, 2011

Let $b$, $r$ and $g$ be the number of blue, red and green balls, respectively. We know that $b + r + g = 100$.

If you draw two balls randomly from the urn, the probability of getting two balls of different colour is $58\%$, and the probability of getting a blue and a green ball is $8\%$ and hence the probability of getting a red ball and a ball of a different colour is $50\%$.

Hence $\frac{r(100 - r) + (100 - r)r}{100 \cdot 99} = \frac{1}{2}$.

From this we get the quadratic equation $-r^2 + 100r - 2475 = 0$, and $r = 45$ or $r = 55$.

Suppose $r = 45$. Then $b + g = 55$, and since the probability of having blue and a green ball is $8\%$, then $\frac{g(55 - g) + (55 - g)g}{100 \cdot 99} = \frac{8}{100}$.

This leads to the quadratic equation $-g^2 + 55g - 396 = 0$ with no integer roots, and hence a contradiction.

The only possibility therefore $r = 55$. (In this case we get $b = 33$ and $g = 12$ or the other way around.)

Let $b$, $r$ and $g$ be the number of blue, red and green balls, respectively. From the probabilities stated we get $\frac{2bg}{100 \cdot 99} = \frac{8}{100}$ and $\frac{2(rg + rb + gb)}{99 \cdot 100} = \frac{58}{100}$, and hence
$$2bg = 8 \cdot 99 \quad \text{and} \quad 2(rg + rb + gb) = 58 \cdot 99.$$
From the first equation we see $11$ divides exactly one of $b$ and $g$, and hence from the second $11$ must also divide $r$. Assume w.l.o.g. that $11$ divides $g$ and let $g = 11g'$. Since $r + b + g = 100$ we have $b \equiv 1 \pmod{11}$. The first equation is now $bg' = 36$, and hence $b = 1$ or $b = 12$. If $b = 12$ we get $g = 11 \cdot 36 > 100$ which is impossible. The only solution is therefore $b = 12$, $g = 33$ (the other way around) and $r = 55$.