A chess club has 8 girls and 6 boys. Two members, Zig and Zag, are fraternal
twins of different genders. If a team of 3 girls and 3 boys is randomly selected
for the district championship, what is the probability that exactly one of the
twins is on the team? Express your answer as a common fraction.
There are $\binom{8}{3}$ ways to choose three girls and $\binom{6}{3}$ ways to choose three boys. There are two cases of interest. Either Zig is chosen and Zag is not, or Zag is chosen and Zig is not. If Zig is chosen and Zag is not, then there are $\binom{7}{2}$ ways to choose the other two girls, and $\binom{6}{3}$ ways to choose the boys. If Zag is chosen and Zig is not, then there are $\binom{7}{3}$ ways to choose the other three girls, and $\binom{6}{2}$ ways to choose the boys. Thus, the probability is $\frac{\binom{7}{2}\binom{6}{3}+\binom{7}{3}\binom{6}{2}}{\binom{8}{3}\binom{6}{3}}$ = $\frac{21(20)+35(15)}{56\cdot20}$ = $\frac{420+525}{1120}$ = $\boxed{\frac{105}{224}}$.
