{{X=55}}} <--- Cherry
{{{100-X=45}}} <--- Grape