This is another good solution from Alan
of Madras College.
Call the number of red balls $r$ and the number of blue balls $b$.
The probability of drawing a red ball is $r/(b+r)$ leaving $b$ blue
balls and $r-1$ red balls and so the probability of drawing a
second red ball is $(r-1)/(b+r-1)$ and the probability of drawing
two red balls is $r(r-1)/(b+r)(b+r-1)$. The question tells us that
this is five times the probability of drawing two blue balls.
By a similar argument, the probability of drawing two blue balls is
$b(b-1)/(b+r)(b+r-1)$.
Also, after one red ball has been drawn the probability of drawing
a blue ball is $b/(b+r-1)$ and, considering that you can draw the
balls in either order, the probability of a red and a blue ball is
$2rb/(b+r)(b+r-1)$. This is six times the probability of drawing
two blue balls.
Hence $$\frac{r(r - 1)}{(b + r)(b + r - 1)} = \frac{5b(b - 1)}{(b +
r)(b + r - 1)}$$ and we know $(b + r)(b + r - 1) \neq 0$.