A box contains $R$ red balls, $B$ blue balls, and no other balls. One ball is removed and set aside, and then a second ball is removed. On each draw, each ball in the box is equally likely to be removed. The probability that both of these balls are red is $\frac{2}{7}$. The probability that exactly one of these balls is red is $\frac{1}{2}$. Determine the pair $(R, B)$.
The box contains $R+B$ balls when the first ball is drawn and $R+B-1$ balls when the second ball is drawn.

Therefore, there are $(R+B)(R+B-1)$ ways in which two balls can be drawn.

If two red balls are drawn, there are $R$ balls that can be drawn first and $R-1$ balls that can be drawn second, and so there are $R(R-1)$ ways of doing this.

Since the probability of drawing two red balls is $\frac{2}{7}$, then $\frac{R(R-1)}{(R+B)(R+B-1)}=\frac{2}{7}$.

If only one of the balls is red, then the balls drawn are either red then blue or blue then red.

There are $R B$ ways in the first case and $B R$ ways in the second case, since there are $R$ red balls and $B$ blue balls in the box.

Since the probability of drawing exactly one red ball is $\frac{1}{2}$, then $\frac{2 R B}{(R+B)(R+B-1)}=\frac{1}{2}$.

Dividing the first equation by the second, we obtain successively

$$
\begin{aligned}
\frac{R(R-1)}{(R+B)(R+B-1)} \cdot \frac{(R+B)(R+B-1)}{2 R B} & =\frac{2}{7} \cdot \frac{2}{1} \\
\frac{R-1}{2 B} & =\frac{4}{7} \\
R & =\frac{8}{7} B+1
\end{aligned}
$$

Substituting into the second equation, we obtain successively

$$
\begin{aligned}
\frac{2\left(\frac{8}{7} B+1\right) B}{\left(\frac{8}{7} B+1+B\right)\left(\frac{8}{7} B+1+B-1\right)} & =\frac{1}{2} \\
\frac{2\left(\frac{8}{7} B+1\right) B}{\left(\frac{15}{7} B+1\right)\left(\frac{15}{7} B\right)} & =\frac{1}{2} \\
\frac{2(8 B+7)}{15\left(\frac{15}{7} B+1\right)} & =\frac{1}{2} \quad(\text { since } B \neq 0) \\
32 B+28 & =\frac{225}{7} B+15 \\
13 & =\frac{1}{7} B \\
B & =91
\end{aligned}
$$

Since $R=\frac{8}{7} B+1$, then $R=105$ and so $(R, B)=(105,91)$.

(We can verify that the given probabilities are correct with these starting numbers of red and blue balls.)